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Consider a Carnot's cycle operating between $T_1=500 \mathrm{~K}$ and $T_2=300 \mathrm{~K}$ producing $1 \mathrm{~kJ}$ of mechanical work per cycle. Find the heat transferred to the engine by the reservoirs.
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As we know that
The efficiency of a Carnot's engine is $\eta=1-\frac{T_2}{T_1}$ where, $T_2$ is temperature of the sink $T_1$ is temperature of the source.
As given that temperature of the source or reservoir $T_1=$ $500 \mathrm{~K}$
Temperature of the sink $T_2=300 \mathrm{~K}$
Work done per cycle $W=1 \mathrm{~kJ}=1000 \mathrm{~J}$
heat transferred to the engine per cycle $Q=$ ?
So, efficiency of a Carnot engine
$(\eta)=1-\frac{\mathrm{T}_2}{\mathrm{~T}_1}=1-\frac{300}{500}=\frac{200}{500}=\frac{2}{5}=0.4$
and $\eta=\frac{W \text { (output work) }}{Q \text { (input work) }}$
Input work $(E)=\frac{W}{\eta}=\frac{1000}{0.4}=2500 \mathrm{~J}$
Output $Q=18500 \mathrm{~J}$
The efficiency of a Carnot's engine is $\eta=1-\frac{T_2}{T_1}$ where, $T_2$ is temperature of the sink $T_1$ is temperature of the source.
As given that temperature of the source or reservoir $T_1=$ $500 \mathrm{~K}$
Temperature of the sink $T_2=300 \mathrm{~K}$
Work done per cycle $W=1 \mathrm{~kJ}=1000 \mathrm{~J}$
heat transferred to the engine per cycle $Q=$ ?
So, efficiency of a Carnot engine
$(\eta)=1-\frac{\mathrm{T}_2}{\mathrm{~T}_1}=1-\frac{300}{500}=\frac{200}{500}=\frac{2}{5}=0.4$
and $\eta=\frac{W \text { (output work) }}{Q \text { (input work) }}$
Input work $(E)=\frac{W}{\eta}=\frac{1000}{0.4}=2500 \mathrm{~J}$
Output $Q=18500 \mathrm{~J}$
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