Question

A heat engine (efficiency=40%) does 20 kJ of mechanical work. How much is the heat supplied to this engine in calories?

Answer 1

efficiency = 40%

$\eta= \frac{work\ done}{heat\ supplied}\\ \\ \Rightarrow 0.4= \frac{20\ kJ}{Heat\ supplied} \\ \\ \Rightarrow Heat\ Supplied= \frac{20}{0.4} =50\ kJ$

$1\ calorie = 4.184\ J\\50 kJ = \frac{50}{4.184}\ kcal=11.95\ kcal$

So 11.95 kcal heat should be supplied.