When a system is taken from state A state B along the path A-C-B, 180 KJ of the heat flows into the system and it does 130 KJ of work (see figure given)
How much heat will flow into the system along the A-D-B if the work done by it along the path is 40 KJ?

Option (A)
40 KJ
Option (B)
60 KJ
90 KJ
135 KJ
Correct Option:
Question Solution:
From first law of thermodynamics
UA + Q = UB + W
UB-UA = Q –W = 180 – 130 = 50kJ
Since internal energy is property (is independent of path function)
UB-UA = Q –W
50 = Q – 40 Þ Q = 90kJ