In a Carnot engine, the temperature of reservoir is $927^{\circ} \mathrm{C}$ and that of…

In a Carnot engine, the temperature of reservoir is $927^{\circ} \mathrm{C}$ and that of $\operatorname{sink}$ is $27^{\circ} \mathrm{C}$. If the work done by the engine when it transfers heat from reservoir to sink is $12.6 \times 10^{6} \mathrm{~J}$, the quantity of heat absorbed by the engine from the reservoir is
  1. $16.8 \times 10^{6} \mathrm{~J}$
  2. $4 \times 10^{6} \mathrm{~J}$
  3. $7.6 \times 10^{6} \mathrm{~J}$
  4. $4.2 \times 10^{6} \mathrm{~J}$

Solution

As we know $\eta=\frac{\mathrm{W}}{\mathrm{Q}_{1}}=1-\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}$
$\Rightarrow \quad \eta=1-\frac{300 \mathrm{~K}}{1200 \mathrm{~K}}=\frac{3}{4}$
$\frac{3}{4}=\frac{\mathrm{W}}{\mathrm{Q}_{1}} \Rightarrow \mathrm{Q}_{1}=\mathrm{W} \times \frac{4}{3} \Rightarrow \mathrm{Q}_{1}=12.6 \times 10^{6} \times \frac{4}{3}$
$Q_{1}=16.8 \times 10^{6} \mathrm{~J}$ ^

Asked in: JEE-TOPICTESTS-CHEMISTRY

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