Question

A heat engine performs 200J of work in each cy-
cle and has efficiency of 20 percent. For each
cycle of operation a) How much heat is absorbed
and b) How much heat is expelled?​

Answer 1

Answer:

Q1=1000J     &      Q2=800J

Explanation:

Given data

W=200J             &         Efficiency  =20%=0.2

To find ??

Q1=?          &         Q2=?

Solution:

E=W/Q1

0.2=200J/Q1

0.2Q1=200J

Q1=200J/0.2

Q1=1000J

As work is  

W=Q1-Q2

Q2=Q1-W

Q2=1000J-200J

Q2=800J