Question:
A Carnot's engine working between $400 \mathrm{~K}$ and $800 \mathrm{~K}$ has a work output of $1200 \mathrm{~J}$ per cycle. The amount of heat energy supplied to the engine from the source in each cycle is :
Correct Option: , 4
Solution:
$\eta=\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\frac{\mathrm{Q}_{2}}{\mathrm{Q}_{1}}=\frac{\mathrm{Q}_{1}-\mathrm{W}}{\mathrm{Q}_{1}}$
$\left(\because \mathrm{W}=\mathrm{Q}_{1}-\mathrm{Q}_{2}\right)$
$\frac{400}{800}=1-\frac{\mathrm{W}}{\mathrm{Q}_{1}}$
$\frac{\mathrm{W}}{\mathrm{Q}_{1}}=1-\frac{1}{2}=\frac{1}{2}$
$\mathrm{Q}_{1}=2 \mathrm{~W}=2400 \mathrm{~J}$
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