A CD player and its battery together do 500 kJ of work.The battery also releases 250 kJ of energy as heat,and the CD player releases 50 kJ as heat due to friction from spinning.What is the change in internal energy of the system,with the system regarded as the battery alone? Assume that the battery does 500 kJ of work on the CD player,which then does the same amount of work on the surroundings.

Question

Quizplus · Accepted Answer

The change in internal energy of the system, with the system regarded as the battery alone, can be calculated using the First Law of Thermodynamics:

ΔU = Q - W

where ΔU is the change in internal energy, Q is the heat added to the system, and W is the work done by the system.

In this case, the battery does 500 kJ of work on the CD player, so the work done by the system is -500 kJ (negative because it is work done BY the system). The battery also releases 250 kJ of energy as heat, and the CD player releases 50 kJ as heat due to friction, so the total heat added to the system is -300 kJ (again negative because it is heat added TO the system).

Therefore,

ΔU = Q - W 
ΔU = (-300 kJ) - (-500 kJ) 
ΔU = 200 kJ

So the change in internal energy of the system, with the system regarded as the battery alone, is +200 kJ.

A CD Player and Its Battery Together Do 500 KJ