91Ó°ÊÓ

Temperatures of the hot and cold reservoirs between which Carnot heat engine operates are: $T_H=600K$and $T_C=300K$respectively. The amount of heat extracted from the hot reservoir (source) is $Q_1=1000J$, the amount of heat rejected to the cold reservoir (sink) is $Q_2$and the amount of work done by the Carnot heat engine is $W_{out}$. Similarly, the temperatures of the hot and cold reservoirs between which Carnot refrigerator operates are: $T{\text{'}}_H=500K$and $T{\text{'}}_C=400K$respectively. The amount of heat extracted from the cold reservoir (source) is $Q_4$, the amount of heat rejected to the hot reservoir (sink) is $Q_3$and the amount of work done by the Carnot refrigerator is $W_{in}$.

Formula Used :

Efficiency of a Carnot engine, $\mathrm{η}=\frac{W_{out}}{Q_1}=\frac{Q_1-Q_2}{Q_1}=\frac{T_H-T_C}{T_H};$

Coefficient of performance of Carnot refrigerator, $\mathrm{β}=\frac{Q_4}{W_{in}}=\frac{Q_4}{Q_3-Q_4}=\frac{T{\text{'}}_C}{T{\text{'}}_H-T{\text{'}}_C}$

Efficiency of the Carnot heat engine: $\mathrm{η}=\frac{Q_1-Q_2}{Q_1}=\frac{T_H-T_C}{T_H}$

$⇒\mathrm{η}=\frac{Q_1-Q_2}{Q_1}=\frac{600-300}{600}=\frac{1}{2}$

or, $\frac{1000-Q_2}{1000}=\frac{1}{2}⇒Q_2=500J$

Thus, $W_{out}=Q_1-Q_2=1000-500=500J$while,

$W_{in}=Q_3-Q_4=W_{out}=500J⇒Q_3=500+Q_4.$Using this relation between $Q_3$and $Q_4$in coefficient of performance, $\mathrm{β}=\frac{Q_4}{W_{in}}=\frac{Q_4}{Q_3-Q_4}=\frac{{T^{\text{'}}}_C}{{T^{\text{'}}}_H-{T^{\text{'}}}_C}⇒\frac{Q_4}{Q_4-Q_3}=\frac{400}{500-100}=4$

or, $\frac{Q_4}{500+Q_4-Q_4}=4⇒Q_4=2000J$

Therefore, $Q_3=500+Q_4=500+2000=2500J$

$Q_3$is greater than$Q_1$

No, the two devices don't violet second law of thermodynamics.