Problem 4: A heat engine performs 200J of work in each cycle and has efficiency of 20%. For each cycle of operation, how much heat is absorbed and how much heat is expelled?

Solution:

Step:1 Overview

Efficiency of a Heat Engine:
Efficieny is a ratio of output to input. Here in heat engine, output is workdone (W) and input is amount of heat supplied by hot body (Q₁).
Hence efficieny of heat engine will be:

E=Output / Input
E=W/Q₁

As we know that workdone is equal to heat supplied to system by hot body (Q₁) minus heat absorbed by cold body (Q₂).

W=Q₁-Q₂
E=(Q₁-Q₂)/Q₁
E=1-Q₂/Q₁

Step:2 Calculation

Given:

W=200J
E=20%=0.2

Required:

Q₁=?
Q₂=?

Solution:

E=W/Q₁
0.2=200J/Q₁
0.2Q₁=200J
Q₁=200J/0.2
Q₁=1000J

As we know that:
W=Q₁-Q₂
Q₂=Q₁-W
Q₂=1000J-200J
Q₂=800J

Q₁=1000J, Q₂=800J (Ans)