Chapter 11: Thermodynamics

Ans: It is provided that,

Water flows at a rate of \[3.0\] litre/min \[=3\times {{10}^{-3}}{{m}^{3}}/\min \]

Density of water, \[\rho ={{10}^{3}}kg/{{m}^{3}}\].

Clearly, mass of water flowing per minute \[=3\times {{10}^{-3}}\times {{10}^{3}}kg/\min =3kg/\min \]

The geyser heats the water, raising the temperature from \[27{}^\circ C\] to \[77{}^\circ C\].

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

Final temperature, \[{{T}_{2}}=77{}^\circ C\]

Thus, rise in temperature,

\[\Delta T={{T}_{2}}-{{T}_{1}}\]

\[\Rightarrow \Delta T=77{}^\circ C-27{}^\circ C\]

\[\Rightarrow \Delta T=50{}^\circ C\]

Now, heat of combustion \[=4\times {{10}^{4}}J/g=4\times {{10}^{7}}J/kg\]

Specific heat of water \[=4.2J/g\,{}^\circ C\]

It is known that total heat used, \[\Delta Q=mc\Delta T\]

\[\Rightarrow \Delta Q=3\times 4.2\times {{10}^{3}}\times 50\]

\[\Rightarrow \Delta Q=6.3\times {{10}^{5}}J/\min \]

Now, consider \[m\]kg of fuel to be used per minute.

Thus, the heat produced \[=m\times 4\times {{10}^{7}}J/\min \]

However, the heat energy taken by water \[=\] heat produced by fuel

Thus, equating both the sides,

\[\Rightarrow 6.3\times {{10}^{5}}=m\times 4\times {{10}^{7}}\]

\[\Rightarrow m=\frac{6.3\times {{10}^{5}}}{4\times {{10}^{4}}}\]

\[\Rightarrow m=15.75\,g/\min \]

Clearly, the rate of consumption of the fuel when its heat of combustion is \[4.0\times {{10}^{4}}J/g\] supposing the geyser operates on a gas burner is \[15.75\,\,g/\min \].

Ans: Provided that,

Mass of Nitrogen, \[m=2.0\times {{10}^{-2}}kg=20g\].

Rise in temperature, \[\Delta T=45{}^\circ C\].

Molecular mass of \[{{N}_{2}}\], \[M=28\]

Universal gas constant, \[R=8.3J\,mo{{l}^{-1}}{{K}^{-1}}\].

Number of moles, \[n=\frac{m}{M}\]

\[\Rightarrow n=\frac{2\times {{10}^{-2}}\times {{10}^{3}}}{28}\]

\[\Rightarrow n=0.714\]

Now, molar specific heat at constant pressure for nitrogen,

\[{{C}_{p}}=\frac{7}{2}R\]

\[\Rightarrow {{C}_{p}}=\frac{7}{2}\times 8.3\]

\[\Rightarrow {{C}_{p}}=29.05J\,mo{{l}^{-1}}{{K}^{-1}}\]

The total amount of heat to be supplied is given by the relation:

\[\Delta Q=n{{C}_{p}}\Delta T\]

\[\Rightarrow \Delta Q=0.714\times 29.05\times 45\]

\[\Rightarrow \Delta Q=933.38J\]

Clearly, the amount of heat to be supplied is \[933.38J\].

Ans: If two bodies at different temperatures \[{{T}_{1}}\]and \[{{T}_{2}}\] are brought in thermal contact, heat is said to flow from the body at higher temperature to the body at lower temperature until an equilibrium is obtained, i.e., until a point at which the temperatures of both the bodies becomes the same. 

The equilibrium temperature turns out to be the same as the mean temperature, which can be denoted as  \[\frac{\left( {{T}_{1}}+{{T}_{2}} \right)}{2}\] only when thermal capacities of the two bodies are the same.

Ans: The coolant in a chemical or nuclear plant must have a high specific heat as it is known that higher the specific heat of the coolant, higher is its heat-absorbing capacity and vice-versa.

Thus, a liquid which has a high specific heat is the best coolant to be utilized in a nuclear or chemical plant. This prevents different parts of the plant from becoming too hot.

Ans: When the car is in motion, the temperature of air inside the tyre rises due to motion of the air molecules. Charle’s law suggests that pressure is directly proportional to the temperature, \[P\propto T\].

Clearly, when the temperature inside a tyre rises, the air pressure inside the tyre would also increase.

Ans:  A harbour town has a more temperate climate than a town located in a desert at the same latitude. 

In a harbour town, the relative humidity is greater than that in a desert town.

Ans: Here, the cylinder is said to be completely insulated from its neighbourhood. As a result, no heat gets exchanged between the system (cylinder) and its neighbourhood. Clearly, the process turns out to be adiabatic.

Now, consider:

Initial pressure inside the cylinder \[={{P}_{1}}\]

Final pressure inside the cylinder \[={{P}_{2}}\]

Initial volume inside the cylinder \[={{V}_{1}}\]

Final volume inside the cylinder \[={{V}_{2}}\]

Ratio of specific heats, \[\gamma =1.4\]

For an adiabatic process, it is known that \[{{P}_{1}}{{V}^{\gamma }}_{1}={{P}_{2}}{{V}_{2}}^{\gamma }\].

Also, the final volume is compressed to half of its initial volume.

\[\Rightarrow {{V}_{2}}=\frac{{{V}_{1}}}{2}\]

Thus,

\[\Rightarrow {{P}_{1}}{{V}^{\gamma }}_{1}={{P}_{2}}{{\left( \frac{{{V}_{2}}}{2} \right)}^{\gamma }}\].

\[\Rightarrow \frac{{{P}_{2}}}{{{P}_{1}}}=\frac{{{V}_{1}}^{\gamma }}{{{\left( \frac{{{V}_{1}}}{2} \right)}^{\gamma }}}\]

\[\Rightarrow \frac{{{P}_{2}}}{{{P}_{1}}}={{2}^{\gamma }}={{2}^{1.4}}=2.639\]

Clearly, the pressure rises by a factor of \[2.639\].

Ans: It is provided that the work done \[\left( W \right)\] on the system when the gas transforms from state A to state B is \[22.3J\].

This is an adiabatic process. Thus, the change in heat is zero.

\[\Rightarrow \Delta Q=0\]

(As the work is done on the system)

Using the first law of thermodynamics,

\[\Delta Q=\Delta U+\Delta W\]

where,

\[\Delta U=\] change in the internal energy of the gas

When the gas transforms from state A to state B via a process, the net heat absorbed by the system is given by

\[\Delta Q=9.35cal=9.35\times 4.19J=39.1765J\]

Heat absorbed can be given by the equation,

\[\Delta Q=\Delta U+\Delta W\]

\[\Rightarrow \Delta W=\Delta Q-\Delta U=39.1765-22.3=16.8765J\]

Clearly, \[16.88J\] of work is done by the system.

Ans: The volume which can be availed by the gas at $1$ atmospheric pressure would be doubled when the stopcock is opened instantly. 

As volume is inversely proportional to pressure, the pressure reduces to one-half times its original value. 

The initial pressure of the gas being \[1\]atm results in the pressure of each cylinder to be \[0.5\]atm.

Ans: Change in the internal energy of any gas occurs when work is done by or on the gas. 

Here, as no work is done by or on the gas and thus, the internal energy of the gas would not change. Clearly, the change in internal energy of the gas is \[0\].

Ans: Change in the temperature of any gas occurs when work is done by or on the gas. 

Here, as no work is done by the gas during the expansion of the gas, the temperature of the gas would not change. Clearly, the change in temperature of the gas is \[0\].

Ans: No, the intermediate states of the system do not lie on the P-V-T surface since the process of free expansion is rapid and cannot be limited.

The intermediate states are non-equilibrium states and do not follow the gas equation. In due course of time, the gas returns to its original state.

Ans: Provided that,

Work done by the steam engine per minute, \[W=5.4\times {{10}^{8}}J\]

Heat supplied from the boiler, \[H=3.6\times {{10}^{9}}J\]

\[Efficiency\,of\,the\,engine=\frac{Output\,Energy}{Input\,Energy}\]

\[\Rightarrow \eta =\frac{W}{H}\]

\[\Rightarrow \eta =\frac{5.4\times {{10}^{8}}}{3.6\times {{10}^{9}}}\]

\[\Rightarrow \eta =0.15\]

Thus, the percentage efficiency of the engine is \[15%\].

Amount of heat wasted \[=3.6\times {{10}^{9}}-5.4\times {{10}^{8}}=30.6\times {{10}^{8}}=3.06\times {{10}^{9}}J\]

Clearly, the amount of heat wasted per minute is \[3.06\times {{10}^{9}}J\].

Ans: It is provided that,

Heat is supplied to the system by an electric heater at a rate of \[100W\]. 

Thus, heat supplied, \[Q=100J/s\]

The system operates at a rate of \[75J/s\]. 

Clearly, work done, \[W=75J/s\]

Using the first law of thermodynamics,

\[Q=U+W\]

where, 

\[U=\]internal energy 

\[\Rightarrow U=Q-W\]

\[\Rightarrow U=100-75\]

\[\Rightarrow U=25J/s\]

\[\Rightarrow U=25W\] 

Clearly, the internal energy of the given electric heater rises at a rate of \[25W\]or \[25J/s\].

Ans: Considering the given linear process in a thermodynamic system, it can be understood that,

Total work done by the gas from \[D\] to \[E\] to \[F\] is equal to the area of \[\Delta DEF\].

Area of \[\Delta DEF\] \[=\frac{1}{2}\times DE\times EF\]

where, 

\[DF=\] Change in pressure 

\[\Rightarrow DF=600N/{{m}^{2}}-300N/{{m}^{2}}\]

\[\Rightarrow DF=300N/{{m}^{2}}\]

Also,

\[FE=\]Change in volume $=5-2=3{{m}^{3}}$

Thus,

Area of \[\Delta DEF\]\[=\frac{1}{2}\times 300\times 3=450J\]  

Clearly, the total work done by the gas from \[D\]to \[E\] to \[F\]is \[450J\].

Ans: Here, the temperature inside the refrigerator can be provided as,

\[{{T}_{1}}=9{}^\circ C=282K\]

Room temperature is given as,

\[{{T}_{2}}=36{}^\circ C=309K\]

Coefficient of performance can be given by the relation,

\[COP=\frac{{{T}_{1}}}{\left( {{T}_{2}}-{{T}_{1}} \right)}\]

\[\Rightarrow COP=\frac{282}{\left( 309-282 \right)}\]

\[\Rightarrow COP=10.44\] 

Clearly, the coefficient of performance of the mentioned refrigerator is \[10.44\].