Expansion Of Liquid 4dt2l

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Thermal Expansion of Liquid We shall consider only the volume expansion of liquid in this chapter. The concept of thermal expansion of length or surface of a liquid is not meaningful as a liquid has no definite shape like that of a solid. Some Common Facts: • The expansion in liquid is usually much more than in a solid for a same rise in temperature; on an average 10 times more. • The rate of expansion of a same liquid sometimes differs greatly in different temperature ranges. Example: The amount of volume expansion of water in the range of 10 0 C − 110 C is quite different from that in the range of 930 C − 94 0 C . • Anomalous expansion of water in the temperature range of 0 0 C − 4 0 C . Apparent and Real Expansion of Liquids:

A liquid is heated while keeping it in a container. There occurs an expansion of the solid container along with the liquid. But the amount of expansion of the container is small compared to that of the liquid. Thus the expansion of a container is not usually noticeable. • When the expansion of liquid is considered ignoring the expansion of the container, it is called apparent expansion. • The real expansion of liquid is calculated by adding the expansion of the part of the container containing liquid (before expansion) with the apparent expansion of liquid. Experiment: The relation among apparent and real expansion of liquid and the expansion of the container can be demonstrated with the help of following experiment:

Fig. to Figure be included Include 3.1 fr A flask made of thick glass is filled with a coloured liquid (as shown in the figure). A cork is fitted through which a narrow glass tube is inserted. There is a scale marked on the tube. Some liquid will rise up inside the tube up to a point which is marked by O on the scale. Now the flask is placed inside a big vessel of boiling water which heats up the flask and as well as the liquid inside. It is observed that the liquid inside the tube first comes down to a position marked by A and later it rises again, goes past O and reaches at a point marked by B. The reason is that there is an expansion of the glass container at the beginning. As the glass is a bad conductor of heat, it takes some time for heat to reach the liquid inside the flask and because of this there is no expansion of liquid at the initial stage. Thus the liquid surface comes down from O to A inside the tube. After a while the heat starts flowing into the liquid for which the liquid expands and it begins to rise inside the tube. The expansion of liquid is more than the expansion of glass. Thus we see the final rise of liquid is at B above the point O.

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As the level of water comes down from O to A for a short time, it appears that the water level directly expands from O to B. The expansion of water level OB, from its initial position O to the final position B is the apparent expansion. AB is the amount of real expansion. According to figure, AB = OB + OA . ∴ The real expansion of the liquid = the apparent expansion of the liquid + the expansion of the container. Coefficients of Apparent and Real expansions of Liquid

We have two different coefficients of expansion for a liquid corresponding to apparent and real expansions. Coefficient of apparent expansion for liquid: The amount of apparent expansion of unit volume of a liquid for a unit change of temperature is called coefficient of apparent expansion of liquid.

If the initial volume of the liquid is V0 and the final apparent volume which is observed to be V ′ due to an increase in temperature ∆T = T − T0 , the apparent expansion of volume of the liquid is ∆V = V ′ − V0 . ∴ The coefficient of apparent expansion of liquid can be written as ∆V γ′= …………………..(1) V0 ∆T Thus we can write, V ′ = V0 [1 + γ ′∆T ] = V0 [1 + γ ′(T − T0 )] . It may be easily understood that the coefficient of apparent expansion of a liquid can not be a characteristic property of a liquid as the apparent expansions will be different when containers of different materials are taken. Coefficient of real expansion for liquid: The amount of real expansion of unit volume of a liquid for a unit change of temperature is called coefficient of real expansion of liquid.

If the initial volume of the liquid is V0 and the final real volume which is calculated to be V due to an increase in temperature ∆T = T − T0 , the real expansion of the volume of liquid is ∆V = V − V0 . ∴ The coefficient of real expansion of liquid is ∆V γ = ……………………..(2) V0 ∆T Thus we can write, V = V0 [1 + γ∆T ] = V0 [1 + γ (T − T0 )] . The coefficient of real expansion is a property of a liquid as the real expansion of volume of a liquid is not dependent on the expansion of the container.

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Note: • The dimension of the coefficient of real or apparent expansion of liquid is dependent only on the temperature difference as can be seen from expressions (1) and (2). For example, the value of the coefficient of real expansion for mercury in Celsius and Fahrenheit scales are 18.18 × 10 −5 / 0C and 10.1 × 10 −5 / 0F . • The coefficient of real expansion is a characteristic property of a liquid. Thus it is more important to know the coefficient of real expansion of a liquid than that for apparent expansion. • The value of the coefficient of linear expansion for a liquid may be slightly different at different temperatures. The value which is usually mentioned is assumed to be an average value within a temperature range.

Table # 1: Coefficient of real expansion of different liquids

Liquid

γ (C 0 ) −1

Mercury Benzene Alcohol Sulphuric acid Chloroform Glycerine Tarpin Oil Olive Oil Water ( 5 0 C − 10 0 C ) Water ( 10 0 C − 20 0 C ) Water ( 20 0 C − 40 0 C ) Water ( 40 0 C − 60 0 C )

18.18 × 10 −5 121 × 10 −5 110 × 10 −5 57 × 10 −5 126 × 10 −5 53 × 10 −5 94 × 10 −5 70 × 10 −5 5.3 × 10 −5 15 × 10 −5 30.2 × 10 −5 45.8 × 10 −5

Relation between the coefficients of apparent and real expansions of liquid

Let us assume the initial volume of a definite amount of liquid in a container is V0 . Due to an increase in temperature ∆T , the apparent volume of the liquid is measured to be V ′ and the real volume is found to be V . ∴ The apparent expansion of the liquid = V ′ − V0 and the real expansion of the liquid = V − V0 . The expansion of the part of the container containing liquid is = V − V ′ . We know, the real expansion of liquid = the apparent expansion of liquid + the expansion of the container. ∴ V − V0 = ( V ′ − V0 ) + ( V − V ′ )

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Dividing both sides by V0 ∆T we find, V − V0 V ′ − V0 V −V ′ = + Or, γ = γ ′ + γ g , V0 ∆T V0 ∆T V0 ∆T where γ g is the coefficient of volume expansion of the material of the container. Therefore, we have the coefficient of real expansion of liquid = the coefficient of apparent expansion of liquid + the coefficient of volume expansion of the material of the container. Relation between the density and the coefficient of real expansion of liquid

As there is an expansion of volume of a liquid due to increase in temperature, the density of it must decrease. The density depends on the coefficient of real expansion at different temperatures. Let us have a liquid of a certain mass m . The initial volume of this liquid is V0 and the density is ρ 0 . The volume and the density are changed to V and ρ as the temperature is increased by ∆T . ρ V We can write, m = V0 ρ 0 = Vρ Or, 0 = …………………(1) ρ V0 If γ be the coefficient of real expansion of the liquid, V V = V0 [1 + γ .∆T ] Or, = 1 + γ .∆T …………………(2) V0 From (1) and (2) we get,

ρ0 = 1 + γ .∆T ρ

Or, ρ 0 = ρ [1 + γ .∆T ] …………………….(3)

We can also write (for small ∆T ),

ρ=

ρ0 −1 = ρ 0 (1 + γ .∆T ) = ρ 0 (1 − γ .∆T ) 1 + γ .∆T

[Q γ is very small]

Thus we see the density of a liquid is decreased due to increase in temperature. The relation (3) can be rewritten as

∴ γ =

ρ0 − 1 = γ .∆T . ρ

ρ0 − ρ ∆ρ =− . The negative sign indicates that the density decreases with ρ .∆T ρ .∆T

increasing temperature. The above can be used to calculate the coefficient of real expansion of a liquid when the density of it is known at two different temperatures. Numerical Problems with Solutions #Example 1

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The relative value of the coefficient of apparent expansion of mercury with respect to glass is 153 × 10 −6 / 0C and the coefficient of real expansion of mercury is 180 × 10 −6 / 0C . Calculate the coefficient of linear expansion of glass. [H.S.] Solution. We know, the coefficient of real expansion of mercury ( γ ) = the coefficient of apparent expansion of mercury ( γ ′) + the coefficient of volume expansion of glass ( γ g ) ∴ 180 × 10 −6 = 153 × 10 −6 + γ g

Or, γ g = (180 − 153) × 10 −6 = 27 × 10 −6

∴ The coefficient of linear expansion of glass, α g =

γg

=

27 × 10 −6 = 9 × 10 −6 / 0C . 3

3 #Example 2 The coefficient of apparent expansion of a liquid is 18 × 10 −5 / 0C when an iron container is used and that is 14.46 × 10 −5 / 0C when an aluminium container is used. If the coefficient of linear expansion of aluminium is 2.38 × 10 −5 / 0C what is the coefficient of linear expansion for iron? Solution. The coefficient of volume expansion of aluminium = 3 × 2.38 × 10 −5 / 0C = 7.14 × 10 −5 / 0C ∴ The coefficient of real expansion for the liquid, γ = 14.46 × 10 −5 + 7.14 × 10 −5 = 21.60 × 10 −5 / 0C . If α be the coefficient of linear expansion of iron then we can write, 21.60 × 10 −5 = 18 × 10 −5 + 3α Or, 3α = 3.6 × 10 −5 ∴ α = 1.2 × 10 −5 = 12 × 10 −6 / 0C . #Example 3 A capillary glass tube of uniform cross-section contains a mercury column of 1 m length in it at 0 0 C . The length of the mercury column increases by 16.5 mm when the temperature is raised to 100 0 C . If the coefficient of real expansion of mercury is 0.000182 / 0C , what is the coefficient of linear expansion of glass? [H.S.] Solution. Let us suppose, the area of cross-section of the glass tube is = A sq.cm. The initial volume of mercury column at 0 0 C = 100 A c.c. The increase in volume of mercury column = 1.65 A c.c. due to the increase in temperature, ∆T = 100 0 C 1.65 A ∴ The coefficient of apparent expansion of mercury, γ ′ = = 1.65 × 10 −4 / 0C . 100 A × 100 We know, the coefficient of real expansion of mercury ( γ ) = the coefficient of apparent expansion of mercury ( γ ′) + the coefficient of volume expansion of glass (γ g ) .

∴ 0.000182 = 0.000165 + γ g Or, γ g = 0.000017 / 0C ∴ The coefficient of linear expansion of glass is

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γg 3

=

6

0.000017 = 0.00000567 = 5.67 × 10 −6 / 0C . 3

#Example 4 The inner volume of a glass flask is V c.c. What quantity of mercury has to be kept inside this so that the upper level of mercury remains same at all temperatures? The coefficient of volume expansion of mercury is 1.8 × 10 −4 / 0C and the coefficient of linear expansion of glass is 9 × 10 −6 / 0C . [H.S. ’85; J.E.E. ‘87] Solution. According to question, the upper level of mercury will remain the same at all temperature only when the volume expansion of mercury becomes equal to the volume expansion of the glass flask. Let us suppose, we take mercury of volume x c.c. For some amount of temperature increase, ∆T we can write, x × 1.8 × 10 −4 × ∆T = V × 3 × 9 × 10 −6 × ∆T V × 27 × 10 −6 3 Or, x = = V −4 20 1.8 × 10 3 Thus th fraction of the volume of the flask has to be filled up by mercury. 20 #Example 5 The coefficients of volume expansion for glass and mercury are 2.4 × 10 −5 / 0C and 1.8 × 10 −4 / 0C . What fraction of a glass vessel has to be filled by mercury so that the mercury level remains same at all temperatures? [H.S. ‘93] Solution. Let us suppose, the volume of mercury = x c.c.; the volume of the vessel = V c.c. According to question, the upper surface of mercury remains same. This means the expansion of mercury is equal to the volume expansion of the glass vessel at any temperature. For any temperature increase ∆T , x × 1.8 × 10 −4 × ∆T = V × 2.4 × 10 −5 × ∆T V × 2.4 × 10 −5 24 2 ∴x= =V × = V −4 180 15 1.8 × 10 2 Hence, a fraction of the glass vessel has to be filled up by mercury. 15 #Example 6 The inner volume of a glass flask is 540 c.c. How much mercury has to be kept in this flask so that the mercury level remains unchanged at all temperatures? The coefficient of real expansion of mercury = 1.8 × 10 −4 / 0C ; the coefficient of volume expansion of glass = 2.5 × 10 −5 / 0C . [H.S. ‘96] Solution. Let us suppose, the volume of mercury = x c.c. According to question, the volume expansion of mercury is equal to the volume expansion of the glass flask at any temperature to keep the upper surface of mercury unchanged.

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For any temperature increase ∆T , x × 1.8 × 10 −4 × ∆T = 540 × 2.5 × 10 −5 × ∆T ∴ x = 75 c.c. Thus we have to keep 75 c.c. of mercury in the flask. #Example 7 The volume of the bulb of a thermometer is 1 c.c. at 0 0 C and the area of cross-section of the attached tube measures 0.1 mm 2 . The bulb is just filled up by mercury at 0 0 C . How far will be the rise of mercury in the tube when the bulb is immersed into boiling water? The coefficient of apparent expansion for mercury = 16 × 10 −5 / 0C . Solution. The rise of mercury in the tube indicates apparent expansion of mercury. The coefficient of apparent expansion, γ ′ = 16 × 10 −5 / 0C , Initial volume of mercury in the bulb = 1 c.c., Temperature increase = 100 0 C . The apparent expansion of mercury = 1 × 16 × 10 −5 × 100 = 16 × 10 −3 c.c. Let us suppose, the mercury rises by x c.c. 0.1 ∴ x× = 16 × 10 −3 Or, x = 16 × 10 −3 × 10 3 Or, x = 16 cm. 100 #Example 8 The volume of water of 1 gm is 1 c.c. at 4 0 C and 1.0169 c.c. at 60 0 C . What is the average coefficient of real expansion of water at those two temperatures? [H.S.(T)] Solution. Here, the volume of water is changed by ∆V = V − V0 = 1.0169 − 1 = 0.0169 c.c. for the

change in temperature, ∆T = 60 − 4 = 56 0 C . ∆V 0.0169 ∴ γ = = = 3.02 × 10 − 4 / 0 C . V0 ∆T 1 × 56 #Example 9 The mass of 10 c.c. water is 9.998 gm and 10 gm at 0 0 C and 4 0 C respectively. Find out the average coefficient of real expansion of water in this temperature range. [H.S.’93] Solution. 9.998 The density of water at 0 0 C is ρ 0 = = 0.9998 gm/c.c., 10 10 the density at 4 0 C is ρ = = 1 gm/c.c. and change in temperature, ∆T = 4 − 0 = 4 0 C . 10 We know, ρ 0 = ρ [1 + γ .∆T ] ρ −ρ 0.9998 − 1 0.0002 ∴γ = 0 = =− = −5 × 10 −5 / 0 C . ρ .∆T 1× 4 4 [Note: The negative sign indicates that the density increases when the temperature is increased in the range 0 0 C to 4 0 C . This is anomalous expansion of water. Note that the mass of a certain volume of water is said to increase as in the question. Actually, the volume of certain mass of water should decrease to have the density to increase.]

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#Example 10 The density of mercury at 0 0 C is 13.5955 gm/cc. What will be its density at 60 0 C if the coefficient of linear expansion of mercury is 0.000061 / 0 C ? [J.E.E.] [Note that, actually, the coefficient of linear expansion of a liquid is meaningless.] Solution. Here, ρ 0 = 13.5955 gm/cc, γ = 3α = 3 × 61 × 10 −6 = 183 × 10 −6 / 0 C .

[

]

∴ ρ 60 = ρ 0 [1 − γ .(60 − 0)] = 13.5955 1 − 60 × 183 × 10 −6 = 13.45 gm/cc.

#Example 11 The density of mercury at 15 0 C is 13.56 gm/cc and the coefficient of real expansion of it is 18 × 10 −5 / 0 C . What will be the mass of 600 c.c. mercury at 130 0 C and what will be volume of 600 gm mercury at this temperature? [H.S. ‘90] Solution. We know, ρ 2 = ρ1 [1 − γ .(t 2 − t1 )] .

Here, ρ1 = 13.56 gm/cc, γ = 18 × 10 −5 / 0 C , t1 = 15 0 C , t 2 = 130 0 C . ∴ ρ 2 = 13.56 × 1 − 18 × 10 −5.(130 − 15) = 13.279 gm/cc ∴ The mass of 600 cc mercury at 130 0 C = 600 × 13.279 = 7967.4 gm, the volume of 600 600 gm mercury = = 45.184 c.c. 13.279

[

]

#Example 12 The density of glass is 2.6 gm/cc at 10 0 C and 2.596 gm/cc at 60 0 C . What is the average coefficient of linear expansion of glass in this temperature interval? [H.S. ‘88] Solution. We know, ρ1 = ρ 2 [1 + γ .∆t ] .

Here, ρ1 = 2.6 gm/cc, ρ 2 = 2.596 gm/cc and ∆t = t 2 − t1 = (60 − 10) = 50 0 C . ρ − ρ 2 2.6 − 2.596 0.004 ∴γ = 1 = = 3.08 × 10 −5 / 0 C = ρ 2 × ∆t 2.596 × 50 2.596 × 50 ∴ The coefficient of linear expansion, α =

γ 3

=

3.08 × 10 −5 = 1.027 × 10 −5 / 0 C . 3

--Determination of the coefficient of apparent expansion of liquid--

The following is one of the experimental methods of measuring the coefficient of apparent expansion. Weight Thermometer method: The weight thermometer is a glass bulb with a narrow and bent neck attached with it (shown in the fig.**) Include Figur Fig. to be included

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The following steps are performed: • The empty weight thermometer is at first weighed making it dry and clean. • Now the narrow pipe of it is immersed into an experimental liquid and the bulb is then heated. Thus the air inside the bulb gets heated and expands; some amount of air comes out through the liquid. After this the bulb is cooled down. Some liquid enters into the bulb through the narrow channel. This way the weight thermometer is repeatedly heated and cooled in order to fill it up with liquid completely. • The temperature of the weight thermometer is brought down to room temperature ( t1 ) keeping the narrow pipe immersed in the liquid. • Now the liquid filled thermometer is weighed after its outer surface is cleanly wiped out. • Next, the bulb of the thermometer is placed inside water taken in a container and the temperature of water is increased to t 2 . Some amount of liquid comes out of the thermometer due to the increase in temperature. • The weight thermometer is then taken away and allowed to cool down. Again the thermometer is weighed after its outer surface is wiped out properly. In this situation, the measured weight is the sum of the empty thermometer and the liquid inside the thermometer at temperature t 2 . Let us suppose, the mass of the empty thermometer = w1 , the mass of the thermometer when it is filled with liquid at temperature t1 = w2 and the mass of the thermometer with liquid at temperature t 2 = w3 . ∴The mass of liquid required to fill up the weight thermometer at temperature t1 is m1 = w2 − w1 . The mass of liquid inside the thermometer at temperature t 2 is m2 = w3 − w1 . Ignoring the expansion of the weight thermometer itself, we can write, the volume of m1 gm liquid at t1 = the volume of m 2 gm liquid at t 2 . m m m ρ ∴ 1 = 2 Or, 1 = 1 , …………………………(1) ρ1 ρ 2 m2 ρ 2 where ρ1 and ρ 2 are densities of the liquid at t1 and t 2 . If γ ′ be the coefficient of apparent expansion of the liquid,

ρ1 = 1 + γ ′.(t 2 − t1 ) ………………(2) ρ2 From equations (1) and (2) we get, ρ1 = ρ 2 [1 + γ ′.(t 2 − t1 )] Or,

m1 − m2 m1 = 1 + γ ′.(t 2 − t1 ) Or, γ ′ = . m2 (t 2 − t1 ) m2 Here, γ ′ is to be called the coefficient of apparent expansion of the liquid since we have not taken into of the expansion of the container that is the thermometer. As the weight of the liquid is measured to determine the coefficient of apparent expansion, this thermometer is called weight thermometer. If the coefficient γ ′ is already

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known, the temperature can also be determined by this. That is why this is called a thermometer. The coefficient of real expansion of liquid (γ ′) can also be determined by this weight thermometer once we know the coefficient of volume expansion (γ g ) of the material (usually, glass) of which the weight thermometer is made of. To find γ g , a liquid of known γ (the coefficient of real expansion) is taken and γ ′ (the coefficient of apparent expansion) is determined by the weight thermometer. Thus γ g can be calculated from the relation: γ = γ ′ + γ g . Now, the weight thermometer can be

used to determine the coefficient of real expansion (γ ) of an unknown liquid by measuring the coefficient of apparent expansion (γ ′) of that liquid and the adding γ g with it: γ = γ ′ + γ g . ---------------Numerical Examples with Solutions-------------

#Example 1 A weight thermometer can contain 304.48 gm mercury at 20 0 C and 301 gm mercury at 100 0 C . What will be the coefficient of volume expansion of the material (glass) of the container of weight thermometer if the coefficient of real expansion of mercury is 0.00018 / 0 C ? Solution. m1 − m2 The coefficient of apparent expansion of mercury, γ ′ = . m2 (t 2 − t1 )

Here, m1 = 304.48 gm, m 2 = 301 gm, and t1 = 20 0 C , t 2 = 100 0 C . 304.48 − 301 3.48 ∴γ′ = = = 0.0001445 / 0 C 301 × (100 − 20) 301 × 80 We know, the coefficient of real expansion of mercury, γ = γ ′ + γ g , where γ g is the coefficient of volume expansion of glass of which the weight thermometer is made of. ∴ γ g = γ − γ ′ = 0.00018 − 0.0001445 = 0.0000355 / 0 C Hence, the coefficient of linear expansion of glass = γ g 0.0000355 = = 0.0000118 = 1.18 × 10 −5 / 0 C . 3 3 #Example 2 The mass of a weight thermometer is 1.8760 gm. This is filled up with mercury at 0 0 C . An amount of 0.2351 gm mercury comes out when this is heated up to 100 0 C . Find the coefficient of apparent expansion of mercury if the thermometer now weighs 16.8625 gm with the residual mercury. Solution.

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The mass of the mercury that comes out = 0.2351 gm. The mass of the residual mercury in the thermometer = 16.8625 − 1.8760 = 14.9865 gm. ∴ The coefficient of apparent expansion of mercury = (mass of mercury that comes out)/ (mass of residual mercury × temperature rise) = 0.2351 = 15.7 × 10 −5 / 0 C . 14.9865 × 100 #Example 3 A weight thermometer contains 100 gm mercury at 20 0 C . How much mercury does come out when this is heated to 80 0 C ? For mercury, γ = 18 × 10 −5 / 0 C and for glass, α g = 9 × 10 −6 / 0 C . Solution. The coefficient of apparent expansion of mercury, γ ′ = the coefficient of real expansion ( γ ) – the coefficient of volume expansion of glass ( γ g ) = 18 × 10 −5 − 9 × 10 −6 × 3 =

15.3 × 10 −5 / 0 C . The mass of mercury at 20 0 C is m1 = 100 gm. Let the mass of mercury at 100 0 C be m 2 . 100 − m2 100 − m 2 ∴ γ′= Or, 15.3 × 10 −5 = Or, m 2 = 99.09 gm. m 2 × (80 − 20) m2 × 60 Thus the mass of mercury that comes out = 100 – 99.09 = 0.91 gm. #Example 4 A weight thermometer is filled up with water. Find the ratio of the mass of water that comes out during heating from 40 0 C and 60 0 C and that due to heating from 60 0 C and 80 0 C . The coefficient of volume expansion of water in the range of 40 0 C and 60 0 C is 45.8 × 10 −5 / 0 C and the coefficient in the range of 60 0 C and 80 0 C is 58.7 × 10 −5 / 0 C . The expansion of the container of the weight thermometer is negligible. Solution. Let the volume of the thermometer be V , the density of water at 60 0 C be ρ 60 and the

density of water at 80 0 C be ρ 80 . The increase in the volume of water due to heating from 40 0 C to 60 0 C is = V × 45.8 × 10 −5 × (60 − 40) = V × 45.8 × 10 −5 × 20 . This is the volume that comes out of the thermometer due to heating from 40 0 C to 60 0 C . The mass of this volume is = V × 45.8 × 10 −5 × 20 × ρ 60 = M 1 (say,). In the same way, we can write, the volume that comes out of the thermometer due to heating from 60 0 C to 80 0 C = V × 58.7 × 10 −5 × (80 − 60) = V × 58.7 × 10 −5 × 20 . The mass of this volume is = V × 58.7 × 10 −5 × 20 × ρ 80 = M 2 (say,).

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M 1 V × 45.8 × 10 −5 × 20 × ρ 60 45.8 ρ 60 ∴ The ratio of the two masses = = = × . M 2 V × 58.7 × 10 −5 × 20 × ρ 80 58.7 ρ 80 Now, we can write, ρ 60 = ρ 80 (1 + 58.7 × 20) = ρ 80 × 1.01174 . Therefore,

M 1 45.8 = × 1.01174 = 0.79 . M 2 58.7

--Determination of the coefficient of real expansion of liquid--

Dulong and Petit’s Method: The coefficient of real expansion of a liquid can be determined by Dulong and Petit’s method as the expansion of the container of liquid plays no role in the calculation. The experimental arrangement is shown in the fig**. Fig. to be Inc included The experimental liquid is taken in the bent tube ABCD as indicated in the figure. The two vertical arms, AB and CD of the tube are placed inside two jacket tubes J 1 and J 2 . Ice cold water is sent through J 1 and steam through J 2 . As a result of this the AB arm cools down and the CD arm gets heated up. Two thermometers T1 and T2 indicate the temperatures inside the jacket tubes which are the temperatures of the liquid columns inside the tubes. The horizontal portion BC of the bent tube is covered with a wet blotting paper so that heat should not flow from one arm to the other due to the difference in temperatures in them. The temperature of AB is less than that of CD. For this reason, the density of liquid in AB is more in comparison. Thus the height of liquid column in AB will be more than that in CD in order to maintain equilibrium. The heights of the liquid columns in the two arms are measured from the horizontal part BC with the help of a cathetometer telescope or a traveling microscope. At equilibrium, for the column of liquid in AB: the height = h1 , the density = ρ1 , the temperature = t1 , for the column of liquid in CD: the height = h2 , the density = ρ 2 , the temperature = t 2 . As the pressure at the bottom of the two arms i.e. at B and C are same in equilibrium we can write, h1 ρ1 g = h2 ρ 2 g [ g = the acceleration due to gravity] h1 ρ1 = h2 ρ 2 Or, We know, ρ1 = ρ 2 [1 + γ .(t 2 − t1 )] ∴ h1 ρ 2 [1 + γ .(t 2 − t1 )] = h2 ρ 2 Or, h1 [1 + γ .(t 2 − t1 )] = h2 h2 − h1 ∴γ = = (the height difference in the liquid columns)/(the height of the cold h1 .(t 2 − t1 ) column × temperature difference in the liquid columns)

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Thus the coefficient of real expansion of liquid is given by pressures in the two liquid columns. The pressure is independent of the area of cross-section of the tubes and for this reason the expansion of the tubes does not affect the calculation of the coefficient of expansion of the liquid. Thus this method is used to measure the coefficient of real expansion of liquid. ---------------Numerical Examples with Solutions------------#Example 1 Two arms of a liquid filled U-tube are kept at 0 0 C and 100 0 C temperatures. If the height of the cold column is of 58 cm and the height of hot column is 1.6 cm more than this, find the coefficient of real expansion the liquid. Solution. The height difference of the two columns of liquid = 1.6 cm., the temperature difference = 100 – 0 =100 0 C . ∴ The coefficient of real expansion, γ = (height difference in liquid columns)/(height of 1.6 cold column× temperature difference) = = 0.0002758 / 0 C . 58 × 100 #Example 2 It is seen in the Dulong and Petit’s experiment that the heights of cold and hot mercury columns are 50.1 cm and 51 cm, respectively. If the temperature of the cold column is 0 0 C , find the temperature of the hot column. The coefficient of real expansion of mercury, γ = 1.79 × 10 −4 / 0 C . Solution. Here, the heights of mercury columns, h1 = 50.1 cm, h2 = 51 cm and the temperature of cold column, t1 = 0 0 C . Let the temperature of the hot column be t 2 . h2 − h1 51 − 50.1 0.9 Or, 1.79 × 10 − 4 = = We know, γ = h1 .(t 2 − t1 ) 50.1 × (t 2 − 0) 50.1 × t 2

0.9 0.9 × 10 4 ∴ t2 = = = 100.35 0 C . −4 50.1 × 1.79 50.1 × 1.79 × 10 #Example 3 The heights of two mercury columns at equilibrium in a U-tube at temperatures 100 0 C and 0 0 C are 76.35 cm and 75 cm, respectively. Determine the coefficient of real expansion of mercury. [H.S. ‘91] Solution. The height and temperature of cold column, h1 = 75 cm, t1 = 0 0 C and the height and temperature of hot column, h2 = 76.35 cm, t 2 = 100 0 C . ∴ The coefficient of real expansion of mercury, h −h 76.35 − 75 γ = 2 1 = = 1.8 × 10 − 4 / 0 C . h1 .(t 2 − t1 ) 75 × (100 − 0)

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-----------Temperature Correction of Barometer Reading-----------

The standard atmospheric pressure is considered equal to the pressure exerted by 76 cm long mercury column at 0 0 C . This standard pressure can be obtained at sea level. We generally take barometer readings at room temperature (standard room temperature is 27 0 C ). Thermal expansion occurs in both the mercury column and the metal scale attached to it. Therefore, the actual atmospheric pressure is found after correcting the reading of mercury column by two ways: (i) the expansion of scale and (ii) the expansion of mercury column. • Correction due to expansion of barometer scale: Let us assume, the metal scale gives correct reading at 0 0 C . Thus the division of 1 cm is correct at 0 0 C . If the coefficient of linear expansion of the material of the scale is α , the correct length for the reading of 1 cm will be (1 + α .t ) cm. Therefore, the correct length for the barometer reading of H t cm at t 0 C will be •

H = H t (1 + α .t ) cm…………….(1) Correction due to expansion of mercury column: If the density of mercury be ρ t at t 0 C , the pressure due to H cm long mercury column is = Hρ t g , g being the acceleration due to gravity. Let this pressure is equal to the pressure due to H 0 cm long mercury column at 0 0 C . Thus H 0 cm will be correct value of barometer reading at this temperature. We can now write, H 0 ρ 0 g = Hρ t g [ ρ 0 = density of mercury at 0 0 C ]

ρt 1 [Q ρ 0 = ρ t (1 + γ .t ) ; γ = the coefficient of = H. ρ0 (1 + γ .t ) real expansion of mercury] Or, H 0 = H .(1 + γ .t ) −1 = H .(1 − γ .t ) [Neglecting the higher order in the binomial expansion] …………………….(2) Or,

H0 = H.

Now combining equations (1) and (2) we find, H 0 = H t (1 + α .t )(1 − γ .t )

[The term αγ .t 2 has been neglected as α Or, H 0 = H t [1 − (γ − α ).t ] and γ both are small]……………….(3)

Thus if the barometer reading is found to be H t (height of mercury column) at t 0 C , the actual reading can be determined from the above formula. ---------------Numerical Examples with Solutions------------#Example 1

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The scale divisions on the brass scale of a barometer are made at 0 0 C . Find the actual height of mercury column when the scale reading is 76 cm at 30 0 C . For mercury, γ = 18 × 10 −5 / 0 C and for brass, α = 18 × 10 −6 / 0 C . Solution. Here, measured height of mercury column, H t = 76 cm, temperature, t = 30 0 C . Given, γ = 18 × 10 −5 / 0 C for mercury and α = 18 × 10 −6 / 0 C for brass. We know, H 0 = H t [1 − (γ − α ).t ] = 76 × 1 − (18 × 10 −5 − 18 × 10 −6 ) × 30

[

[

]

]

= 76 × 1 − (18 − 1.8) × 10 −5 × 30 = 76 − 76 × 16.2 × 10 −5 × 30 = 76 − 0.36936 = 75.63064 cm.

Apparent Loss of Weight of a Solid Body ------Immersed in a Liquid at Different Temperatures-----

When a solid body is immersed in a liquid, it displaces liquid equal to its own volume. There is an apparent loss of weight of the solid body due to this. The amount of apparent loss of weight of the solid body is different at different temperatures due to the following reasons: • There is a change in volume of the solid body due to thermal expansion (Volume increases when temperature is increased.). • There is a change in density of the liquid due to temperature rise (Density usually decreases as temperature is increased.). Let us assume the weight of a solid body in air to be W . The weight of this is W1 when it is immersed in a liquid at temperature t1 . ∴The apparent weight loss of this solid body = W − W1 = M 1 . If the volume of the solid body at temperature t1 is V1 then the volume of liquid displaced by the solid body = V1 . Now we can write, M 1 = V1 ρ1 , where ρ1 is density of the liquid at temperature t1 . Similarly, we can write M 2 = V2 ρ 2 , where M 2 , V2 and ρ 2 are corresponding quantities at a different temperature t 2 . Now if γ be the coefficient of real expansion of the liquid and γ S be the coefficient of volume expansion of the material of the solid body, we can write

ρ 2 = ρ1 [1 − γ (t 2 − t1 )] = ρ1 (1 − γ .t ) , V2 = V1 (1 + γ S .t )

[ t = t 2 − t1 , say]

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∴ M 2 = V1 (1 + γ S .t ) × ρ1 (1 − γ .t ) = V1 ρ1 (1 + γ S .t )(1 − γ .t ) Or, M 2 = M 1 [1 − (γ − γ S ).t ]

[Neglecting the γγ S t 2 term]

The coefficient of real expansion of liquid (γ ) is in general much greater than the coefficient of volume expansion of solid (γ S ) : γ > γ S . If we assume t 2 > t1 , the difference in temperature t = (t 2 − t1 ) > 0 . ∴We have, M 2 > M 1 and thus we can write, W2 > W1 . Therefore, the apparent weight of a solid body increases due to the increase in temperature of the liquid in which the solid body is immersed. ---------------Numerical Examples with Solutions------------#Example 1 A metal bob weighs 50 gm in air. The metal bob weighs 45 gm when it is immersed in a liquid at temperature 25 0 C . The weight of the same becomes 45.1 gm when the temperature of the liquid is increased to 100 0 C . Find the coefficient of volume expansion of the liquid if the coefficient of linear expansion of the metal is 12 × 10 −6 / 0 C . [I.I.T] Solution. The apparent loss of weight of the metal bob at 25 0 C is M 1 = 50 − 45 = 5 gm. The apparent loss of weight of the bob at 100 0 C is M 2 = 50 − 45.1 = 4.9 gm. Temperature difference, t = t 2 − t1 = 100 − 25 = 75 0 C . The coefficient of volume expansion of the metal, γ S = 3 × 12 × 10 −6 / 0 C = 36 × 10 −6 / 0 C . If γ be the coefficient of real expansion of liquid, we can write, M 2 = M 1 [1 − (γ − γ S ).t ] Or, 4.9 = 5 × 1 − (γ − 36 × 10 −6 ) × 75

[

]

0.1  4.9  1 Or, γ − 36 × 10 −6 = 1 − = = 2.67 × 10 − 4 × 5 75 5 × 75   −6 Or, γ = 36 × 10 + 2.67 × 10 −4 = (0.36 + 2.67 ) × 10 −4 Or, γ = 3.03 × 10 −4 / 0 C . #Example 2 A glass rod weighs 90 gm in air. The weight of the glass rod is 49.6 gm in a liquid at 12 0 C . The weight of this is 51.9 gm in the same liquid at 97 0 C . Determine the coefficient of real expansion of the liquid. (The coefficient of volume expansion of glass = 2.4 × 10 −5 / 0 C .) [H.S. ‘94] Solution.

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Let at 12 0 C temperature, the volume of the glass rod be V1 and the density of liquid be ρ1 ; the mass of the displaced liquid (when the rod is immersed in it) at this temperature is = 90 – 49.6 = 40.4 gm. 40.4 ∴V1 = ……………….(1)

ρ1

0

At 97 C , the mass of displaced liquid at is = 90 – 51.9 = 38.1 gm. If the volume of the glass rod at this temperature be V2 and the density of liquid be ρ 2 , we can write, 38.1 V2 = ……………….(2)

ρ2

From (1) and (2) we have,

For the glass rod,

V2 ρ1 38.1 = × …………..(3) V1 ρ 2 40.4

[

V2 = V1 1 + 2.4 × 10 −5 × (97 − 12)

]

V2 = 1 + 2.4 × 10 −5 × 85 = 1.00204 ……………….(4) V1 If the coefficient of real expansion of liquid be γ , we can write,

Or,

ρ1 = 1 + γ × 85 ……………….(5) ρ2 Combining (3), (4) and (5) we have, 38.1 1.00204 × 40.4 Or, 1 + 85γ = 1.00204 = (1 + γ × 85) × 40.4 38.1 1.00204 × 40.4 2.383 Or, 85γ = −1 = 38.1 38.1 −4 0 ∴ γ = 7.4 × 10 / C (approx). ρ1 = ρ 2 [1 + γ × 85] Or,

#Example 3 The weight of a solid body is 50 gm in a liquid at 25 0 C and the weight is 52 gm in the same liquid at 75 0 C . The coefficient of linear expansion of the material of the solid body = 6.6 × 10 −6 / 0 C and the coefficient of volume expansion of the liquid = 7.3 × 10 −4 / 0 C . What is the real weight of the body? [J.E.E. ‘90] Solution. Let the real weight of the solid body be M gm. The loss of weight of the body at 25 0 C is = M 1 = M − 50 gm, the loss of weight of the body at 75 0 C is = M 2 = M − 52 gm and the temperature difference, t = t 2 − t1 = 75 − 25 = 50 0 C . The coefficient of volume expansion of the material of the solid body, γ S = 3 × 6.6 × 10 −6 / 0 C = 19.8 × 10 −6 / 0 C . Now, M 2 = M 1 [1 − (γ − γ S ).t ] , where γ is the coefficient of real expansion of the liquid.

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∴ ( M − 52) = ( M − 50) 1 − (7.3 × 10 −4 − 19.8 × 10 −6 ) × 50 = ( M − 50) 1 − (7.3 − 0.198) × 10 −4 × 50 = ( M − 50) 1 − 7.102 × 10 −4 × 50 = ( M − 50) × 0.96449 Or, M − 52 = M × 0.96449 − 48.2245 Or, M (1 − 0.96449) = 52 − 48.2245 3.7755 Or, M = = 106.32 gm (approx.). 0.03551

]

]

#Example 4 A solid sphere, whose diameter is 7 cm and mass 266.5 gm, floats on a liquid. The sphere starts sinking when the liquid is heated to 35 0 C . If the density of the liquid is 1.527 gm/cc at 0 0 C , find the coefficient of volume expansion of the liquid. Neglect the thermal expansion of the sphere. [I.I.T.] Solution. 3

4 7 Volume of the sphere = π ×   c.c. 3 2 3

4 7 ∴ The volume of displaced liquid at 35 C is = π ×   c.c. If the density of the liquid 3 2 0

3

4 7 at 35 C is ρ , the mass of the displaced liquid = π ×   × ρ gm. 3 2 0

3

4 7 π ×   × ρ = 266.5 [the condition for floating] 3 2 1.527 Also, ρ 0 = ρ [1 + γ (35 − 0)] = ρ (1 + 35.γ ) ⇒ ρ = 1 + 35.γ 0 where the density of the liquid at 0 C is ρ 0 = 1.527 gm/cc, and γ is the coefficient of volume expansion of the liquid. Hence we have, We can now write,

3

4 7 π ×   × 1.527 × (1 + 35.γ ) −1 = 266.5 3 2 Or, Or,

4 × 3.14 × 7 3 × 1.527 = 1.0285 3 × 2 3 × 266.5 1.0285 − 1 γ = = 8.14 × 10 − 4 / 0 C . 35

(1 + 35.γ ) =

#Example 5 A metal piece weighs 46 gm in air. The weight of this is 30 gm when immersed in a liquid at temperature 27 0 C and of specific gravity 1.24. The weight of the piece is found to be 30.5 gm when the temperature of the liquid is increased to 42 0 C . The specific gravity of the liquid is 1.20 at 42 0 C . Determine the coefficient of linear expansion of the metal. [H.S. ‘98] Solution.

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At 27 0 C , the mass of the displaced liquid = 46 – 30 = 16 gm; the volume of this 16 c.c. displaced liquid is = 1.24 16 ∴ The volume of the metal piece at 27 0 C is V1 = c.c. 1.24 Similarly, the mass of the displaced liquid at 42 0 C is = 46 – 30.5 = 15.5 gm; the volume 16 c.c. of this displaced liquid at 42 0 C is = 1.20 15.5 ∴ The volume of the metal piece at 42 0 C is V2 = c.c. 1.20 Now, we can write, V2 = V1 [1 + γ .(42 − 27)], where γ = the coefficient of volume expansion of the metal. Thus we have, 15.5 16 [1 + γ × 15] = 1.20 1.24 15.5 × 1.24 1 + 15.γ = = 1.001 Or, 16 × 1.20 1.001 − 1 0.001 ∴ = γ = 15 15 γ 0.001 = 2.2 × 10 −5 / 0 C. Thus the coefficient of linear expansion of the metal, α = = 3 45 ---------------Anomalous Expansion of Water-------------

Generally, the volume of a liquid increases and the density decreases when it is heated. But water does not behave in the same way in a certain temperature range. There is an anomaly. When a certain mass of water is heated from 0 0 C to 4 0 C , the volume of it keeps decreasing instead of increasing. From 4 0 C above, the volume increases and so the density decreases again like any other liquid. Thus the volume of a certain mass of water is minimum and hence the density of water is maximum at 4 0 C . In the same way, water behaves like any other liquid when it is cooled up to 4 0 C . But when cooled down from 4 0 C to 0 0 C , the volume of water is seen to increase. So it is observed, water behaves differently in the temperature range of 0 0 C to 4 0 C . Such a different kind of expansion of water in the temperature range of 0 0 C to 4 0 C is called anomalous expansion of water. In the following, it is demonstrated graphically how the density of water and the volume of one gm of water change with temperature. Fig. to be included

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From the above discussions and from the graphs we have the following conclusions: 1. The volume of water keeps decreasing when the temperature is raised from 0 0 C to 4 0 C . This expansion is contrary to the usual thermal expansion. 2. Above 4 0 C , water expands when heated like any other liquid or solid materials. 3. The volume of a given mass of water is smaller at 4 0 C than at any other temperature. Thus the density is maximum at 4 0 C (3.98 0 C , to be precise). Note, just below 0 0 C when water becomes solid ice, the volume is considerably larger (and density smaller). Thus in reality we see, ice floats on water, frozen water pipes crack in very cold climates. 4. The coefficient of volume expansion of water is negative in the temperature range of 0 0 C to 4 0 C . 5. The density at and around 4 0 C is very close to 1. For this reason , the density of water at 4 0 C is assumed to be unity. ------------Experimental Study of Anomalous Expansion of Water------------

Anomalous expansion of water can be demonstrated by a dilatometer or volume thermometer (see fig.**). Fig. to be included 1 The th fraction of the volume of a dilatometer is filled with mercury. The coefficient of 7 volume expansion of mercury is seven times compared to the coefficient of volume expansion of glass. Thus for any increase in temperature, the expansion in mercury and the expansion in glass will be equal. For this reason the volume of the other part of the dilatometer will not change. Next, the dilatometer is filled with distilled water up to a level around the middle of the tube. Now, the dilatometer bulb with the tube is placed inside ice-cold water at 0 0 C . After a while the water surface inside the tube attains a certain fixed level. The reading of this water level can be noted from the attached scale. After this the temperature of the cold water is increased from 0 0 C to 4 0 C while keeping the dilatometer inside this. Now the water level is seen to come down rather than going up. This shows that the volume of water decreases as the temperature is increased from 0 0 C to 4 0 C . Again after 4 0 C , the water level inside the tube starts rising which indicates that the volume of water increases due to the increase in temperature like any other liquid. ---Hope’s Experiment to Show that Water has Maximum Density at 4 0 C ---

Hope’s experiment shows that the density of water is maximum at 4 0 C . The simple experimental arrangement of scientist Hope is shown in the adjacent figure** where AB

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is a long metallic cylinder. Two thermometers T1 and T2 are inserted through two holes one around the top and other around the bottom. There is a container J surrounding the middle part of the cylinder which contains a freezing mixture of ice and salt. The temperature of this mixture is in between -20 0 C and -15 0 C . There is an outlet attached to the container to drain out the water from melting ice. Fig. to be included The cylinder is filled up with pure water at approximately 10 0 C . The readings in the two thermometers will be equal before keeping the freezing mixture in the container J . As the freezing mixture is kept in the container J , the middle portion of the cylinder begins to cool down slowly. In this situation, the readings in the two thermometers T1 and T2 have to be noticed. The temperature shown by T2 will come down slowly and reach 4 0 C where it remains fixed. During this time the temperature reading in T1 remains close to 10 0 C . After a while the reading in T1 also comes down slowly and then rapidly reaches to 0 0 C . If we wait for more time, we may see floating ice on the water surface in the cylinder but the thermometer reading in T2 remains fixed in 4 0 C even then. The reading of T2 is seen to come down a little less than 4 0 C when the experiment is carried out for a long time. Explanation: The water in the mid part of the cylinder begins to cool down at first due to the freezing mixture kept in the container J around this. As a result of this the density of water at that position increases and heavier water comes down towards the bottom. The warmer water at the bottom rises up as its density is less. This again cools down with the of freezing mixture and comes down again. Thus convection current is set up in the lower portion of the cylinder and the reading in the T2 thermometer continuously goes down. This continues until the temperature of water at the bottom attains 4 0 C . During this time the temperature of water at the upper portion does not change and thus the thermometer reading in T1 shows no change. After a while the reading in T1 begins to come down. This is because the temperature of water in the of the freezing mixture now comes down below 4 0 C and thus the density of this water decreases. However, this water can not come down as it is lighter. On the other hand, it can not go up as the water there is warmer and lighter. As a result of this the temperature of water at this middle portion reaches 0 0 C and then the water freezes to become ice in the of freezing mixture. Ice floats on the surface of water. For this the water on the upper portion of the cylinder soon reaches 0 0 C . Thus the thermometer T1 shows 0 0 C reading. But the temperature of water at the bottom remains fixed at 4 0 C and thus the reading in thermometer T2 remains fixed at 4 0 C . We know that the liquid of highest density should remain at the bottom. As the temperature of water at the bottom remains fixed at 4 0 C , we conclude that water has highest density at 4 0 C .

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The temperature of water at the bottom becomes slightly less than 4 0 C after a long time as some heat is slowly transmitted to the upper layers due to convection. --------Effect of Anomalous Expansion of Water on Marine Life-------

The anomalous behaviour of water has an important effect on plants and animals in lakes and oceans. In colder regions on earth in winter, the surface water of ocean or lake cools first and flows down to the bottom because of its greater density. The whole water body continues to cool through convection until it reaches 4 0 C . Then the surface water is cooled further, becomes less dense compared to the water at the bottom layers and this remain at the top. The water at the bottom remains at 4 0 C being most dense. Successive layers of lower temperatures than 4 0 C are formed from bottom to top (see an accompanying picture, fig**). The surface water then freezes to become ice in with cold air. The ice floats as it is less dense than water. As the surface is covered with ice, it slows further ice formation since ice is a much poorer conductor of heat than water. Thus water at the bottom of deep lakes or oceans in colder regions on earth does not easily become ice which allows marine life to continue at the bottom. Fig. to be included Note: • Very deep bodies of water are not ice-covered in the coldest of winters, because all the water in the lake must be cooled to 4 0 C before lower temperatures to be reached. Often the span of a winter is not enough for this. If only some of water is at 4 0 C , it lies at the bottom. • Because of water’s high specific heat and poor conductivity of heat, the bottom of deep lakes and oceans in cold regions is a constant 4 0 C all the year round. • If water would behave like most other substances, this would contract continuously on cooling and freezing. In this case water bodies would start freezing from bottom up, convections due to density differences would continuously carry warmer water up to the surface, efficient cooling would take place and the lakes or oceans would freeze solid as a whole more easily. This would then destroy all plants, fishes and other animal life in water in very cold climates. ---------------Numerical Problems with Solutions------------#Example 1 The cross-section of the capillary tube of a glass thermometer is A0 and the volume of the

mercury bulb is V0 at 0 0 C . If the bulb is filled with mercury at 0 0 C , find the reading of the thermometer at t 0 C . Given, the coefficient of linear expansion of glass = α and the [J.E.E. ‘94] coefficient of volume expansion of mercury = β .

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Solution. If the volume of the glass bulb be V g at t 0 C ,

V g = V0 (1 + γ .t ) = V0 (1 + 3α .t ) , the coefficient of volume expansion of glass, γ = 3α . Volume of mercury at t 0 C , Vm = V0 (1 + β .t ) . Therefore, the volume of mercury that comes out of the glass bulb at t 0 C , ∆V = Vm − V g = V0 (1 + β .t ) − V0 (1 + 3α .t ) = V0 ( β − 3α ).t The area of cross-section of the capillary tube at t 0 C , At = A0 (1 + 2α .t ) . Let the length of the mercury column be l at t 0 C . Then we can write, V ( β − 3α ).t V0 ( β − 3α ).t At × l = V0 ( β − 3α ).t Or, l = 0 = At A0 (1 + 2α .t ) V V Or, l = 0 ( β − 3α ).t.(1 + 2α .t ) −1 = 0 ( β − 3α ).t.(1 − 2α .t ) A0 A0 V ∴ l = 0 ( β − 3α ).t = l 0 ( β − 3α ) , where l 0 is the length of mercury column at 0 0 C . A0 The length l of the mercury column in the capillary tube represents the temperature reading at t 0 C . #Example 2 There is some mercury in a container of volume 1 litre. It is seen that the volume of air inside this container remains constant at different temperatures. What is the volume of mercury in the container? The coefficient of linear expansion of the material of the container = 9 × 10 −6 / 0 C and the coefficient of real expansion of mercury = 1.8 × 10 −4 / 0 C . [I.I.T.] Solution. Suppose, the inner volume of the container be V c.c. and consider an arbitrary increase in temperature, ∆t 0 C . The condition for the volume of air inside the container to remain constant due to the rise in temperature by ∆t 0 C is the following. The increase in volume of mercury = the increase in the volume of container. Or, V × ∆t × 1.8 × 10 −4 = 1000 × ∆t × 3 × 9 × 10 −6 27 × 10 −6 Or, V = = 150 c.c. 1.8 × 10 − 4 #Example 3 A barometer with a brass scale reads 75.34 cm at 25 0 C . The scale divisions are made at 20 0 C . What will be the reading at 0 0 C ? The coefficient of real expansion of mercury is 18 × 10 −5 / 0 C and the coefficient of linear expansion of brass is 18 × 10 −6 / 0 C . Solution. The actual reading of the barometer scale should be more than the apparent reading due to the expansion of the barometer scale.

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If the actual reading is h and the apparent reading is H , we can write, h = H (1 + α .t ) = 75.34 × 1 + 18 × 10 −6 × (25 − 20) Or, h = 75.34 × (1 + 18 × 10 −6 × 5) = 75.34 × (1 + 9 × 10 −5 ) = 75.34 × 1.00009 = 75.35 cm (approx) So, 75.35 cm will be the correct barometer reading at 25 0 C as this is the actual length of mercury column at this temperature. The actual length of mercury column at 0 0 C , H 0 = 75.35 × [1 − 18 × 10 −5 × 25] = 75.35 × (1 − 0.0045) = 75.35 × 0.9955 = 75.01 cm. Therefore the actual barometer reading at 0 0 C should be 75.01 cm. However, the barometer scale will not show this correct reading at 0 0 C as the scale gives correct reading only at 20 0 C . Hence the barometer reading at 0 0 C will be = 75.01 × 1 + 18 × 10 −6 × 20 = 75.01 × (1 + 36 × 10 −5 ) = 75.01 × 1.00036 = 75.04 cm.

[

]

[

]

#Example 4 The coefficient of volume expansion of benzene and wood are respectively, 1.2 × 10 −3 / 0 C and 1.5 × 10 −4 / 0 C ; the densities of them at 0 0 C are 900 kg/m 3 and 800 [J.E.E.] kg/m 3 respectively. At which temperature wood will sink into benzene? Solution. Wood will get immersed into benzene at some temperature when the density of wood will be equal to the density of benzene. Let the required temperature be t 0 C and then the density be ρ . ∴ We can write, 900 = ρ 1 + 1.2 × 10 −3 × t (1) −4 800 = ρ 1 + 1.5 × 10 × t (2) and Dividing (1) by (2) we have, 9 1 + 1.2 × 10 −3 × t = 8.8 1 + 1.5 × 10 − 4 × t 9 − 8.8 t= Or, t = 21.7 0 C . Or, −3 −4 8.8 × 1.2 × 10 − 9 × 1.5 × 10

( (

(

) )

) (

)

#Example 5 A metal piece of density 8 gm/cc is hanged with a mass less string from a wooden hook. The tension in the string is 56 gm-weight. If the whole arrangement is immersed in a liquid at 40 0 C what will be the tension in the string? The surrounding temperature during this time is 20 0 C . The specific gravity of the liquid is 1.24 at 20 0 C ; the coefficients of volume expansion of the liquid and the metal are respectively, 4 × 10 −5 / 0 C and 8 × 10 −4 / 0 C . [J.E.E. ‘88] Solution.

Fundamental Physics -I by Dr. Abhijit Kar Gupta (email: [email protected]) The volume of the metal piece at 20 0 C is V1 =

25

56 = 7 c.c. 8

The volume of the metal piece at 40 0 C , V2 = V1 1 + 8 × 10 −4 × 20 = 7 × (1 + 16 × 10 −3 ) = 7 × 1.016 c.c. We can write, the volume of the displaced liquid is = V2 ∴ The mass of the displaced liquid is = V2 × ρ , where ρ is the density of the liquid at 40 0 C . 1.24 1.24 1.24 = = gm/cc We can write, ρ = −5 −4 1.0008 1 + 4 × 10 × 20 1 + 8 × 10 1.24 ∴ The mass of the displaced liquid = 7 × 1.016 × = 8.81 gm. 1.0008 Thus the buoyancy = 8.81 gm-weight. Therefore, the resultant tension in the string = 56 – 8.81 = 47.19 gm-weight.

[

]

(

)

#Example 6 A body floats in water at 4 0 C while keeping 0.98 fraction of its volume immersed. At what temperature will the body sink fully? Given, the volume expansion coefficient of water = 3.3 × 10 −4 / 0 C . The volume expansion of the body is to be neglected. Solution. Let the volume of the body = V cc. and the density of water at 4 0 C is = ρ1 gm/cc. Suppose, the required temperature = t 0 C and the density of water at that temperature = ρ 2 gm/cc. [Note that the required temperature must be greater than the initial temperature to have the body fully immersed: t > 4 ]

We can write,

ρ1 1 = …………..(1) ρ 2 0.98 The densities at two temperatures are related by, ρ1 ρ1 = ρ 2 [1 + 3.3 × 10 −4 × (t − 4)] Or, = [1 + 0.00033 × (t − 4)] ………..(2) ρ2 From (1) and (2) we can write, [1 + 0.00033 × (t − 4)] = 1 0.98 1 − 0.98 0.02 0.00033 × (t − 4) = = Or, 0.98 0.98 0.02 Or, t = 65.84 0 C . Or, t = 4 + 0.98 × 0.00033 V × 0.98 × ρ1 = V × ρ 2 Or,

#Example 7

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A solid body floats in water at 0 0 C while keeping 98% of its volume immersed. The body gets fully immersed when the temperature is 25 0 C . Find the coefficient of volume expansion of the liquid. Given, the coefficient of volume expansion of the solid = 2.6 × 10 −6 / 0 C Solution. Let at 0 0 C temperature, the volume of the body = V0 , the density of the liquid = ρ 0 and at 25 0 C , the volume of the body = V , the density of the liquid = ρ . We can write, V = V0 1 + 2.6 × 10−6 × 25 = V0 1 + 65 × 10 −6 = V0 × 1.000065 If the coefficient of volume expansion of the liquid be γ , ρ 0 = ρ [1 + γ × 25] ∴ Following the condition of floating, we can write, V0 × 0.98 × ρ [1 + 25.γ ] = V × ρ = V0 × 1.000065 × ρ = 0.98 × [1 + 25.γ ] = 1.000065 Or, 1.000065 Or, 25.γ = − 1 = 1.0205 − 1 = 0.0205 0.98 0.0205 = 8.2 × 10 − 4 / 0 C . Or, γ = 25

[

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[

]

#Example 8 There is 0.4 cc mercury in a mercury thermometer at 0 0 C . The diameter of the capillary tube of the thermometer is 0.5 mm. How much should be the length of the scale in order to measure temperature from 0 0 C to 100 0 C ? The coefficient of apparent expansion of mercury = 1.7 × 10 −4 / 0 C . [J.E.E.] Solution. The volume expansion of mercury from 0 0 C to 100 0 C is = 0.4 × 1.7 × 10 −4 × 100 = 0.0068 c.c. 2

 0.05  The area of cross-section of the capillary tube = π    2  2 = π (0.025) = 0.000625.π sq. cm. 0.0068 ∴ The required length of scale = = 3.46 cm. 0.000625 × 3.14 #Example 9 Two thermometers A and B are made of same glass material and same liquid is used in them. The bulbs of both the thermometers are spherical. The inner diameter of the thermometer A is 7.5 mm and radius of the cross-section of this is 1.25 mm. For the thermometer B the same quantities are respectively, 6.2 mm and 0.9 mm. Find the ratio of the length of one degree of the two thermometers. [J.E.E.] Solution. Let the length of one degree in the thermometers A and B are respectively, x and y .

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For the thermometer A, the volume expansion of mercury in the bulb for raising 1 0 C temperature is = the volume of x cm length in the capillary tube. 3

4  0.75  2 Or, π   × γ ′ × 1 = x × π (0.125) 3  2 

[ γ ′ = coeff. of apparent expansion of the liquid] …………….(1)

Similarly, for the thermometer B, 3

4  0.62  2 π ……………..(2)  × γ ′ × 1 = y × π (0.09) 3  2  Dividing (1) by (2) we get, (0.375)3 = x × (0.125)2 Or, x = (0.375)3 × (0.09)2 = 0.92 (approx.). y (0.31)3 y (0.09)2 (0.31)3 (0.125)2 #Example 10 A container is filled with 500 gm water and 1000 gm mercury. An amount of 3.52 gm water flows out of this container when 21200 calorie heat is given to it. Determine the coefficient of volume expansion of mercury. Neglect the volume expansion of the container. Given, the coefficient of volume expansion of water = 1.5 × 10 −4 / 0 C , the density of mercury = 13.6 gm/cc, density of water = 1 gm/cc and the specific heat of [I.I.T.] mercury = 0.03 cal/gm 0 C . Solution. 3.52 = 3.52 c.c. The volume of water that flows out of the container = 1 The volume expansion of water and mercury together in the container is = the volume of water that flows out = 3.52 c.c. Suppose, the temperature is increased by ∆t 0 C . We can write for this temperature increase, heat taken by mercury + heat taken by water = 21200 Or, 1000 × 0.03 × ∆t + 500 × 1 × ∆t = 21200 [specific heat for water = 1 cal/gm 0 C ] Or, ∆t = 40 0 C 500 ∴ The volume expansion of water = × 1.5 × 10 − 4 × 40 = 3 c.c. 1 ∴ The volume expansion of mercury = 3.52 – 3 = 0.52 c.c. If the coefficient of volume expansion of mercury is γ , we can write, 1000 0.52 × 13.6 × γ × 40 = 0.52 Or, γ = = 1.768 × 10 − 4 / 0 C . 13.6 1000 × 40 #Example 11 A glass bulb is filled with 350 gm of mercury at 0 0 C . If some steel balls are kept inside, the bulb is filled with 265 gm of mercury at 0 0 C . When the bulb is heated to 100 0 C , 5 gm mercury comes out. Determine the coefficient of linear expansion of mercury. Given, the coefficient of real expansion of mercury = 18 × 10 −5 / 0 C . Ignore the expansion of glass.

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Solution. Let the density of mercury at 0 0 C is = ρ 0 and the density at 100 0 C is = ρ100 . 350 The volume of mercury at 0 0 C = and the volume of mercury when the steel balls

ρ0

are inserted at 0 0 C is =

265

ρ0

.

The volume of mercury that comes out at 100 0 C is =

∴ The volume of the steel balls at 0 0 C is =

350

ρ0

At 100 0 C , the volume expansion of mercury = expansion of the steel balls =

85

ρ0

× γ S × 100

−

265

ρ0

265

ρ0

5

ρ100 =

85

ρ0

.

× 18 × 10 −5 × 100 and the volume

[ γ S = the coefficient of volume expansion

of steel]. Now, we can write, the volume expansion of mercury + the volume expansion of steel balls = volume of mercury that comes out. 5 265 85 ∴ × 18 × 10 −5 × 100 + × γ S × 100 =

ρ0

ρ0

ρ100 ρ Or, 265 × 18 × 10 −5 × 100 + 85 × γ S × 100 = 0 × 5 ρ100 ρ100 (1 + 18 × 10 −5 × 100) ρ Or, 53 × 18 × 10 −5 × 100 + 17 × γ S × 100 = 0 = ρ100 ρ100 −5 −5 Or, 53 × 18 × 10 × 100 + 17 × γ S × 100 = 1 + 18 × 10 × 100 Or, 17 × γ S × 100 = 1 − 52 × 18 × 10 −5 × 100 Or, 17 × γ S = 0.01 − 52 × 18 × 10 −5 = 0.01 − 0.00936 = 0.00064 Or, γ S =

0.00064 = 37.65 × 10 −6 / 0 C 17

Hence the coefficient of linear expansion of steel, α S =

γS 3

= 12.55 × 10 −6 / 0 C .

#Example 12 Two thermometers of same type, one filled with mercury and another with alcohol at 0 0 C . If the length of one degree in mercury thermometer is l and the length of one l γ − 3α . The coefficient of real degree in alcohol thermometer is l ′ , show that = l ′ γ 1 − 3α expansion of mercury = γ ; the coefficient of linear expansion of glass = α ; the coefficient of real expansion of alcohol = γ 1 . Solution.

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Let the area of cross-section of the capillary tube of both the thermometers is A and the volume of the mercury bulb for the thermometers = V . The coefficient of volume expansion of glass = 3α . ∴ The coefficient of apparent expansion of mercury = γ − 3α and the coefficient of apparent expansion of alcohol = γ 1 − 3α . In the mercury thermometer, due to 1 0 increase in temperature, the apparent expansion of volume of mercury = the volume of mercury column of length l in the capillary tube. ∴ V (γ − 3α ) × 1 = l × A ………………..(1) Similarly, in the alcohol thermometer, due to 1 0 increase in temperature, the apparent expansion of volume of alcohol = the volume of the column of alcohol of length l ′ in the capillary tube. ∴ V (γ 1 − 3α ) × 1 = l ′ × A ……………….(2) Dividing (1) by (2), l × A V (γ − 3α ) l (γ − 3α ) Or, = . = l ′ × A V (γ 1 − 3α ) l ′ (γ 1 − 3α ) #Example 13 Some amount of liquid is there in a cylindrical container made of some material having coefficient of linear expansion α . When the container is heated, the level of liquid inside the container is seen to remain unchanged. What is the coefficient of volume expansion of the liquid? [I.I.T.] Solution. Suppose, the height of the level in the container = x and the area of cross-section of the cylinder = A . The level of liquid remains unchanged for some temperature rise when the expansion of the volume of the container is equal to the expansion of volume of the liquid. Thus for a change in temperature, ∆t 0 C we can write, [ γ = coefficient of volume expansion of the liquid] xA × 3α × ∆t = xA × γ × ∆t Or, γ = 3α . #Example 14 The weight of a body is W0 in air. The apparent weight of this when immersed in a liquid 0

of temperature t1 C is W1 and the apparent weight of the body when immersed in the same liquid is W2 . What is the coefficient of volume expansion of the liquid if the coefficient of volume expansion of the material of the body is γ . [I.I.T.] Solution. 0

The weight of the displaced liquid at t1 C is = W0 − W1 . ∴ W0 − W1 = V1 ρ1 g , ………………………..(1) where V1 = Volume of liquid at t1 C and ρ1 = density of liquid at t1 C . 0

0

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0

Similarly, the weight of the displaced liquid at t 2 C is = W0 − W2 . ∴ W0 − W2 = V2 ρ 2 g , ………………………..(2) where V2 and ρ 2 are volume and density of liquid , respectively at t 2 C . Dividing (1) by (2) we get, W0 − W1 V1 ρ1 g V1 ρ1 = = . W0 − W 2 V 2 ρ 2 g V 2 ρ 2 The volumes and the densities at two temperatures are related by the following. V2 = V1 [1 + γ .(t 2 − t1 )] ; ρ1 = ρ 2 [1 + γ l .(t 2 − t1 )] , where γ l = the coefficient of volume expansion of liquid. W − W1 1 + γ l .(t 2 − t1 ) 1 + γ l .t ∴ 0 [ t = t 2 − t1 ] = = W0 − W2 1 + γ .(t 2 − t1 ) 1 + γ .t W − W1 Or, 1 + γ l .t = (1 + γ .t ) × 0 W0 − W 2 W − W1 (W0 − W1 )γ .t Or, γ l .t = 2 + W0 − W2 W0 − W 2 (W0 − W1 )γ W2 − W1 Or, γ l = . + W0 − W 2 (W0 − W2 ).t 0

Discussions of a few questions Q.1 The level of liquid in a container comes down initially and then rises up when the liquid filled container is heated suddenly. Explain this. Ans. Initially the container receives heat when the liquid filled container is heated. Thus the volume of the container expands at first. The temperature of the liquid does not change much during this time and so the volume of it remains unchanged. Hence the liquid level in the container comes down at the beginning. When the heating is continued, the liquid in the container gets heated eventually and its volume expands. The expansion of a liquid is greater than the expansion of a solid in general. Thus the liquid level is seen to rise again after a while. Q.2 Will the coefficient of apparent expansion of mercury be the same when it is kept in a glass vessel or in a copper vessel? If not, in which case will it be greater? The coefficient of volume expansion of glass and copper are respectively, 25 × 10 −6 / 0 C and 50 × 10 −6 / 0 C . Ans. Thus the coefficient of apparent expansion of mercury will not be same in case of glass vessel and copper vessel. The coefficient of apparent expansion of a liquid depends on the material of the container. We know, γ = γ ′ + γ g or, γ ′ = γ − γ g , where γ ′ is the coefficient of apparent

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expansion of liquid, γ is the coefficient of real expansion of the liquid and γ g is the coefficient of volume expansion of the material of the container. The coefficient γ is fixed for a liquid. The coefficient γ g is greater for copper than glass. Thus the apparent coefficient of expansion of a liquid in a copper vessel is less than that in a glass vessel. Q.3 A hollow iron sphere floats in water at 10 0 C while completely immersed. Will the sphere float as before or sink when the temperature of both the water and the sphere is increased to 50 0 C ? Ans. Let at 10 0 C , the volume of the iron sphere = V1 and density of water = ρ1

The apparent loss of weight of the sphere in water at 10 0 C is = V1 ρ1 . If at 50 0 C , the volume of the iron sphere = V2 and density of water = ρ 2 then the apparent loss of weight of the sphere is = V2 ρ 2 . Now, if the coefficient of volume expansion of iron be γ and the coefficient of real expansion of water be γ w , we can write, V2 = V1 [1 + γ .(50 − 10)] = V1 (1 + 40.γ ) and ρ 2 = ρ1 (1 − 40.γ w ) . ∴ V2 ρ 2 = V1 ρ1 (1 + 40.γ ) . (1 − 40.γ w ) = V1 ρ1 [1 − 40.(γ w − γ )] [approx.] As γ w > γ [Q the coefficient of expansion of water is greater than that of iron], we have V2 ρ 2 < V1 ρ1 . This means, the apparent loss of weight of the iron sphere at 50 0 C is less

than the apparent loss of weight of that at 10 0 C . Therefore, the iron sphere will sink in water at 50 0 C . Q.4 What will happen if water is used instead of mercury in a thermometer to measure temperature in a range of 0 0 C to 10 0 C ? Ans. If the volume expansion of a liquid is regular within some temperature range then the liquid can be used as a standard material to measure temperature. The volume of mercury increases when temperature is increased and the expansion rate is almost same for any temperature. However, the volume expansion of water is a different kind. The volume expansion of water in the temperature range of 0 0 C to 4 0 C is anomalous. When a certain mass of water is heated from 0 0 C to 4 0 C , the volume of it decreases (the density increases) and attains minimum at 4 0 C . Again the volume increases if heated beyond 4 0 C . The volume of water at 0 0 C and at 8 0 C will almost be the same. Thus the reading of the thermometer will be almost same at 0 0 C and at 8 0 C . Therefore, there will be problem if water is used in a thermometer to measure temperature between 0 0 C and at 10 0 C . Q.5 A wooden block is floating in water at 0 0 C keeping its V volume above water. If now the temperature of water is increased slowly from 0 0 C to 20 0 C , what will be the [I.I.T.; J.E.E. ‘86] change in volume V ?

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Ans. We know, the buoyancy or the upward force on a floating body on a liquid depends on the density of that liquid. If the density of the liquid is less, the body is less immersed which means the part of the body that is above liquid is more. When the temperature of water is increased from 0 0 C to 4 0 C , the density of it keeps increasing and becomes maximum at 4 0 C . During this time the upward force due to displaced water increases as the density increases. Thus the volume V , the part of the wooden block above water will increase. The volume V will be maximum at 4 0 C . Again, the density of water decreases as its temperature is increased from 4 0 C to 20 0 C . Thus the upward force on the wooden block due to water also decreases and as a result the block keeps sinking and V decreases. Q.6 A lake is covered with ice. The temperature of the air is -15 0 C . What will be (i) the temperature of water just below the ice and (ii) the highest temperature at the bottom of the lake? [I.I.T.] Ans. (i) We know, the minimum temperature of water in with ice should be 0 0 C since the water will turn ice if the temperature goes down below 0 0 C . Thus the temperature of water in the lake just below the ice will be 0 0 C . (ii) The maximum temperature of water at the bottom of the lake can be 4 0 C . This is because the density of water is maximum at 4 0 C . Thus water of this temperature will occupy the bottom of the lake. Detailed discussions: Suppose, the initial temperature of water was 10 0 C and the temperature of air was 15 0 C . The temperature of the surface of water in the lake keeps decreasing for which the density of upper layer of water keeps increasing. Thus the colder water from the upper layer of water slowly goes down towards the bottom and warmer and lighter water from down comes up. This way through convection, the temperature of the whole water body keeps decreasing. This convection continues until the temperature becomes 4 0 C . The density of water is maximum at 4 0 C and so the water of the surface when cools down below this temperature, it can not come down. Next, the water cools further by conduction. The upper layer of water in with very cold air cools more and eventually ice is formed. Since the ice is a bad conductor of heat, the layer of ice once formed on the surface, slows down the process of conversion of whole water body into ice. Q.7 What kind of differences can be noticed when mercury and water are heated from 0 [H.S. ‘90] C? 0 Ans. When mercury is heated from 0 C its volume keeps on increasing. But when water is heated from 0 0 C , the volume keeps decrea to 4 0 C and the volume becomes minimum at 4 0 C . When water is heated above 4 0 C , its volume increases like that of mercury.

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33

Q.8 Water is poured into a glass up to its edge at 4 0 C . What will happen in the following cases when (i) water is heated and (ii) water is cooled? [H.S. ’93; I.I.T.] Ans. (i) We know, the volume of some amount of water is minimum at 4 0 C . If now the temperature of water is increased above 4 0 C , the volume increases. Thus the water will spill out of the glass when heated above 4 0 C . (ii) Water has anomalous expansion below 4 0 C . The volume of water also increases if the temperature is decreased below 4 0 C . Thus water will spill out of glass again when it is cooled below 4 0 C . Q.9 A beaker is filled with a liquid of density 1.5 gm/cc and a piece of ice is floating on it. Will the liquid spill when the ice melts? [I.I.T.] Ans. The liquid will spill when ice melts. Let the mass of ice = m gm. When the piece of ice melts, m gm liquid is displaced. Thus the volume of displaced m c.c. liquid is V1 = 1.5 When m gm ice melts, the volume of that water is V2 = m c.c. As V2 < V1 , the surplus water will be up on the liquid and spill out of the beaker. Q.10 Water is filled up to the edge of a glass. An ice block is floating on water in this situation [fig.** (a)] . Comment on the change in water level in the following cases when ice melts. (i) if the water in the glass is at 0 0 C , [J.E.E. ‘96] 0 (ii) if the water in the glass is at 4 C , (iii) if the water in the glass is at a temperature greater than 4 0 C . [J.E.E. ‘85] Ans. The change in volume of the glass is ignored in this discussion.

Fig. to be included (i) In this case, there will not be any change of water level in the glass [fig**(b)]. The temperature of displaced water by the ice block is at 0 0 C as this is the temperature of water in the glass. The block of ice slowly melts taking heat from its surroundings and is converted to water at 0 0 C . As the temperature of water is 0 0 C and the temperature of water from melted ice is also 0 0 C , the volume of displaced water by ice and the volume of water from melted ice are same. Thus there will be no change of water level in the glass. (ii) In this case, the water will overflow out of the glass [see fig.**(c)] The temperature of displaced water by the ice block is at 4 0 C as this is the temperature of water in the glass. The ice block melts by taking heat partly from water and party from air. The temperature of water from melted ice is 0 0 C . We know, the density of water is

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34

maximum at 4 0 C . Thus the volume of water from melted ice at 0 0 C will be more than the volume of displaced water at 4 0 C . Also, the ice-block will take heat from water at 4 0 C for melting. For the above two reasons the temperature of water at the end becomes less than 4 0 C and its volume increases. Thus water will overflow out of the glass. (iii) In this case, the water level comes down to some extent [see fig**(d)] . According to question, the temperature of water is more than 4 0 C . Let us think that the temperature of water is something much above 4 0 C (say, more than standard room temperature, 27 0 C ). Thus the ice-block takes heat from water and also from the surrounding air to melt. The temperature of water from melted ice is 0 0 C . We can say, the density of water at 0 0 C is more than the density of water at a temperature much above 4 0 C . Thus the volume of displaced water will be more than the volume of water from melted ice. Also, the resulting temperature of water in the glass in with ice comes down to some extent as it gives heat to the ice block. The volume of water reduces as the temperature of it comes down. For the above two reasons, the water level comes down. Q.11 Let us suppose a substance is hanged from the left end of a balance bar and then it is immersed in a liquid while the weights are placed in the right pan of the balance for weighing. Will the measure differ if the liquid with the substance in it is heated up? Ans. If the liquid is heated with the substance in it, the left side of the balance bar wherefrom the substance is hanged, will tilt downwards. To obtain the balance more weights will be required. At the balanced condition, the mass of the substance is equal to the weights on the right pan. When the liquid is heated along with the solid substance, the densities of both of them will decrease. The decrease in density in the liquid will be much more than in the solid substance. Thus there will be a decrease in the upward force on the substance due to the displaced liquid. The substance will appear heavier now. Q.12 Which of the following graphs correctly represents the relationship of the density of water with temperature? [I.I.T.] Ans.

Fig. to be included Incl The graphical in (a) is the correct one as we know that the density of water increases with temperature in the range of 0 0 C to 4 0 C and it is maximum at 4 0 C . The nature of the graph reveals the anomalous expansion in water.

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Q.13 Show that the variation in volume, ∆V = γ .V0 .∆T due to the variation in temperature, ∆T implies the variation in density, ∆ρ = −γ .ρ .∆T . What is the meaning of the minus sign? Ans. m m We can write, ∆ρ = ρ − ρ 0 = − , where ρ 0 and ρ are the initial and final densities V V0 respectively, of some liquid; V0 and V are volumes of the liquid of some mass m .

Now we have, ∆ρ =

m(V0 − V ) ∆V m.∆V  m  = γ .∆T . =− =  .γ .∆T , where we use V0 VV0 VV0 V 

∴ ∆ρ = ρ .γ .∆T The negative sign indicates that the density decreases with increasing temperature.

Q.14 Show that the length of a mercury column in a uniform glass tube due to raising of temperature t can be written as l = l 0 [1 + (γ − 2α ).t ] , where γ is the coefficient of volume expansion of mercury and α is the coefficient of linear expansion of glass. Ans. The increase in length of the mercury column is governed by the volume expansion of mercury due to change in temperature t. We can write, V = V0 (1 + γ .t ) , where V0 and V are the initial and final volumes of mercury. For the area of cross-section of the tube, A = A0 (1 + 2α .t ) . We can write, V = A.l ∴ A0 (1 + 2α .t ) × l = V0 (1 + γ .t ) V0 .(1 + γ .t ).(1 + 2α .t ) −1 = l 0 [1 + (γ − 2α ).t ] . A0 [Note that here the expansion of the glass tube in the length direction is ignored.]

Or, l =

Questionnaire Very Short Questions:

Mark: 1 (Answer in one or two words)

1. Which one is the characteristic property of a liquid – the coefficient of apparent [γ ] expansion, γ ′ or the coefficient of real expansion γ ? 2. A glass is filled with water at normal temperature. What will happen if the temperature is increased? [Water overflows] 0 3. A glass is filled with water at 0 C . What will happen if the temperature is increased? [Water level goes down] 4. At which temperature is the density of water maximum? [4 0 C ] 5. How the density of water changes when the temperature of water at room temperature is increased? [Density decreases]

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6. How the density of water changes when the temperature of water is increased from 20C ? [Density increases] (Fill in the blanks) 1. The real expansion of a liquid = the apparent expansion of liquid + the expansion of the------[The container] 0 0 2. The coefficient of real expansion of water in between 0 C and 4 C is -------------

[Negative] 3. Generally, the thermal expansion of a liquid is --------than the thermal expansion of a solid. [More] 4. Generally, the thermal expansion of a liquid is ----than the thermal expansion of a gas. [Less] 5. Dulong and Petit’s method measures the coefficient of………expansion of liquid. [Real] 6. Weight thermometer is used to measure the coefficient of …………expansion of liquid. [Apparent] (Multiple choice type)

If water is heated from 0 0 C to 50 0 C , the density of water (a) remains the same (b) keeps on decreasing (c) keeps on increasing (d) first increases and then decreases. [(d)] 2. In a cold country because of anomalous expansion of water in between 0 0 C and 4 0 C (a) all the water in a water body becomes ice (b) water does not turn ice (c) the aqua animals survive (d) Oxygen intake in water increases. [(c)] 1.

3. The relation between the coefficient of apparent expansion and the coefficient of real expansion is (a) the coefficient of apparent expansion = the coefficient of real expansion (b) the coefficient of real expansion < the coefficient of apparent expansion (c) the coefficient of real expansion > the coefficient of apparent expansion (d) sometimes the coefficient of real expansion is more than the coefficient of apparent expansion and sometimes the otherwise which depends on the kind of liquid. [(c)] 4. How many coefficients of expansion are used in case of a liquid? (a) 1 (b) 2 (c) 3 (d) 4

[(b)]

5. The unit of the coefficient of real expansion and that of the coefficient of apparent expansion are (a) both / 0 C (b) both m 3 / 0 C (c) / 0 C for γ and m 3 / 0 C for γ ′ (d) m 3 / 0 C for γ and / 0 C for γ ′ . [(a)]

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37

6. The coefficient of real expansion of a liquid is γ and the coefficient of linear expansion of the material of the container is α . For which of the conditions there will be no change in the apparent expansion of the liquid? [(b)] (a) γ = α (b) γ = 3α (c) γ < α (d) γ > α 7. If a liquid having the coefficient of volume expansion γ is heated in a container

γ

whose material has the coefficient of linear expansion , the level of the liquid will 3 (a) rise (b) fall (c) remain same (d) remain almost same [(c)] 8. The coefficient of volume expansion of mercury is 18 × 10 −5 / 0 C and the coefficients of linear expansion for copper and glass are 17 × 10 −6 / 0 C and 9 × 10 −6 / 0 C respectively. If mercury is first kept in a copper container and then in a glass container, (a) the coefficient of apparent expansion for mercury will be less in the first case (b) the coefficient of apparent expansion for mercury will be more in the first case (c) the coefficient of real expansion for mercury will be less in the first case (d) the coefficient of real expansion for mercury will be more in the first case. [(a)] 9. The coefficient of volume expansion of mercury is 18 × 10 −5 / 0 C . How much temperature has to be increased in order to increase the volume by 0.2%? (a) 22.2 0 C (b) 11.1 0 C (c) 5.55 0 C (d) 2.22 0 C

[(b)]

10. The coefficient of volume expansion of liquid is (a) positive for all liquids (b) negative for all liquids (c) negative for water in between 0 0 C and 4 0 C and positive for all other cases (d) positive for water in between 0 0 C and 4 0 C and negative for all other cases. [(c)] 11. Two containers are filled with water at same temperature. Water overflows while one container being heated and other being cooled. The initial temperature of water at both the containers is (a) 273 0 C (b) 273 K (c) 277 0 C (d) 277 K [(d)] 12. The coefficient of real expansion of mercury is γ and the coefficient of linear expansion of glass is α . The coefficient of expansion of a mercury column in a uniform glass tube is γ γ − 3α γ − 2α (c) (d) γ − 2α [(d)] (a) (b) 3 3 3 13. A brass scale is attached to a barometer scale to measure the height of mercury column. The coefficient of real expansion of mercury is γ and the coefficient of linear expansion of brass is α . The rate of increase of barometer reading is (a) γ − 3α (b) γ + α (c) γ − α (d) γ − 2α [(c)] 14. A container is partially filled with a liquid at 0 0 C . What is the condition that the rest volume of the rest of the container remains same?

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(a) The expansion of the entire container = the real expansion of the liquid inside the container (b) The expansion of the empty part of the container = the real expansion of the liquid inside the container (c) The expansion of the entire container = the apparent expansion of the liquid inside the container (d) The expansion of the empty part of the container = the apparent expansion of the liquid inside the container. [(a)] 15. A glass flask contains some mercury. The volume of air inside the flask remains same at different temperatures. What is the volume of mercury when the volume of the flask is 1 litre? The coefficient of volume expansion of mercury is 18 × 10 −5 / 0 C and the coefficient of linear expansion of glass is 9 × 10 −6 / 0 C . (a) 50 cc (b) 100 cc (c) 150 cc (d) 200 cc [(c)] 16. The apparent weight of a solid body while immersed in water at 20 0 C is W1 . The apparent weight is W2 at 40 0 C . In this case (a) For different materials, it can be W2 > W1 or W2 < W1 (b) Always W2 = W1 (c) Always W2 < W1 (d) Always W2 > W1 [(d)] 17. The apparent weight of a solid body while immersed in a liquid is W1 at temperature t1 and W2 at temperature t 2 . If the coefficients of volume expansion of the liquid and the solid material are γ and γ S respectively, quantity W2 − W1 is proportional to (γ − γ S ) (t − t ) 1 (c) 2 1 (d) (a) (γ − γ S )(t 2 − t1 ) (b) (t 2 − t1 ) (γ − γ S ) (γ − γ S )(t 2 − t1 ) 18. The apparent weight of a metal body is 100 gm while immersed in water at 0 0 C and 100.5 gm at 50 0 C . What will be the apparent weight at 20 0 C ? (a) 100.1 gm (b) 100.2 gm (c) 100.3 gm (d) 100.4 gm [(b)] Short Questions:

Marks: 2

1. Determine the relation between the coefficients of apparent and real expansions of liquid. 2. Show how the density of a liquid changes with temperature. 3. Establish the relation between the coefficient of real expansion of a liquid and its density. [H.S. ‘96] 4. A solid body is immersed in a liquid. How does the apparent weight of the body changes when the temperature is raised? 5. What is the anomalous expansion of water? [H.S. ‘01] 6. Discuss the effect of anomalous expansion of water on the marine life. [H.S. ’05, ’01, ’97, ’93 ] 7. How does the property of anomalous expansion of water help the aquatic animals to survive? [H.S. ‘99] 8. On what factors the two coefficients of expansion of liquid depend? 9. What are the units of the coefficients of apparent and real expansions?

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10. Are the values of the coefficients of apparent and real expansions constant? 11. What is the relation between the apparent and real expansion coefficients in Fahrenheit and Celsius scales? 12. If the coefficients of volume expansion of mercury and glass would be equal, the thermometer would not work. Explain this. 13. If the bulb of a thermometer is immersed in hot water, it is seen that the mercury column comes down at first and then goes up again. What is the reason behind this? 14. Will the coefficient of apparent expansion of mercury be same when it is kept in a glass container and in a copper container? 15. What kind of difference may be noted if mercury and water both are heated separately from 0 0 C ? 16. A hollow iron ball floats in water at 10 0 C while being completely immersed. If now the temperature of both the ball and water is increased to 50 0 C , will the ball keep floating like this or sink completely? 17. A wooden block floats in water keeping its V volume above the water surface. If now the temperature of water is increased from 0 0 C to 20 0 C , how does V change? 18. What will happen if water is used in thermometer in place of mercury to measure temperatures in the range of 0 0 C to 10 0 C ? 19. A substance, immersed in a liquid, is hanged from the left end of a balance bar and the weights are kept in the right pan to obtain balance. Now what will be the change when the liquid along with the substance is heated? 20. A piece of ice is floating in a liquid of density 1.5 gm/cc which fills a beaker. Will the liquid spills when the ice melts? 21. Ice is formed over a lake. The temperature of surrounding air is -15 0 C . What will be the temperature of water (i) just below the ice and (ii) at the bottom of the lake? 22. ‘The barometer reading taken at room temperature needs correction’ – explain this. 23. Show that the change in density of a liquid due to the increase in temperature ∆t is ∆ρ = − ργ∆t ( γ = the coefficient of real expansion of the liquid). 24. Why is the name ‘weight thermometer’? 25. Can one determine the coefficient of real expansion of a liquid by weight thermometer? 0 0 26. The density of a liquid is d 1 at t1 C and d 2 at t 2 C . Determine the coefficient of [H.S. ‘06] volume expansion of the liquid in of d 1 , d 2 , t1 and t 2 . Medium level Questions:

Marks: 4

1. Explain with diagram why the coefficient of real expansion and the coefficient of apparent expansion are referred in case of a liquid. [4] 2. Define the coefficient of apparent expansion and the coefficient of real expansion of a liquid. Find the relationship between them. [H.S. ’02, ’00, ’98, ’96, ‘93] [2+2] 3. What do you mean by anomalous expansion of water? Explain with graphs how the density of water and the volume of 1 gm of water vary with temperature in the range of 0 0 C and 10 0 C . [H.S. ’00, ’99, ’97, ’95, ’93] [2+1+1] 4. How can the anomalous expansion of water be experimentally demonstrated? [4]

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5. The density of water is maximum at 4 0 C - prove this with a simple experiment. [4] 6. What is proved by Hope’s experiment? Discuss the experiment with a sketch. [1+3] 7. A glass is completely filled with water at 4 0 C . Explain with reason what happens in the following cases: (i) the water is heated; (ii) the water is cooled. [2+2] 8. Which coefficient of liquid is determined by the weight thermometer? Describe the method. [1+3] 9. Describe the method by Dulong and Petit to determine the coefficient of real expansion. Explain why the coefficient of real expansion can be measured by this method. [3+1] 0 10. The brass scale of a Fortin Barometer is standardized at 0 C . If the barometer height is found to be H t at t 0 C what is the height at 0 0 C considering the expansion of brass scale and mercury? [4] 11. What kind of corrections needed for a barometer reading due to thermal expansion? [2+2] 12. Give definitions of the coefficients of real and apparent expansion of a liquid. Determine the relation between the two coefficients. [H.S. ‘06] [2+2] Numerical Problems: (Short Numerical Problems)

Marks: 2

1. If the coefficient of real expansion of mercury is 18.7 × 10 −6 / 0 C and the coefficient of linear expansion of glass is 0.9 × 10 −5 / 0 C find the coefficient of apparent expansion of mercury with respect to glass. [Ans. 1.6 × 10 −5 / 0 C ] 2. The coefficient of apparent expansion of a liquid is 13.2 × 10 −5 / 0 C when copper container is used and that is 15.5 × 10 −5 / 0 C when a glass container is used. For copper, [Ans. 9 × 10 −6 / 0 C ] α = 1.7 × 10 −5 / 0 C . Determine α for glass. 3. The coefficient of apparent expansion of a liquid in a brass container is 14.6 × 10 −5 / 0 C . What is the coefficient of apparent expansion of the liquid with respect to iron? For brass, α = 18 × 10 −6 / 0 C and for iron, α = 12 × 10 −6 / 0 C . [Ans. 16.4 × 10 −5 / 0 C ] 4. What fraction of the inner volume of a glass flask has to be filled with mercury so that the rest of the volume remains same at all temperatures? The coefficient of real expansion of mercury = 1.8 × 10 −4 / 0 C and the coefficient of volume expansion of glass = 2 ] 2.4 × 10 −5 / 0 C . [Ans. 15 5. The volume of the bulb of a thermometer is 1 cc. Each division of a degree on the scale has to be made 5 mm long. What will be the cross-section of the thermometer tube? The coefficient of apparent expansion of mercury with respect to glass = 1.6 × 10 −4 / 0 C . [Ans. 0.032 sq. mm] 6. There is 1 cc mercury in the bulb of a thermometer and up to zero mark in the tube. If the inner radius of the thermometer tube is 0.3 mm what will be the length of a division

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of 1 0 C ? The coefficient of real expansion = 18 × 10 −5 / 0 C and the coefficient of linear expansion of glass = 8 × 10 −6 / 0 C . [Ans. 2.21 mm] 7. The area of inner cross-section of a glass tube is 0.004 sq. cm and one end of this tube is attached to a bulb whose volume is 10 cc. The bulb is filled up with a liquid at 10 0 C . The coefficient of apparent expansion of the liquid is16 × 10 −5 / 0 C . How far the liquid rises in the tube if the bulb is heated up to 50 0 C ? [Ans. 16 cm.] 0 0 8. The mass of 10 cc water at 0 C and at 4 C is 9.998 gm and 10 gm, respectively. Find the coefficient of expansion of water in this temperature interval. [H.S. ‘93] [Ans. − 5 × 10 −5 / 0 C ] 1 th of its volume. Show that the volume of 9. A container is filled with glycerin up to 10 empty space does not change with temperature. For the material of the container, α = 18 × 10 −6 / 0 C ; for glycerin, γ = 54 × 10 −5 / 0 C . 10. A glass tube of uniform cross-section contains a mercury column of 1 m length. The length of the column increases by 16.5 mm when the temperature is raised to 100 0 C . If the coefficient of real expansion of mercury is 18.2 × 10 −5 / 0 C , find the coefficient of linear expansion of glass. [Ans. 8.36 × 10 −6 / 0 C ] 11. The bulb of a thermometer contains 0.45 cc of mercury. The radius of the bore of thermometer tube is 3.49 × 10 −2 mm. How far apart are the degree marks on the thermometer? The coefficient of apparent expansion of mercury with respect to glass is 1.55 × 10 −4 / 0 C . [Ans. 2 mm] 0 12. The volume of mercury in a thermometer is 0.5 cc at 0 C . The distance between its lowest and highest degree marks is 15 cm. Find the radius of cross-section of the thermometer tube. The coefficient of linear expansion of glass = 8 × 10 −6 / 0 C and the coefficient of volume expansion of mercury = 18.1 × 10 −5 / 0 C . [Ans. 0.013 cm] 0 13. The density of mercury at 20 C is 13.50 gm/cc. How much temperature of it has to be increased so that its density may change by 2%? The coefficient of real expansion of [Ans. 131.11 0 C ] mercury = 1.8 × 10 −4 / 0 C . 14. The volume of a glass flask is 600 cc. What amount of a liquid has to be kept in it so that volume of the empty space of the flask may remain the same at all temperatures? The coefficient of real expansion of the liquid = 1.8 × 10 −4 / 0 C and the coefficient of linear expansion of glass = 0.7 × 10 −5 / 0 C . [H.S. ‘04] [Ans. 70 cc] 0 15. A glass vessel contains air at 67 C . How much temperature has to be increased 1 keeping air pressure fixed so that part of air comes out? (Neglect the expansion of 3 glass.) [H.S.(XI) ‘06] [Ans. 180.3 0 C ] (Medium Level Numerical Problems)

Marks: 4

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1. If the density of mercury is 13.56 gm/cc at 15 0 C , how much is the mass of 600 cc of mercury at 130 0 C ? What will be the volume of 600 gm of mercury at that temperature? [The coefficient of volume expansion of mercury is 18 × 10 −5 / 0 C .] [Ans. 7.968 kg, 45.18 cc] 0 2. The density of mercury at 0 C is 13.6 gm/cc. What is the volume of 100 gm of 1 0 mercury at 100 0 C ? The coefficient of volume expansion of mercury is / C. 5550 [H.S. ‘06] [Ans. 7.49 cc.] Numerical Problems for advanced Students:

1. A weight thermometer contains 150 gm of mercury at 0 0 C . When the thermometer is immersed into boiling water, 2.27 gm of mercury flows out. What is the coefficient of apparent expansion of mercury? [Ans. 15.3 × 10 −5 / 0 C ] 2. A weight thermometer weighs 20 gm while empty; the weight of this is 120 gm when it is filled with mercury at 0 0 C . The weight thermometer weighs 118.4 gm when it is heated up to 100 0 C . What is the coefficient of apparent expansion of mercury? If the coefficient of real expansion of mercury is18 × 10 −5 / 0 C , what is the coefficient of linear expansion of the material of the container? [Ans. 16.2 × 10 −5 / 0 C ; 6 × 10 −6 / 0 C ] 3. A weight thermometer contains 500 gm of mercury at 0 0 C . An amount of 10 gm of mercury flows out when it is heated. If the coefficient of apparent expansion of mercury is 15.5 × 10 −5 / 0 C , how much has been the increase in temperature? [Ans. 131.67 0 C ] 4. An empty weight thermometer weighs 40 gm. When this is filled with mercury at 10 0 C , the weight becomes 540 gm. How much mercury would come out if this is heated to 60 0 C ? The coefficient of real expansion of mercury = 18 × 10 −5 / 0 C and the coefficient of linear expansion of glass = 10 × 10 −6 / 0 C . [Ans. 3.72 gm] 5. In Dulong and Petit’s experiment, the U-tube contains liquid whose temperature is 0 0 C and 100 0 C at two arms. The height of the liquid column in the cold arm is 50 cm and the height of liquid column is 10 mm more in the other arm. If the coefficient of linear expansion of the tube is 1 × 10 −5 / 0 C what is the coefficient of apparent expansion of the liquid? [Ans. 19 × 10 −5 / 0 C ] 6. The brass scale of a barometer gives correct reading at 0 0 C . If the reading is 75 cm at 40 0 C what is the atmospheric pressure? The coefficient of linear expansion of brass = 1.8 × 10 −5 / 0 C and the coefficient of volume expansion of mercury = 18.2 × 10 −5 / 0 C . [Ans. 74.51 cm]

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7. A glass rod weighs 90 gm in air. The weight of this is 49.6 gm in a liquid at 12 0 C . The apparent weight of the glass rod is 51.9 gm in the same liquid at 97 0 C . Find the coefficient of real expansion of the liquid. The volume expansion coefficient of glass = 2.4 × 10 −5 / 0 C . [Ans. 7.3 × 10 −4 / 0 C ] 8. A solid body weighs 50 gm in air. The weight of this in a liquid is 45 gm at 25 0 C and 45.1 gm at 100 0 C . If the coefficient of linear expansion of the material of the body is 12 × 10 −6 / 0 C , find the coefficient of real expansion of the liquid. [Ans. 3.11 × 10 −4 / 0 C ] 9. A specific gravity bottle is filled with a liquid of m gm at 0 0 C . Find the mass of liquid that comes out when it is heated up to t 0 C . The coefficient of real expansion of the liquid = γ and the coefficient of the linear expansion for the bottle = α . (γ − 3α )mt ] [Ans. 1 + γt 10. Determine how much glycerin a specific gravity bottle may contain if it contains 50 gm of glycerin at 30 0 C . Suppose, the coefficient of real expansion of glycerin = 0.00051 / 0 C and the coefficient of volume expansion of glass = 0.00003 / 0 C . [Ans. 48.38 gm] 11. The volume of the bulb of a thermometer is 0.4 cc and the length of the capillary tube is 10 cm and the diameter is 0.4 mm; the bulb is filled with mercury at 0 0 C . What is the highest temperature than can be measured by this thermometer? For mercury, γ = 18 × 10 −5 / 0 C and for glass, α = 9 × 10 −6 / 0 C . [Ans. 205.3 0 C ] 12. The mean value of the coefficient of real expansion of mercury is 18.2 × 10 −5 / 0 C . The density of mercury at 20 0 C is 13.55 gm/cc. For what higher temperature the error in the value of density will be 1%? [Ans. 74.94 0 C ] 13. The brass scale of a barometer is graded correctly at 15 0 C . At what temperature no correction will be required for the barometer reading? For brass, α = 19 × 10 −6 / 0 C and the coefficient of real expansion of mercury = 18 × 10 −5 / 0 C . [Ans. -1.8 0 C ] 14. A uniformly graded glass tube having uniform bore contains mercury from 0 0 C to 100 0 C . How much temperature has to be increased to make the mercury column touch the 102 mark? The coefficient of volume expansion of mercury = 18 × 10 −5 / 0 C ; the coefficient of linear expansion of glass = 9 × 10 −6 / 0 C . [Ans. 130.7 0 C ] 15. Aniline does not get dissolved in water. There is a drop of aniline at the bottom of water at 10 0 C . At which temperature, the aniline comes up to the surface? The density of water is 0.9997 gm/cc and the density of aniline is 1.025 gm/cc at 10 0 C ; the coefficient of real expansion of aniline is 84 × 10 −5 / 0 C and the coefficient of real expansion of water is 43.9 × 10 −5 / 0 C . [Ans. 64.9 0 C ]

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16. The density of mercury is 13.55 gm/cc at 20 0 C . If this mercury is heated, then at which temperature the value of density deviates by 1%? The coefficient of real expansion [Ans. 74.94 0 C ] of mercury = 18.2 × 10 −5 / 0 C . 17. A weight thermometer contains m gm of a liquid at 0 0 C . Prove that mγ .t (1 − γ .t ) gm of the liquid will come out if the thermometer is heated to t 0 C . The coefficient of apparent expansion of the liquid = γ . 3 part of this volume is 18. The inner volume of a glass vessel is V cc at 20 0 C . The 20 occupied by mercury and the rest is filled with water. The temperature is now increased to 40 0 C . Find the volume of water that overflows; the fraction of the initial volume of the vessel. Given, the coefficient of linear expansion of glass = 9 × 10 −6 / 0 C ; the coefficient of real expansion of mercury = 18 × 10 −5 / 0 C and the coefficient of real expansion of water = 30.2 × 10 −5 / 0 C . [Ans. 0.00513] 19. A solid body floats in water at 50 0 C while fully immersed. If the temperature of the liquid is brought down to 0 0 C , what fraction of the volume of the body will be inside the liquid now? The coefficient of volume expansion of body = 0.3 × 10 −5 / K and the coefficient of volume expansion of the liquid = 8 × 10 −5 / K . [Ans. 0.997] 20. The mass of a steel made submarine is 10 7 gm. When the temperature of the sea is 20 0 C , the submarine is filled by 10 5 gm of water to sink. How much less water is required to fill when the sea temperature is 30 0 C ? Given, the coefficient of volume expansion of the sea water = 2 × 10 −4 / 0 C and the coefficient of linear expansion of steel = 1.2 × 10 −5 / 0 C . [Ans. 1.616 × 10 4 gm] 21. An aluminium sphere of diameter 10 cm floats while remaining just immersed in glycerin of density 1.26 gm/cc at 20 0 C . If now the temperature of glycerin is raised to 100 0 C , what will be the change in apparent weight of the sphere? The coefficient of volume expansion of the glycerin = 5 × 10 −4 / 0 C and the coefficient of linear expansion of aluminum = 2.5 × 10 −6 / 0 C . [Ans. 23.4 gm] 22. If a mercury thermometer is completely immersed in boiling water it shows a reading of 100 0 C . Then the thermometer tube is taken out of water and kept in such a way that the average temperature of the tube from the 0 0 C mark upwards remains at 20 0 C . This time the thermometer shows a reading of 98.8 0 C . Determine the coefficient of apparent expansion of mercury. [Ans. 15.2 × 10 −5 / 0 C ] 23. A 30 cm long piece of iron having a square cross-section floats in mercury at 0 0 C . How much will it sink along the length when the temperature of mercury is increased to 100 0 C ? The density of mercury at 0 0 C is 13.6 gm/cc; the density of iron at 0 0 C is 7.6

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gm/cc; the coefficient of real expansion of mercury = 1.82 × 10 −4 / 0 C and the coefficient of volume expansion of iron = 3.51 × 10 −5 / 0 C . [Ans. 0.27 cm] 0

24. The apparent weight loss of a solid body when immersed in a liquid at t1 C is = M 1

and at t 2 C it is = M 2 . Show that M 2 = M 1 [1 − (γ − γ S ).t ] ; γ = the coefficient of real expansion of the liquid and γ S = the coefficient of volume expansion of the solid. 0

25. A vessel contains 5.450 gm of water at 0 0 C . The rest of the vessel is filled with paraffin. When the water in the vessel turns ice, 0.620 gm of paraffin comes out. If the specific gravity of paraffin at 20 0 C is 0.800 and the coefficient of real expansion of it is 0.00090 / 0 C find the specific gravity of ice. [The specific gravity of water at 0 0 C is =1.] [Ans. 0.917] 26. A water filled weight thermometer is being heated. Find the ratio of the mass of water that comes out while heated between 20 0 C and 50 0 C and that between 50 0 C and 80 0 C . The coefficient γ for water in the temperature interval of 20 0 C and 50 0 C is 38 × 10 −5 / 0 C and that in between 50 0 C and 80 0 C is 54 × 10 −5 / 0 C . The expansion of thermometer can be neglected. [Ans. 0.711] 27. A weight thermometer is heated after it is filled with 100 gm of mercury at 0 0 C . What is the ratio of emergent masses of mercury due to heating between 0 0 C and 40 0 C and between 40 0 C and 80 0 C ? For mercury, γ = 18 × 10 −5 / 0 C and for glass, α = 3 × 10 −6 / 0 C . [Ans. 1.006] 28. If the pressure remains constant, then for the increase of temperature ∆t show that the change in the height of liquid in it is ∆h = γ .h.∆t . Here γ is the coefficient of volume expansion of the liquid. 29. A L-shaped glass tube of uniform bore has its two ends sealed and the length of the longer arm is twice the length of the smaller arm. The tube is hanged from the point where the tube bends. The smaller arm is filled with a liquid for which the coefficient of expansion is 3 × 10 −3 / 0 C . The mass of this liquid is 9 times to the mass of the smaller arm. The whole arrangement is cooled by 100 0 C . Show that if the smaller arm makes tan α 1 157 = . angles α 1 and α 2 with vertical at two temperatures, then tan α 2 130 30. The arrangement in the figure** consists of four vertical glass tubes; these vertical tubes are ed with each other by horizontal tubes. The height of two tubes B and C at the middle is 49 cm. The two outer tubes A and D are exposed to atmosphere. The temperature of A and C is 95 0 C ; the temperature of B and D is 5 0 C . The heights of liquid in A and D are 52.8 cm and 51 cm respectively, above the base line. Find the coefficient of thermal expansion of the liquid. [Ans. 2 × 10 −4 / 0 C ] Fig. to be included

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31. A closed glass bulb contains some amount of mercury. If the unoccupied part of the bulb is 1.00 cc at 0 0 C . 0.98 cc at 40 0 C , find the volume of mercury. Given, the coefficient of linear expansion of glass is 7.2 × 10 −6 / 0 C and the coefficient of volume expansion of mercury is 1.81 × 10 −4 / 0 C . [Ans. 3.27 cc] 32. The diameter of a aluminium sphere is 10 cm. This is immersed in glycerin at 20 0 C . What will be the change in the apparent weight of the sphere if the glycerin is heated to 100 0 C ? The density of glycerin at 20 0 C is 1.26 gm/cc and for glycerin, γ = 0.0005 / 0 C and for aluminum, α = 0.000025 / 0 C . [Ans. 22.44 gm]

The End of this chapter