Q4 of 51 Page 61

Refer to figure. Let ΔU1 and ΔU2 be the change in internal energy in processes A and B respectively, ΔQ be the net heat given to the system in process A + B and ΔW be the net work done by the system in the process A + B.


As seen from the figure initial state of A is the final state of B and vice-versa.


So, the change in internal energy of process A and B will be equal in magnitude but with a negative sign.


So ΔU1=-ΔU2


∴ ΔU1+ΔU2=0.

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2

In a process on a system, the initial pressure and volume are equal to the final pressure and volume.

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A system can be taken from the initial state p1, V1 to the final state p2, V2 by two different methods. Let ΔQ and ΔW represent the heat given to the system and the work done by the system. Which of the following must be the same in both the methods?

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The internal energy of an ideal gas decreases by the same amount as the work done by the system.

 1

A thermally insulated, closed copper vessel contains water at 15°C. When the vessel is shaken vigorously for 15 minutes, the temperature rises to 17°C. The mass of the vessel is 100 g and that of the water is 200 g. The specific heat capacities of copper and water are 420 J kg–1 K–1 and 4200 J kg–1 K–1 respectively.

Neglect any thermal expansion.


(a) How much heat is transferred to the liquid-vessel system?


(b) How much work has been done on this system?


(c) How much is the increase in internal energy of the system?