Gujarati
Hindi
11.Thermodynamics
normal

The efficiency of a carnot engine is $0.6$.It rejects total $20 \ J$ of heat. The work done by the engine is   .... $J$

A

$40$

B

$50$

C

$20$

D

$30$

Solution

$0.6=\frac{\text { work done }}{Q_{\text {input }}}=\frac{Q_{\text {input }}-Q_{\text {reject }}}{Q_{\text {input }}}=1-\frac{Q_{r}}{Q_{i}}$

$0.6=1-\frac{20}{\mathrm{Q}_{\mathrm{i}}}$

$\mathrm{Q}_{\mathrm{i}}=50$

$\mathrm{W}=\mathrm{Q}_{\mathrm{i}}-\mathrm{Q}_{\mathrm{r}}=30 \mathrm{J}$

Standard 11
Physics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.