Question

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

Answer 1

Given:

Temperature of source, $T_1=27°C=(27+273)K=300K$

Temperature of sink, $T_2=227°C=(227+273)K=500K$

Heat, $Q_1=100cal$

To find:

Work done in one cycle, $W=?$

We know efficiency of an engine can be found by the following formula,

$\eta =\frac{T_1-T_2}{T_1}=\frac{500-300}{500}=0.4$

Again, efficiency is also equal to following formula,

$\eta =\frac{Q_1-W}{Q_1}⟹0.4=\frac{100-W}{100}⟹40=100-W⟹W=60cal$

Hence the work done in one cycle us 60cal.