KPK 11th Class Physics Chapter 10 Thermodynamics Short Questions Answers

Thermal equilibrium will exist b/w two bodies, if heat loss by one body is absorbed by the 2nd body. Due to very high temperature, fusion reactions takes place at sun surface and as a result tremendous amount of heat energy is produced. The earth is not a good absorber. It absorbs some part of the heat energy loss by sun. also during night, the earth radiate some heat back into the atmosphere. That is why the earth is not in thermal equilibrium with the sun.

As the block is heated, all the dimensions increases, including the diameter of the hole. A useful analogy to thermal expansion is that of photographic enlargement. Every linear dimension of the object undergoes the same fractional change. Thus the increase in temperature will expand the hole and will not reduce it.

A thermometer registers its own temperature. When a thermometer is in thermal contact with an object, the temperature of which we wish to measure, energy will flow b/w the two until their temperature becomes equal and thermal equilibrium is established. At this stage, the temperature of thermometer will gives us the temperature of the body. A thermometer placed in direct sun light is in more contact with the surrounding air, so it will read the temperature of the air.

According to Graham’s law of diffusing, the rate of diffusion of a certain gas is inversely proportional to its molecular mass and density i.e. smaller the molecular mass and density, greater will be the rate of diffusion and vice versa. As the hydrogen is lighter than oxygen i.e. its molecular mass and density is lower than that of oxygen, so the pressure in a gas cylinder containing hydrogen will leak more quickly than if it is containing oxygen.

A the air conditioner is placed in the middle of the room, so the absorption and rejection of heat energy take place in the same room. As a result there will be no change in the temperature of the room.

a) When a sealed thermos flask bottle full of hot coffee is shaken, then the K.E of the molecules of the coffee increase and thus the temperature of the coffee will increase.

b) When a sealed thermos flask bottle full of hot coffee is shaken, the K.E and temperature of molecules of the coffee increase. As such activity is on adiabatic process, so the internal energy will increase.

When an object is heated, not all the energy it absorbs goes into increasing the velocity of the molecules. Some part of supplied heat energy is used against frictional effects among molecules and some part is used to increase the P.E of molecules of the object.

When an automobile is driven for a while, work is done against the frictional effects b/w try and road. As a result, heat energy is produced due to which the temperature of tyre and are inside it increase. As a result the collision of air molecules with each other and with the walls of the tyre increases and hence the pressure of air inside the try increases.

The air inside cycle tube is at high pressure and high temperature. When the valve is removed, the air rushes out from the tube. The outside atmospheric pressure is less than the air pressure inside the tube. As a result, the temperature of air falls down and hence becomes cool .

A refrigerator is a device which operates in the revers order of a heat energy. That is it extract heat from a cold body and transfer it to a hot body. No refrigerator is 100% efficient. Thus it exhausts more heat into the room than it extracts from it. Therefore, the net effect is an increase in the temperature of the room and it cannot be cooled by opening the door of the refrigerator inside the room.

1. I) it does not indicate the direction of heat transfer.
2. ii) it does not tells us anything about the conditions under which heat can be converted into work.

 iii) it given no information about frictional effects.

1. iv) it does not tells us that why the whole of heat energy cannot be converted into mechanical work continuously.
2. v) it gives no information about the principle of heat energy and refrigerator.
3. vi) it gives no information about thermal equilibrium and entropy of the system.

No it is not possible to construct a heat engine which is free from thermal pollution, because some part of heat is used to do work against frictional effects. 

The 2nd law of thermo synamic also tells that it is impossible to construct a machine which take heat from a hot body and convert whole of it into useful work with out having a sink [cold body] to which some part of heat is transfer during the operation. Thus the efficiency of all heat engines can never be 100%. It is always less than 100% due to thermal pollution.

No, it is not necessary for two systems to have the same amount of K.E. when they are in thermal equilibrium with each other.

For example, if a thermometer is placed in thermal contact in a pond of water until thermal equilibrium is reached. Then it is not necessary for both systems i.e. thermometer and pond of water to have the same amount of K.E, although both are in thermal equilibrium with each other.

It is true that work can be converted into heat completely. For example, when we rub our hands together, we do work which is complete it converted into heat. But it is not possible to convert heat completely into useful work, because some part of the heat is used against frictional forces.

When a certain process is performed, time must be consumed i.e. we say that how much time will be taken during the completion of a certain process. Similarly, we explain the law of increase of entropy by giving the reference of time i.e. we say that the entropy of the universe either remains constant or increases with the passage of time. That is why, we called entropy as “Arrow of time”.

1. A) We know that, ∆Q=nc∆T=>c=∆Q/n∆T……………………..(1)

Now in case of adiabatic process, no heat enters the system or leave the system, so we put, ∆Q=0 in eq 1,

C= 0/n∆T    =>   c=0

Thus the specific heat of a gas is zero in adiabatic process.

1. b) in case isothermal process, the temperature of the system remains constant, so put ∆T=0 in eq , we get,

  c= ∆Q/nxo=∆Q/0=∞

• C=∞

Thus the specific heat of a gas is infinitein isothermal process. 

1. c) When we remove the heat from a system, then the heat is taken as negative. But by doing so, there will be also fall in temperature in such a case. So according to eq(1), there will be no effect on the sign of ‘c’ i.e. the specific heat of a gas always remains positive, not negative.

Temperature of hot body =  T₁=227°c

• T₁= (227+273) k=500K

Temperature of cold body = T₂=27°c

• T₂= (27+273) K=> T₂=300K

We know that the efficiency of heat energy is given by,Thus the maximum possible efficiency is 40% which is less than the claimed efficiency i.e. 45%. Thus the claim of the inventor is not valid.