A Carnot engine having efficiency frac{1}{6} | vedclass

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Question

(B) The efficiency $\eta$ of a Carnot engine is given by $\eta = 1 - \frac{T_C}{T_H}$.For the first case, $\eta_1 = \frac{1}{6} = 1 - \frac{T_C}{T_H}$

Vedclass · Accepted Answer

$384 	ext{ K}, 320 	ext{ K}$

Vedclass · Answer

(B) The efficiency $\eta$ of a Carnot engine is given by $\eta = 1 - \frac{T_C}{T_H}$.For the first case, $\eta_1 = \frac{1}{6} = 1 - \frac{T_C}{T_H}$, which implies $\frac{T_C}{T_H} = 1 - \frac{1}{6} = \frac{5}{6}$, so $T_C = \frac{5}{6} T_H$.For the second case, the sink temperature becomes $T_C' = T_C - 64$. The new efficiency is $\eta_2 = \frac{1}{3} = 1 - \frac{T_C - 64}{T_H}$.This implies $\frac{T_C - 64}{T_H} = 1 - \frac{1}{3} = \frac{2}{3}$, so $T_C - 64 = \frac{2}{3} T_H$.Substituting $T_C = \frac{5}{6} T_H$ into the second equation: $\frac{5}{6} T_H - 64 = \frac{2}{3} T_H$.Rearranging gives $(\frac{5}{6} - \frac{4}{6}) T_H = 64$, which means $\frac{1}{6} T_H = 64$, so $T_H = 384 	ext{ K}$.Then $T_C = \frac{5}{6} 	imes 384 = 5 	imes 64 = 320 	ext{ K}$.Thus, $T_H = 384 	ext{ K}$ and $T_C = 320 	ext{ K}$.

List-$I$	List-$II$
$A$. $T_1 = 500 \text{ K}, T_2 = 300 \text{ K}$	$i$. $0.2$
$B$. $T_1 = 500 \text{ K}, T_2 = 350 \text{ K}$	$ii$. $0.3$
$C$. $T_1 = 800 \text{ K}, T_2 = 400 \text{ K}$	$iii$. $0.4$
$D$. $T_1 = 450 \text{ K}, T_2 = 360 \text{ K}$	$iv$. $0.5$

$A$ Carnot engine having efficiency $\frac{1}{6}$ operates between the source temperature $T_H$ and the sink temperature $T_C$. Its efficiency increases to $\frac{1}{3}$ when $T_C$ is decreased by $64 \text{ K}$. The temperatures $T_H$ and $T_C$ are respectively:

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$A$ Carnot engine has an efficiency of $40\%$. The temperature of its sink is $300 \ K$. To increase the efficiency by $50\%$ of its original efficiency while keeping the sink temperature constant,by how many $K$ must the source temperature be increased?

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