Question

An ideal engine which works between ${27}^{0}C$ and ${227}^{0}C$ takes 100 calorie of heat in one cycle.Calculate the work done in one cycle.

Answer 1

We know,

Heat input, $Q=100cal$

$T_1=27°C=(27+273)K=300K$

$T_2=227°C=(227+273)K=500K$

Efficiency of engine, $\eta =1-\frac{T_1}{T_2}=\frac{\text{Work done}}{\text{Heat input}}$

$⟹\text{Work done}=\frac{Q\times (T_2-t_1)}{T_2}=\frac{100\times (500-300)}{500}=40cal$