Question

In a 2 stage process Q1 = 5 J, Q2 = 10 J, if the change in internal energy in first stage is 10 J and the total change in internal energy is 20 J, what is the work do done in second stage?

(a) 0

(b) 5 J

(c) 10 J

(d) – 5 J

I have been asked this question during an interview.

This interesting question is from Energy Balances for Closed, Unsteady-State Systems topic in portion Introduction to Energy Balances for Processes without Reaction and Calculation of Enthalpy Changes of Basic Chemical Engineering

Answer 1

Right choice is (a) 0

To explain I would say: Change in internal energy in second stage = 20 – 10 =10 J, Work done in second stage = 10 – 10 = 0.