Question

A system releases 125 kJ of heat while 104 kJ of work is done on the system. Calculate the change in internal energy.

Answer 1

Given:

The heat released, Q=-125 kJ

The work done on the system, W=104 kJ

To find:

The change in the internal energy.

Step-by-step explanation:

From the 1st law of thermodynamics, the change in the internal energy is given by,

`\Delta U=Q+W`

On substituting the known values,

`\begin{gathered} \Delta U=-125*10^3+104*10^3 \\ =-21\text{ kJ} \end{gathered}`

Final answer:

Thus the change in the internal energy is -21 kJ.