# 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)]$$\\ Where, T is the temperature when the resistance of lead is 123.4\Omega$$\\$ $123.4 = 101.6 [ 1 + 1.92 * 10^{-3} ( T - 273.15)]\\ 1.214 = 1 + 1.92 * 10^{-3} (T - 273.15)\\ \dfrac{0.214}{1.92*10^{-3}} \\ =T-273.15\\ \therefore T=384.61K$

##### 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.