Thermal Properties of Matter

Class 11 NCERT Physics

NCERT

1   The triple points of neon and carbon dioxide are $24.57 K$ and $216.55 K$ respectively. Express these temperatures on the Celsius and Fahrenheit scales.

Solution :

Kelvin and Celsius scales are related as:$\\$ $T_C=T_K-273.15....(i)$$\\$ Celsius and Fahrenheit scales are related as:$\\$ $T_F=\dfrac{9}{5}T_C+32....(ii)$$\\$ $\underline{For \ neon:}\\ T_K=24.57 K\\ \therefore T_C=24.57-273.15=-248.58^oC\\ T_F=\dfrac{9}{5}T_C+32\\ =\dfrac{9}{5}(-248.58)+32\\ =415.44^o F\\ \underline{For \ carbon \ dioxide:}\\ T_k=216.55 K\\ \therefore T_C=216.55-273.15=-56.60^oC.\\ T_F=\dfrac{9}{5}T_C+32\\ =\dfrac{9}{5}(-56.60)+32\\ =-69.88^o C$

2   Two absolute scales A and B have triple points of water defined to be $200$ A and $350$ B. What is the relation between $T_A$ and $T_B $?

Solution :

Triple point of water on absolute scale A, $T_1 =200$ A .$\\$ Triple point of water on absolute scale B, $T_2 = 350$ B Triple$\\$ point of water on Kelvin scale, $T_K = 273.15 K$$\\$ The temperature $273.15 K$ on Kelvin scale is equivalent to $200$ A on absolutes scale A.$\\$ $T_1 = T_K\\ 200 A = 273.15 K\\ \therefore A=\dfrac{273.15}{200}$$\\$ The temperature $273.15 K$ on Kelvin scale is equivalent to $350$ B on absolute scale B.$\\$ $T_2 = T_K\\ 350 B=273.15\\ \therefore B=\dfrac{273.15}{350}$$\\$ $T_A$ is triple point of water on scale $A.$$\\$ $T_ B$ is triple point of water on scale $B.$$\\$ $\therefore \dfrac{273.15}{200}*T_A\\ =\dfrac{273.15}{350}*T_B\\ T_A=\dfrac{200}{350}T_B$$\\$ Therefore, the ratio $T_A : T_B$ is given as $4 : 7.$

3   The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law $R = R_o[ 1 +\alpha (T - T_o )]$ The resistance is $101.6 \Omega$ at the triple-point of water $273.16 K,$ and $165.5 \Omega$ at the normal melting point of lead $(600.5 K)$. What is the temperature when the resistance is $123.4 \Omega$?

Solution :

It is given that:$\\$ $R = R_o[ 1 +\alpha (T - T_o )].......(i)$$\\$ Where,$\\$ $R_0$ and $T_0$ are the initial resistance and temperature respectively $R$ and $T$ are the final resistance and temperature respectively $\alpha $ is a constant$\\$ At the triple point of water, $T_0 = 273.15 K$$\\$ Resistance of lead, $R_0 = 101.6 \Omega $$\\$ At normal melting point of lead, $T = 600.5 K$$\\$ Resistance of lead, $R = 165.5 \Omega$$\\$ Substituting these values in equation (i), we get:$\\$ $R = R_o[ 1 +\alpha (T - T_o )]\\ 165.5 = 101.6 [ 1 + a ( 600.5 - 273.15 )]\\ 1.629 = 1 +\alpha( 327.35)\\ \therefore \alpha =\dfrac{0.629}{327.35}\\ =1.92*10^{-3}K(-1)$$\\$ For resistance,$R_1=123.4\Omega\\ R_1=R_0[1+\alpha(T-T_0)]$$\\$ Where, $T$ is the temperature when the resistance of lead is $123.4\Omega $$\\$ $123.4 = 101.6 [ 1 + 1.92 * 10^{-3} ( T - 273.15)]\\ 1.214 = 1 + 1.92 * 10^{-3} (T - 273.15)\\ \dfrac{0.214}{1.92*10^{-3}} \\ =T-273.15\\ \therefore T=384.61K$

4   Answer the following:$\\$ (a) The triple-point of water is a standard fixed point in modern thermometry. Why? What is wrong in taking the melting point of ice and the boiling point of water as standard fixed points (as was originally done in the Celsius scale)?$\\$ (b)There were two fixed points in the original Celsius scale as mentioned above which were assigned the number $0^oC$ and $100^oC$ respectively. On the absolute scale, one of the fixed points is the triple-point of water, which on the Kelvin absolute scale is assigned the number $273.16 K.$ What is the other fixed point on this (Kelvin) scale? The absolute temperature (Kelvin scale) $T$ is related to the temperature $t_c$ on the Celsius scale by $\\$ $t_c = T - 273.15$$\\$ Why do we have $273.15$ in this relation, and not $273.16$?$\\$ (d)What is the temperature of the triple-point of water on an absolute scale whose unit interval size is equal to that of the Fahrenheit scale?

Solution :

(a)The triple point of water has a unique value of $273.16 K.$ At particular values of volume and pressure, the triple point of water is always $273.16 K.$ The melting point of ice and boiling point of water do not have particular values because these points depend on pressure and temperature.$\\$ (b)The absolute zero or $0 K$ is the other fixed point on the Kelvin absolute scale.$\\$ (c)The temperature $273.16 K$ is the triple point of water. It is not the melting point of ice. The temperature $0^oC$ on Celsius scale is the melting point of ice. Its corresponding value on Kelvin scale is $273.15 K.$ Hence, absolute temperature (Kelvin scale) $T$ , is related to temperature $t_c$ , on Celsius scale as:$\\$ $t_c = T - 273.15$$\\$ (d)Let $T_F$ be the temperature on Fahrenheit scale and $T_K$ be the temperature on absolute scale. Both the temperatures can be related as:$\\$ $\dfrac{T_F-32}{180}=\dfrac{T_K-273.15}{100}....(i)$$\\$ Let $T_{ F1}$ be the temperature on Fahrenheit scale and $T_{K1}$ be the temperature on absolute scale. Both the temperatures can be related as:$\\$ $\dfrac{T_{F1}-32}{180}=\dfrac{T_{KI}-273.15}{100}...(ii)$$\\$ It is given that:$\\$ $T_{K1}-T_K=1 K$$\\$ Subtracting equation (i) from equation (ii), we get:$\\$ $\dfrac{T_{F1}-T_F}{180}=\dfrac{T_{K1}-T_K}{100}=\dfrac{9}{5}$$\\$ Triple point of water $= 273.16 K$$\\$ $\therefore $ Triple point of water on absolute scale $\\$ $=273.16*\dfrac{9}{5}=491.69 K$

5   Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made:$\\$ $\begin{array}{|c|c|c|c|} \hline \textbf{Temperature} & \textbf{Pressure thermometer A} &\textbf{Pressure thermometer B} \\ \hline \text{Triple-point of water } &1.250 * 10^ 5 \ \ Pa & 0.200 *10^5 \ \ Pa \\ \hline \text{ Normal melting point of sulphur} &1.797 * 10^5 \ \ Pa & 0.287 * 10^5 \ \ Pa \\ \hline \end{array}$$\\$ (a)What is the absolute temperature of normal melting point of sulphur as read by thermometers A and B?$\\$ (b)What do you think is the reason behind the slight difference in answers of thermometers A and B? (The thermometers are not faulty). What further procedure is needed in the experiment to reduce the discrepancy between the two readings?

Solution :

(a)Triple point of water, $T = 273.16 K$$\\$ At this temperature, pressure in thermometer $A, P_A =1.250 * 10^5 \ \ Pa$$\\$ Let $T_1$ be the normal melting point of sulphur.$\\$ At this temperature, pressure in thermometer $A, P_1 = 1.797 * 10^5 \ \ Pa$ $\\$ According to Charles’ law, we have the relation:$\\$ $ \dfrac{P_A}{T}=\dfrac{P_1}{T_1}\\ \therefore T_1=\dfrac{P_1T}{P_A}=\dfrac{1.797*10^5*273.16}{1.25*10^5}\\ =392.69K$$\\$ Therefore, the absolute temperature of the normal melting point of sulphur as read by thermometer $A$ is $392.69 K.$$\\$ At triple point $273.16 K,$ the pressure in thermometer$\\$ $B,P_B=0.200*10^5 Pa$$\\$ At temperature $T_1 $, the pressure in thermometer $B, P_2 =0.287 * 10^5 \ \ Pa$$\\$ According to Charles’ law, we can write the relation:$\\$ $\dfrac{P_B}{T}=\dfrac{P_1}{T_1}\\ \dfrac{0.200*10^5}{273.16} =\dfrac{0.287*10^5}{T_1}\\ \therefore T_1=\dfrac{0.287*10^5}{0.200*10^5}*273.16\\ =391.98K$$\\$ Therefore, the absolute temperature of the normal melting point of sulphur as read by thermometer $B$ is $391.98 K.$$\\$ (b)The oxygen and hydrogen gas present in thermometers $A$ and $B$ respectively are not perfect ideal gases. Hence, there is a slight difference between the readings of thermometers $A$ and $B.$ To reduce the discrepancy between the two readings, the experiment should be carried under low pressure conditions. At low pressure, these gases behave as perfect ideal gases.

6   A steel tape 1m long is correctly calibrated for a temperature of $27.0^oC$. The length of a steel rod measured by this tape is found to be $63.0 cm$ on a hot day when the temperature is $45.0 ^o C$ . What is the actual length of the steel rod on that day? What is the length of the same steel rod on a day when the temperature is $27.0^o C $? Coefficient of linear expansion of steel =$ 1.20 * 10^{-5} K^{-1}$.

Solution :

Length of the steel tape at temperature $T = 27 ^o C , l = 1 m =100 cm$$\\$ At temperature $T_1 = 45^o C$ , the length of the steel rod,$ l_ 1 = 63 cm$$\\$ Coefficient of linear expansion of steel, $\alpha= 1.20 * 10^{ -5} K ^{-1}$$\\$ Let $l_ 2$ be the actual length of the steel rod and $l '$ be the length of the steel tape at $45 ^o C $.$\\$ $l ' = l +al ( T_ 1 - T )\\ \therefore l'=100+1.20*10^{-5}*100(45-27)\\ =100.0216 cm$$\\$ Hence, the actual length of the steel rod measured by the steel tape at $45^o C $ can be calculated as: $\\$ $l_2=\dfrac{100.0216}{100}*63=63.0136 cm$$\\$ Therefore, the actual length of the rod at $45.0^oC$ is $ 63.0136 cm$ .Its length at $ 27.0^oC$ is 63.0 cm $$\\$

7   A large steel wheel is to be fitted on to a shaft of the same material. At $27^o C$ , the outer diameter of the shaft is $ 8.70 cm $ and the diameter of the central hole in the w heel is $ 8.69 cm. $ The shaft is cooled using ‘dry ice’. At what temperature of the shaft does the wheel slip on the shaft? Assume coefficient of linear expansion of the steel to be constant over the required temperature range : $\alpha_\text{ steel} = 1.20 * 10^{-5} K^{-1}$ .

Solution :

The given temperature, $T = 27 ^oC$ can be written in Kelvin as: $27 + 273 = 300 K$$\\$ Outer diameter of the steel shaft at $T, d _1 = 8.70 cm$$\\$ Diameter of the central hole in the wheel at $T, d_ 2 = 8.69 cm$$\\$ Coefficient of linear expansion of steel, $\alpha_{ steel} = 1.20 * 10^{-5 }K^{-1}$$\\$ After the shaft is cooled using ‘dry ice’, its temperature becomes $T_1$ .$\\$ The wheel will slip on the shaft, if the change in diameter, $\Delta d = 8.69 -8.70\\ = - 0.01 cm$$\\$ Temperature $T_ 1$ , can be calculated from the relation:$\\$ $\Delta d=d_1\alpha_{steel}(T_1-T)\\ =8.70*1.20*10^{-5}(T_1-300)\\ (T_1-300)=95.78\\ \therefore T_1=204.21K\\ =204.21-273.16\\ =-68.95^oC$$\\$ Therefore, the wheel will slip on the shaft when the temperature of the shaft is $-69 ^o C$ .

8   A hole is drilled in a copper sheet. The diameter of the hole is $4.24 cm$ at $27.0 ^oC$ . What is the change in the diameter of the hole when the sheet is heated to $227^o C$ ? Coefficient of linear expansion of copper $= 1.70 * 10^{-5} K^{-1} .$

Solution :

Initial temperature, $T _1 =27.0^o C$$\\$ Diameter of the hole at $T_1 , d _1 = 4.24 cm$$\\$ Final temperature, $T_2 = 227^o C$$\\$ Diameter of the hole at $T _2 = d_2$$\\$ Co-efficient of linear expansion of copper, $\alpha_{Cu} = 1.70 * 10^{-5} K^{-1}$$\\$ For co-efficient of superficial expansion $\beta $, and change in temperature $\Delta T$, we have the relation:$\\$ $\dfrac{\text{Change in area}(\Delta A)}{\text{Original area(A)}} =\beta \Delta T$$\\$ $\dfrac{(\pi \dfrac{d^2_2}{4}-\pi \dfrac{d^2_1}{4})}{(\pi \dfrac{d^2_1}{4})}\\ \dfrac{\Delta A}{A}\\ \therefore \dfrac{\Delta A}{A}=\dfrac{d^2_2-d^2_1}{d^2_1}\\ \text{But} \beta =2a\\ \dfrac{d^2_2}{d^2_1}-1=2\alpha(T_2-T_1)\\ \therefore \dfrac{d^2_2-d^2_1}{d^2_1}=2\alpha\Delta T\\ \dfrac{d^2_2}{(4.24)^2}=2*1.7*10^{-5}(227-27)+1\\ d^2_2=17.98*1.0068=18.1\\ \therefore d_2=4.2544 cm$$\\$ Change in diameter $=d_2-d_1=4.2544-4.24=0.0144 cm$$\\$ Hence, the diameter increases by $ 1.44 * 10^{-2} cm $$\\$

9   A brass wire $1.8 m$ long at $27^oC$ is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of $-39 ^oC $, what is the tension developed in the wire, if its diameter is $ 2.0 mm $ ? Co-efficient of linear expansion of brass $= 2.0 * 10^{-5} K^{-5 }; $Young’s modulus of brass $= 0.91*10^{11 }Pa .$$\\$

Solution :

Initial temperature, $T_1 = 27^o C$$\\$ Length of the brass wire at $T _1 , l = 1.8 m$$\\$ Final temperature,$ T _2 =- 39^oC$$\\$ Diameter of the wire, $d = 2.0 mm = 2 * 10^{-3} m$$\\$ Tension developed in the wire = F$\\$ Coefficient of linear expansion of brass, $\alpha =2.0 * 10^{-5} K^{-1}$$\\$ Young’s modulus of brass, $Y = 0.91 * 10^{11} Pa$$\\$ Young’s modulus is given by the relation:$\\$ $Y=\dfrac{\text{Stress}}{\text{Strain}}=\dfrac{\dfrac{F}{A}}{\dfrac{\Delta L}{L}}$$\\$ $\Delta L=\dfrac{F*L}{A*L}.....(i)$$\\$ Where,$\\$ F = Tension developed in the wire$\\$ A = Area of cross-section of the wire.$\\$ L = Change in the length, given by the relation:$\\$ $L=\alpha L(T_2-T_1)....(ii)$$\\$ Equating equations (i) and (ii), we get:$\\$ $\alpha L(T_2-T_1)=\dfrac{FL}{\pi (\dfrac{d}{2})^2*Y}\\ F=\alpha(T_2-T_1)\pi Y(\dfrac{d}{2})^2\\ F=2*10^{-5}*(-39-27)*3.14*0.91*10^{11}*(\dfrac{2*10^{-3}}{2})^2\\ =-3.8*10^2N$$\\$ (The negative sign indicates that the tension is directed inward.) Hence, the tension developed in the wire is $3.8 *10^2 N $.

10   A brass rod of length $ 50 cm $ and diameter $ 3.0 mm $ is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at $ 250^o C$ , if the original lengths are at $ 40.0 ^o C $ ? Is there a ‘thermal stress’ developed at the junction? The ends of the rod are free to expand (Co-efficient of linear expansion of brass $= 2.0 * 10^{-5} K^{-1}$ , steel $= 1.2*10^{-5} K ^{-1} )$

Solution :

Intial Temperature, $T_ 1 = 40^o C$$\\$ Final temperature, $T_ 2 = 250 ^o C$$\\$ Change in temperature $\Delta T = T_ 2 - T _1 = 210 ^o C$$\\$ Length of the brass rod at $T_ 1 , l_ 1 = 50 cm$$\\$ Diameter of the brass rod at $T _1 , d_ 1 = 3.0 mm$$\\$ Length of the steel rod at $T_ 2 , l_ 2 = 50 cm$$\\$ Diameter of the stell rod $T_ 2 , d _2 = 3.0 mm $$\\$ Coefficient of linear expansion of brass, $\alpha _1 = 2.0 * 10^{-5} K ^{-1}$$\\$ Coefficient of linear expansion of steel, $\alpha_2 = 1.2 * 10^{-5} K^{-1}$$\\$ For the expansion in the brass rod, we have:$\\$ $\dfrac{\text{Change in length($\Delta l_1$)}}{\text{Original length($l_1$)}}=\alpha_1 \Delta T$$\\$ $\therefore \Delta l_1=50*(2.1*10^{-5})*210\\ =0.2205 cm$$\\$ For the expansion in the steel rod, we have:$\\$ $\dfrac{\text{Change in length }(\Delta l_2)}{\text{Original length}(l_2)}=\alpha_2\Delta T\\ \therefore \Delta l_2=50*(1.2*10^{-5})*210\\ =0.126 cm $$\\$ Total change in the lengths of brass and steel, $\Delta l =\Delta l_ 1 +\Delta l _2\\ =0.2205 + 0.126\\ =0.346 cm$$\\$ Total change in the length of the combined rod = $0.346 cm$$\\$ Since the rod expands freely from both ends, no thermal stress is developed at the junction.