Question

If work done on a system is 20 J and a heat of 15 J is taken out, what is the change in internal energy of the system?

(a) 5 J

(b) 10 J

(c) 20 J

(d) 30 J

I have been asked this question during an interview.

The doubt is from Energy balance for Different type of systems topic in section Miscellaneous of Chemical Process Calculation

Answer 1

The change in internal energy ($\Delta U$) can be calculated using the first law of thermodynamics, which states:

$$\Delta U=Q-W$$

Where:

• $Q$ is the heat added to the system (positive if heat is added, negative if heat is taken out).
• $W$ is the work done on the system (positive if work is done on the system, negative if work is done by the system).

Given:

• Work done on the system, $W=20\text{J}$ (positive, because work is done on the system).
• Heat taken out of the system, $Q=-15\text{J}$ (negative, because heat is removed from the system).

Now, applying the values:

$$\Delta U=(-15\text{J})-(20\text{J})=-35\text{J}$$

So, the change in internal energy of the system is:

−35 J\boxed{-35 \, \text{J}}−35J​

However, since none of the options match this result exactly, it may be useful to double-check the given problem conditions or if there was a typo in the question or choices. Based on standard thermodynamics, this would be the answer for the change in internal energy.