Question

A tank containing air is stirred by a paddle wheel. The work input to the wheel is 14000 kJ and heat transferred to the surrounding from the tank is 5000 kJ. The change in internal energy of the system (air) is:

1. -9000 kJ
2. +9000 kJ
3. +19000 kJ
4. -19000 kJ

Answer 1

Correct Answer - Option 2 : +9000 kJ

Cobcept:

In this case, the volume is constant.

δQ = -5000 kJ, and Work done into the system = -14000 kJ.

As per First Law of Thermodynamics:

δQ = dU + δW

-5000 = dU + (-14000)

dU = 9000 kJ.

Due to this change in internal energy, the system temperature will be raised but this is due to the work done on the system (by paddle wheel).

Here the boundary is rigid, the work done by the system is not changing its control volume, therefore the work done by the system (pdV work) is zero.