- Home
- Standard 11
- Physics
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
normal