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Question

(A) The efficiency $\eta$ of a Carnot engine is given by the formula:$\eta = 1 - \frac{T_2}{T_1}$where $T_1$ is the source temperature and $T_2$ is th

Vedclass · Accepted Answer

$A-iii, B-ii, C-iv, D-i$

Vedclass · Answer

(A) The efficiency $\eta$ of a Carnot engine is given by the formula:$\eta = 1 - \frac{T_2}{T_1}$where $T_1$ is the source temperature and $T_2$ is the sink temperature.Calculating the efficiency for each case:$(A)$ $T_1 = 500 	ext{ K}, T_2 = 300 	ext{ K} \Rightarrow \eta = 1 - \frac{300}{500} = 1 - 0.6 = 0.4$ (Matches $iii$)$(B)$ $T_1 = 500 	ext{ K}, T_2 = 350 	ext{ K} \Rightarrow \eta = 1 - \frac{350}{500} = 1 - 0.7 = 0.3$ (Matches $ii$)$(C)$ $T_1 = 800 	ext{ K}, T_2 = 400 	ext{ K} \Rightarrow \eta = 1 - \frac{400}{800} = 1 - 0.5 = 0.5$ (Matches $iv$)$(D)$ $T_1 = 450 	ext{ K}, T_2 = 360 	ext{ K} \Rightarrow \eta = 1 - \frac{360}{450} = 1 - 0.8 = 0.2$ (Matches $i$)Thus,the correct matching is $A-iii, B-ii, C-iv, D-i$.

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

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