At a certain temperature T(K), during a process, 500 J is absorbed by the system and work of 200 J is done by the system. Then change in internal energy of the system is:

To find the change in internal energy (ΔU) of a system we use the first law of thermodynamics.

$\mathrm{\Delta U}=q-W$

where:

• $q$ is the heat absorbed by the system (positive when heat enters the system).
• $W$ is the work done **by** the system (positive when the system does work on the surroundings).

From the problem statement:

• Heat absorbed: $q=\; +500 J$
• Work done by the system: $W=\; +200 J$

Insert these values into the equation:

$\mathrm{\Delta U}=\; 500 J\; -\; 200 J$

Perform the subtraction:

$\mathrm{\Delta U}=\; 300 J$

Therefore, the internal energy of the system increases by 300 J.