Question

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

Answer 1

Given that,

A system dissipates 12 J of heat into the surroundings.

28 J of work is done on the system.

To find,

The internal energy of the system.

Solution,

The first law of thermodynamics is used here. According to this law,

$\Delta E=Q-W$

Q is heat and W is work done

Here,

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

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

ATQ,

$\Delta E=(-12)-(-28)\\\\=-12+28\\\\=16\ J$

So, the change of internal energy of the system is 16 J.