Question

a heat engine is operating between 500k to 300k and it absorbs 10 kcal of heat from 500k reservoir reversibly per cycle. calculate the work done(in kcal) per cycle.

Answer 1

Answer:

$T_{1} =300k$

$T_{2} =500k$

$T_{a} =10kcal$

η=$\frac{W}{Q_{a} }$

η=1-$\frac{T_{1} }{T_{2} }$

η=1-0.6

=0.4

so work done is

0.4=W/10kcal

W=4kcal