Thermodynamics

Class 11 NCERT Physics

NCERT

1   A geyser heats water flowing at the rate of $3.0$ litres per minute from $27^oC$ to $77^oC$ . If the geyser operates on a gas burner, what is the rate of consumption of the fuel if its heat of combustion is $4.0 * 10^4 J / g$ ?

Solution :

Water is flowing at a rate of $3.0$ litre/min.$=3*10^{-3}m^3 $$\\$ Density of water ,$\rho =10^3 kg / m^3$$\\$ $\therefore $ mass of water flowing per minute =$3*10^{-3}m^3 /min\\ =3kg /min$$\\$ The geyser heats the water, raising the temperature from $27^oC$ to $ 77^oC.$$\\$ Initial temperature,$T_1=27^oC$$\\$ Final temperature,$T_2=77^oC$$\\$ $\therefore $ Rise in temperature , $Delta T=T_2-T_1$$\\$ $=77-27=50^o C$$\\$ Heat of combustion$=4*10^4 J/g\\ =4*10^7 J/Kg$$\\$ Specific heat of water,$4.2 J/g^oC$$\\$ Total heat used,$\Delta Q=mc \Delta T$$\\$ $3*4.2*10^3*50\\ =6.3*10^5 J/min.$$\\$ Let $m$ kg of fuel is utilized per minute. $\therefore $ heat product =$m*4*10^7J/m$$\\$ But, heat energy taken by water = heat produced by fuel$\\$ $6.3*10^5=m*4*10^7\\ =\dfrac{6.3*10^5}{(4*10^4)}=15.75 g/min$

2   What amount of heat must be supplied to $2.0 *10 -2 kg$ of nitrogen (at room temperature) to raise its temperature by $45^oC$ at constant pressure? (Molecular mass of $N_2 =28; R = 8.3 J mol^{-1} K ^{-1} .)$

Solution :

Mass of nitrogen,$m=2.0*10^{-2}kg=20g$$\\$ Rise in temperature,$\Delta T=45^oC$$\\$ Molecular mass of $N_2,M=28$$\\$ Universal gas constant,$R=8.3 J mol^{-1} K^{-1}$$\\$ Number of moles,$n=m/M$$\\$ $=(2*10^{-2}*10^3)/28\\ =0.714$$\\$ Molar specific heat at constant pressure for nitrogen,$C_p=(7/2)R$$\\$ $=(7/2)*8.3\\ =29.05 J mol^{-1} K^{-1}$$\\$ The total amount of heat to be supplied is given by the relation:$\\$ $\Delta Q=nC_p\Delta T\\ =0.714*29.05*45\\ 933.38 J$$\\$ Therefore, the amount of heat to be supplied is $933.38 J.$

3   Explain why$\\$ (a) Two bodies at different temperatures $T_1$ and $T_2$ if brought in thermal contact do not necessarily settle to the mean temperature$(T_1 +T_2) / 2.$$\\$ (b) The coolant in a chemical or a nuclear plant (i.e., the liquid used to prevent the different parts of a plant from getting too hot) should have high specific heat.$\\$ (c) Air pressure in a car tyre increases during driving.$\\$ (d) The climate of a harbour town is more temperate than that of a town in a desert at the same latitude.

Solution :

(a) When two bodies at different temperatures $T_1$ and $T_2$ are brought in thermal contact, heat flows from the body at higher temperature to the body at lower temperature till equilibrium is achieved, i.e., the temperatures of both the bodies becomes equal. The equilibrium temperature equal to the mean temperature $(T_1+T_2) / 2$ only when thermal capacities of the two bodies are equal.$\\$ (b) The coolant in a chemical or nuclear plant should have a high specific heat. This is because higher the specific heat of the coolant, higher is its heat-absorbing capacity and vice-versa. Hence, a liquid having a high specific heat is the best coolant to be used in a nuclear or chemical plant. This would prevent different parts of the plant from getting too hot.$\\$ (c) When the car is in motion, the temperature of air inside the tyre increases due to motion of the air molecules. According to Charle’s law, pressure is directly proportional to the temperature, $P \propto T $. Hence, if the temperature inside a tyre increases, then the air pressure inside the tyre will also increase .$\\$ (d) A harbor town has a more temperate climate than a town located in a dessert at the same latitude. This is because in a harbour town, the relative humidity is more than in a desert town.

4   A cylinder with a movable piston contains $3$ moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas is compressed to half its original volume?

Solution :

The cylinder is completely insulated from its surroundings. As a result, no heat is exchanged between the system (cylinder) and its surroundings. Thus, the process is adiabatic.$\\$ 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$$\\$ For an adiabatic process, we have:$\\$ $P_1V_1^\gamma =P_2V_2^\gamma$$\\$ The final volume is compressed to half of its initial volume.$\\$ $\therefore V_2=V_1/2\\ P_1V_1^\gamma=P_2(V_1/2)^\gamma\\ P_2/P_1=V_1^\gamma/(V_1/2)^\gamma\\ =2^\gamma = 2^{1.4}=2.639$$\\$ Hence, the pressure increases by a factor of $2.639.$

5   In changing the state of a gas adiabatically from an equilibrium state $A$ to another equilibrium state $B$, an amount of work equal to $22.3 J$ is done on the system. If the gas is taken from state $A$ to $B$ via a process in which the net heat absorbed by the system is $9.35$ cal, how much is the net work done by the system in the latter case? (Take $1$ cal = $4.19 J$)

Solution :

The work done $(W)$ on the system while the gas changes from state $A$ to state $B$ is $22.3 J.$This is an adiabatic process. Hence, change in heat is zero.$\\$ $\therefore \Delta Q=0\\ \Delta W=-22.3 J $(Since the work is done on the system)$\\$ From the first law of thermodynamics, we have:$\\$ $\Delta Q=\Delta U+\Delta W$$\\$ where,$\\$ $\Delta U=$Change in the internal energy of the gas$\\$ $\therefore \Delta U=\Delta Q-\Delta W\\ =-(-22.3 J)\\ \Delta U=+22.3 J$$\\$ When the gas goes from state $A$ to state $B$ via a process, the net heat absorbed by the system is:$\\$ $\Delta Q=9.35 cal\\ =9.35*4.19\\ =39.1765 J$$\\$ Heat absorbed,$\Delta Q=\Delta U+\Delta Q$$\\$ $\therefore \Delta W=\Delta Q-\Delta U.\\ =39.1765-22.3\\ =16.8765 J$$\\$ Therefore, $16.88 J $ of work is done by the system.