Chapter 10: Thermal Properties of Matter

Ans: The relation between Kelvin and Celsius scales:

${{T}_{C}}={{T}_{K}}-273.15$

The relation between Celsius and Fahrenheit scales:

${{T}_{F}}=\frac{9}{5}{{T}_{c}}+32$

For neon:

${{T}_{K}}=24.57K$ 

$\Rightarrow {{T}_{C}}=24.57-273.15=-248.58{}^\circ C$ 

As ${{T}_{F}}=\frac{9}{5}{{T}_{C}}+32$ 

$\Rightarrow {{T}_{F}}=\frac{9}{5}\left( -248.58 \right)+32$ 

$\Rightarrow {{T}_{F}}=-415.44{}^\circ F$ 

For carbon dioxide:

${{T}_{K}}=216.55K$ 

$\Rightarrow {{T}_{C}}=216.55-273.15=-56.60{}^\circ C$ 

As ${{T}_{F}}=\frac{9}{5}{{T}_{C}}+32$

$\Rightarrow {{T}_{F}}=\frac{9}{5}\left( -56.60 \right)+32$ 

$\Rightarrow {{T}_{F}}=-69.88{}^\circ F$ 

Therefore, the temperatures on the Celsius scale and Fahrenheit scale are $-248.58{}^\circ C$  and $-415.44{}^\circ F$ respectively for neon and $-56.60{}^\circ C$ and $-69.88{}^\circ F$ respectively for carbon dioxide.

Ans: Given that,

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 an absolute scale A.

${{T}_{1}}={{T}_{K}}$ 

$\Rightarrow 200A=273.15K$ 

$\Rightarrow A=\frac{273.15}{200}$ 

The temperature 273.15 K on Kelvin scale is equivalent to 350 B on absolute scale B.

${{T}_{2}}={{T}_{K}}$ 

$\Rightarrow 350B=273.15$ 

\[\Rightarrow B=\frac{273.15}{350}\] 

${{T}_{A}}$ is the triple point of water on scale A.

${{T}_{B}}$ is the triple point of water on scale B.

Therefore,

$\frac{273.15}{200}\times {{T}_{A}}=\frac{273.15}{350}\times {{T}_{B}}$ 

$\Rightarrow {{T}_{A}}=\frac{200}{350}{{T}_{B}}$ 

Therefore, the ratio ${{T}_{A}}$ : ${{T}_{B}}$ is given as $4$: $7$.

Ans: It is given that:

$R={{R}_{0}}\left[ 1+\alpha \left( T-{{T}_{0}} \right) \right]$ \[\] …

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 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 $=600.5K$ 

Resistance of lead, R$=165.5$$\Omega $ 

Substituting these values in, we get:

$R={{R}_{0}}\left[ 1+\alpha \left( T-{{T}_{0}} \right) \right]$

$\Rightarrow 165.5=101.6\left[ 1+\alpha \left( 600.5-273.15 \right) \right]$ 

$\Rightarrow 1.629=1+\alpha (327.35)$ 

$\Rightarrow \alpha =\frac{0.629}{327.35}=1.92\times {{10}^{-3}}$ 

For resistance, ${{R}_{1}}=123.4$ $\Omega $ 

$R={{R}_{0}}\left[ 1+\alpha \left( T-{{T}_{0}} \right) \right]$

where, T is the temperature when the resistance of lead is 123.4$\Omega $

$\Rightarrow 123.4=101.6\left[ 1+1.92\times {{10}^{-3}}\left( T-273.15 \right) \right]$

$\Rightarrow 1.214=1.192\times {{10}^{-3}}\left( T-273.15 \right)$ 

$\Rightarrow \frac{0.214}{1.92\times {{10}^{-3}}}=\left( T-273.15 \right)$ 

$\Rightarrow T=384.61K$ 

Clearly, the temperature when the resistance is 123.4$\Omega $ is $384.61K$.

Ans: The triple point of water has a unique value of $273.16$ K.

The melting point of ice and boiling point of water do not have particular values because these points depend on pressure and temperature.

1. There were two fixed points in the original Celsius scale as mentioned above which were assigned the numbers $0{}^\circ C$ and $100{}^\circ C$ 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?

Ans: The absolute zero or 0 K is itself the other fixed point on the Kelvin absolute scale.

1. The absolute temperature (Kelvin scale) T is related to the temperature ${{t}_{C}}$  the Celsius scale by ${{t}_{C}}=T-273.15$.

Why do we have 273.15 in this relation, and not 273.16?

Ans: The temperature 273.16 K is the triple point of water; not the melting point of ice.

The temperature $0{}^\circ C$  on Celsius scale is the melting point of ice. 

Its corresponding value on the Kelvin scale is 273.15 K.

Thus, absolute temperature (Kelvin scale) T, is related to temperature ${{t}_{C}}$  on Celsius scale as: ${{t}_{C}}=T-273.15$.

1. 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?

Ans: Let ${{T}_{F}}$  be the temperature on Fahrenheit scale and ${{T}_{K}}$ be the temperature on absolute scale. Both the temperatures have the relation:

$\frac{{{T}_{F}}-32}{180}=\frac{{{T}_{K}}-273.15}{100}$ \[\] …

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:

$\frac{{{T}_{F1}}-32}{180}=\frac{{{T}_{K1}}-273.15}{100}$     …

It is given that:

${{T}_{K1}}-{{T}_{K}}=1K$ 

Subtracting equation (a) from equation (b), we get:

$\frac{{{T}_{F1}}-{{T}_{F}}}{180}=\frac{{{T}_{K1}}-{{T}_{K}}}{100}=\frac{1}{100}$

$\Rightarrow {{T}_{F1}}-{{T}_{F}}=\frac{1\times 180}{100}=\frac{9}{5}$

Triple point of water $=273.16K$ 

Therefore, triple point of water on absolute scale $=273.16\times \frac{9}{5}=491.69K$. 

Ans: Given that,

Triple point of water, T = 273.16 K 

At this temperature, pressure in thermometer A, \[{{P}_{A}}=1.250\times {{10}^{5}}\] Pa

 Let ${{T}_{1}}$  be the normal melting point of Sulfur. 

At this temperature, pressure in thermometer A, ${{P}_{1}}=1.797\times {{10}^{5}}$Pa

According to Charles’ law, we have the relation:

$\frac{{{P}_{A}}}{T}=\frac{{{P}_{1}}}{{{T}_{1}}}$ 

$\Rightarrow {{T}_{1}}=\frac{{{P}_{1}}T}{{{P}_{A}}}=\frac{1.797\times {{10}^{5}}\times 273.16}{1.25\times {{10}^{5}}}$ 

$\Rightarrow {{T}_{1}}=392.69K$ 

Therefore, the absolute temperature of the normal melting point of Sulfur as read by thermometer A is 392.69 K.

Now, at triple point 273.16 K, the pressure in thermometer B,  \[{{P}_{B}}=0.200\times {{10}^{5}}\]Pa

At temperature ${{T}_{1}}$ , the pressure in thermometer B, \[{{P}_{2}}=0.287\times {{10}^{5}}\]Pa

According to Charles’ law, we can write the relation: 

$\frac{{{P}_{B}}}{T}=\frac{{{P}_{1}}}{{{T}_{1}}}$

$\Rightarrow {{T}_{1}}=\frac{0.287\times {{10}^{5}}\times 273.16}{0.200\times {{10}^{5}}}=391.98K$

Therefore, the absolute temperature of the normal melting point of Sulfur as read by thermometer B is 391.98 K.

1. 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?

Ans: Oxygen, as well as hydrogen gas contained in thermometers A and B respectively, are not perfect ideal gasses.

There is a minor difference between the readings of thermometers A and B.

To lessen the discrepancy between the two readings, the experiment must be carried under low pressure conditions so that gasses can behave as ideal gas.

Ans: Given that,

Length of the steel tape at temperature $27.0{}^\circ C$, $l=1m=100cm$ 

At temperature  ${{T}_{1}}=45{}^\circ C$,  the length of the steel rod, ${{l}_{1}}=63cm$ 

Coefficient of linear expansion of steel, $\alpha =1.20\times {{10}^{-5}}{{K}^{-1}}$ 

 Let ${{l}_{2}}$  be the actual length of the steel rod and ${l}'$ be the length of the steel tape at  $45{}^\circ C$.

\[{l}'=l+al\left( {{T}_{1}}-T \right)\] 

\[\Rightarrow {l}'=100+1.20\times {{10}^{-5}}\times 100\left( 45-27 \right)\]

\[\Rightarrow {l}'=100.0216cm\] 

Hence, the actual length of the steel rod measured by the steel tape at $45{}^\circ C$ can be calculated as: 

\[{{l}_{2}}=\frac{100.0216}{100}\times 63=63.0136cm.\]

Therefore, the actual length of the rod at $45{}^\circ C$  is 63.0136 cm. Its length at $27.0{}^\circ C$ is 63.0 cm.

Ans: Here,

The given temperature, $T=~27{}^\circ C$  can be written in Kelvin as \[27+273=300\text{ }K\].

Outer diameter of the steel shaft at T,  ${{d}_{1}}=8.70cm$ 

Diameter of the central hole in the wheel at T,  ${{d}_{2}}=8.69cm$  

Coefficient of linear expansion of steel, $:{{\alpha }_{steel}}=1.20\times {{10}^{-5}}{{K}^{-1}}$

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, \[\Delta d=8.698.70=0.01\]cm

Temperature ${{T}_{1}}$ , can be calculated from the relation: 

\[\Delta d={{d}_{1}}{{\alpha }_{steel}}\left( {{T}_{1}}-T \right)\] 

\[\Rightarrow \Delta d=8.70\times 1.20\times {{10}^{-5}}\left( {{T}_{1}}-300 \right)\] 

$\Rightarrow \left( {{T}_{1}}-300 \right)=95.78$ 

$\Rightarrow {{T}_{1}}=204.21K$ 

Or ${{T}_{1}}=204.21-273.16=-68.95{}^\circ C$ 

Clearly, the wheel will slip on the shaft when the temperature of the shaft is approximately $-69{}^\circ C.$ 

Ans: Given that,

Initial temperature, ${{T}_{1}}=27.0{}^\circ C$ 

Diameter of the hole at ${{T}_{1}},{{d}_{1}}=4.24cm$ 

Final temperature,  ${{T}_{2}}=227{}^\circ C$ 

Diameter of the hole at ${{T}_{2}}={{d}_{2}}$ 

Coefficient of linear expansion of copper, ${{\alpha }_{Cu}}=1.70\times {{10}^{-5}}{{K}^{-1}}$ 

For coefficient of superficial expansion $\beta $, and change in temperature  ∆T, we have the relation: 

$\frac{change\text{ }in\text{ }area\left( \Delta A \right)}{original\text{ }area\left( A \right)}=\beta \Delta T$ 

$\Rightarrow \frac{\left( \pi \frac{d_{2}^{2}}{4}-\pi \frac{d_{1}^{2}}{4} \right)}{\left( \pi \frac{d_{1}^{2}}{4} \right)}=\frac{\Delta A}{A}$ 

$\Rightarrow \frac{\Delta A}{A}=\frac{d_{2}^{2}-d_{1}^{2}}{d_{1}^{2}}$

However, $\beta =$ $2\alpha $ 

$\Rightarrow \frac{d_{2}^{2}}{d_{1}^{2}}-1=2\alpha \left( {{T}_{2}}-{{T}_{1}} \right)$ 

$\Rightarrow \frac{d_{2}^{2}-d_{1}^{2}}{d_{1}^{2}}=2\alpha \Delta T$ 

$\Rightarrow \frac{d_{2}^{2}}{{{\left( 4.24 \right)}^{2}}}=2\times 1.7\times {{10}^{-5}}\left( 227-27 \right)+1$ 

$\Rightarrow d_{2}^{2}=17.98\times 1.0068=18.1$ 

$\Rightarrow {{d}_{2}}=4.2544cm$ 

Change in diameter  =${{d}_{2}}-{{d}_{1}}=4.2544-4.24=0.0144cm$.

Clearly, the diameter increases by $1.44\times {{10}^{-2}}cm.$ 

Ans: Given that,

Initial temperature,  ${{T}_{1}}=27{}^\circ C$ 

Length of the brass wire at \[{{T}_{1}},l=1.8m\] 

Final temperature,  ${{T}_{2}}=-39{}^\circ C$

Diameter of the wire,  $d=2.0mm=2\times {{10}^{-3}}m$ 

Tension developed in the wire = F 

Coefficient of linear expansion of brass,$\alpha =2.0\times {{10}^{-5}}{{K}^{-1}}$

Young’s modulus of brass,  $Y=0.91\times {{10}^{11}}Pa$ 

Young’s modulus is given by the relation:

$Y=\frac{Stress}{Strain}=\frac{\frac{F}{A}}{\frac{\Delta L}{L}}$ 

$\Rightarrow \Delta L=\frac{F\times L}{A\times L}$              … 

Where, F = Tension developed in the wire

A = Area of cross-section of the wire

L = Change in the length, given by the relation:

$L=\alpha \left( {{T}_{2}}-{{T}_{1}} \right)$          …

Equating equations (a) and (b), we get:

$\alpha L\left( {{T}_{2}}-{{T}_{1}} \right)=\frac{FL}{\pi {{\left( \frac{d}{2} \right)}^{2}}\times Y}$ 

$\Rightarrow F=\alpha \left( {{T}_{2}}-{{T}_{1}} \right)\pi Y{{\left( \frac{d}{2} \right)}^{2}}$

$\Rightarrow F=2\times {{10}^{5}}\times \left( -39-27 \right)\times 3.14\times 0.91\times {{10}^{11}}\times {{\left( \frac{2\times {{10}^{-3}}}{2} \right)}^{2}}$

$\Rightarrow F=-3.8\times {{10}^{2}}N$ 

(The negative sign indicates that the tension is directed inward.) 

Clearly, the tension developed in the wire is $3.8\times {{10}^{2}}N$.

Ans: Given that,

Initial Temperature, ${{T}_{1}}=40{}^\circ C$

Final temperature,  ${{T}_{2}}=40{}^\circ C$

Change in temperature $={{T}_{2}}-{{T}_{1}}=210{}^\circ C$ 

Length of the brass rod at ${{T}_{1}}$, ${{l}_{1}}=50cm$ 

Diameter of the brass rod at ${{T}_{1}}$, ${{d}_{1}}=3.0mm$

Length of the steel rod at ${{T}_{2}}$, ${{l}_{2}}=50cm$

Diameter of the steel rod ${{T}_{2}}$, ${{d}_{2}}=3.0mm$

Coefficient of linear expansion of brass, ${{\alpha }_{1}}=2.0\times {{10}^{-5}}{{K}^{-1}}$ 

Coefficient of linear expansion of steel, ${{\alpha }_{2}}=1.2\times {{10}^{-5}}{{K}^{-1}}$ 

For the expansion in the brass rod, we have:

$\frac{change\text{ }in\text{ length}\left( \Delta {{l}_{1}} \right)}{original\text{ length}\left( {{l}_{1}} \right)}={{\alpha }_{1}}\Delta T$

$\Rightarrow \Delta {{l}_{1}}=50\times \left( 2.1\times {{10}^{-5}} \right)\times 210=0.2205cm$ 

For the expansion in the steel rod, we have:

$\frac{change\text{ }in\text{ length}\left( \Delta {{l}_{2}} \right)}{original\text{ length}\left( {{l}_{2}} \right)}={{\alpha }_{2}}\Delta T$

$\Rightarrow \Delta {{l}_{2}}=50\times \left( 1.2\times {{10}^{-5}} \right)\times 210=0.126cm$

Total change in the lengths of brass and steel, $\Delta l=\Delta {{l}_{1}}+\Delta {{l}_{2}}$ 

$\Rightarrow \Delta l=0.2205+0.126=0.346cm$

Clearly, the total change in the length of the combined rod = 0.346 cm.

As the rod expands freely from both ends, no thermal stress is developed at the junction.