Question

A system is provided 50 kJ energy and work done on the system is 100 J. The change in internal energy is:

• A. 150 J
• B. 50 kJ
• C. 50.1 kJ
• D. 49.9 kJ

Answer 1

The correct option is C 50.1 kJ

A system is provided 50 kJ energy and work done on the system is 100 J. The change in internal energy is 50.1 kJ.
The calculations are as shown blow.
$q=\Delta U+(-W)$
or $\Delta U=q+\left(W\right)$ where $W$ is work done on the system
$=50\times {10}^3+100=50100J=50.1kJ$