A Carnot engine operating between temperature | vedclass

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Question

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

Vedclass · Accepted Answer

$150 	ext{ K}, 120 	ext{ K}$

Vedclass · Answer

(A) The efficiency $\eta$ of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$, where $T_1$ is the source temperature and $T_2$ is the sink temperature.Given, $\eta_1 = 0.2 = 1 - \frac{T_2}{T_1} \implies \frac{T_2}{T_1} = 0.8 \implies T_2 = 0.8 T_1$ (Equation $1$).When $T_2$ is lowered by $45 	ext{ K}$, the new sink temperature is $T_2' = T_2 - 45$. The new efficiency is $\eta_2 = 0.5$.So, $0.5 = 1 - \frac{T_2 - 45}{T_1} \implies \frac{T_2 - 45}{T_1} = 0.5 \implies T_2 - 45 = 0.5 T_1$ (Equation $2$).Substitute $T_2 = 0.8 T_1$ from Equation $1$ into Equation $2$:$0.8 T_1 - 45 = 0.5 T_1$$0.3 T_1 = 45$$T_1 = \frac{45}{0.3} = 150 	ext{ K}$.Now, find $T_2$ using Equation $1$:$T_2 = 0.8 	imes 150 = 120 	ext{ K}$.Thus, the temperatures are $T_1 = 150 	ext{ K}$ and $T_2 = 120 	ext{ K}$.

$A$ Carnot engine operating between temperatures $T_1$ and $T_2$ has an efficiency of $0.2$. When $T_2$ is lowered by $45 \text{ K}$, its efficiency becomes $0.5$. The temperatures $T_1$ and $T_2$ are respectively:

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