Question

A system dissipates 12 J of heat into the surroundings; meanwhile, 28 J of work is done on the system. What is the change of the internal energy of the system?

Answer 1

+16 J

Step-by-step explanation:

We can solve the problem by using the 1st law of thermodynamics:

`\Delta U = Q-W`

where

`\Delta U` is the change of the internal energy of the system

Q is the heat (positive if supplied to the system, negative if dissipated by the system)

W is the work done (positive if done by the system, negative if done by the surroundings on the system)

In this case we have:

Q = -12 J is the heat dissipated by the system

W = -28 J is the work done ON the system

Substituting into the equation, we find the change in internal energy of the system:

`\Delta U=-12 J-(-28 J)=+16 J`