Question

In process A, 48 J of work are done on the system and 72 J of heat are added to the system. Find the change in the system's internal energy.

In process B, the system does 48 J of work and 72 J of heat are added to the system. What is the change in the system's internal energy?

In process C, the system's internal energy decreases by 129 J while the system performs 129 J of work on its surroundings. How much heat was added to the system?

Answer 1

From 1st law of thermodynamics

$dQ = dU + dW$

where dQ is the heat supplied to the system

dU is the change in internal energy of the system

dW is work done by the system.

Process A:

dQ = 72

dW = -48 since work is done on the system

Hence, substituting the values

$72 = dU - 48\\ \Rightarrow dU = 120\:J$

Process B:

dQ = 72

dW = 48

$72 = dU + 48\\ \Rightarrow dU = 72 -48 = 24\:J$

Process C:

$dQ = -129 + 129\\ \Rightarrow dQ = 0\:J$

dU is negative, since U decreases.

Heat added to the system 0J.