A Carnot engine with efficiency 50% takes hea | vedclass

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(C) The efficiency 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 temperatur

Vedclass · Accepted Answer

$1000$

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(C) The efficiency 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.Initially, $\eta_1 = 50\% = 0.5$ and $T_1 = 600	ext{ K}$.$0.5 = 1 - \frac{T_2}{600} \implies \frac{T_2}{600} = 0.5 \implies T_2 = 300	ext{ K}$.The efficiency is increased by $20\%$, so the new efficiency $\eta_2 = 50\% + 20\% = 70\% = 0.7$.Using the same sink temperature $T_2 = 300	ext{ K}$, we find the new source temperature $T_1'$:$0.7 = 1 - \frac{300}{T_1'} \implies \frac{300}{T_1'} = 1 - 0.7 = 0.3$.$T_1' = \frac{300}{0.3} = 1000	ext{ K}$.

$A$ Carnot engine with efficiency $50\%$ takes heat from a source at $600\text{ K}$. To increase the efficiency by $20\%$, keeping the temperature of the sink the same, the new temperature of the source will be: (in $\text{ K}$)

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