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 CD player 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 is given by:
ΔU = Q - W
where Q is the heat added to the system and W is the work done by the system.
In this case, we can consider the CD player as the system, and use the information given to find the values of Q and W.
Q = -250 kJ (the negative sign indicates heat lost by the system)
W = 500 kJ (as the battery does 500 kJ of work on the CD player, and the CD player does the same amount of work on the surroundings)
ΔU = Q - W = -250 kJ - 500 kJ = -750 kJ
However, we need to subtract the additional 50 kJ of heat released by the CD player, due to friction.
Therefore, the final value of ΔU is:
ΔU = -750 kJ + 50 kJ = -700 kJ
This value is equivalent to F1F1F1S1F1F1F1050 kJ.

A CD Player and Its Battery Together Do 500 KJ