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A Carnot engine operates between two reservoirs at temperatures of $\mathrm{500 \mathrm{~K}}$ and $\mathrm{300 \mathrm{~K}}$ . It absorbs $\mathrm{800 \mathrm{~J}}$ of heat from the hot reservoir in each cycle. Calculate the work done by the engine, the heat rejected to the cold reservoir, and the efficiency of the engine in each cycle.
Option: 1
$\mathrm{320 \mathrm{~J}, 480 \mathrm{~J}}$

Option: 2
$\mathrm{720 \mathrm{~J}, 320 \mathrm{~J}}$

Option: 3
$\mathrm{480 \mathrm{~J}, 320 \mathrm{~J}}$

Option: 4
$\mathrm{800 \mathrm{~J}, 200 \mathrm{~J}}$

Answers (1)

best_answer

Given: Hot reservoir temperature $\mathrm{\left(T_h\right)=500 \mathrm{~K}}$

Cold reservoir temperature $\mathrm{\left(T_c\right)=300 \mathrm{~K}}$

Heat absorbed $\mathrm{\left(Q_h\right)=800 \mathrm{~J}}$

Step 1: Calculate the work done by the engine:

The maximum efficiency of a Carnot engine is given by the equation:

$\mathrm{ \operatorname{Efficiency}(\eta)=1-\frac{T_c}{T_h} }$

So, $\mathrm{\eta=1-\frac{300 \mathrm{~K}}{500 \mathrm{~K}}=0.4}$

The work done (W) by the engine is given by:

$\mathrm{ \begin{gathered} W=\eta \cdot Q_h \\ W=0.4 \cdot 800 \mathrm{~J}=320 \mathrm{~J} \end{gathered} }$

Step 2: Calculate the heat rejected to the cold reservoir:

Since the engine operates in a cycle, the heat rejected $\mathrm{\left(Q_c\right)}$ is equal to the heat absorbed $\mathrm{\left(Q_h\right)}$ minus the work done (W) :

$\mathrm{ \begin{gathered} Q_c=Q_h-W \\ Q_c=800 \mathrm{~J}-320 \mathrm{~J}=480 \mathrm{~J} \end{gathered} }$

Posted by

Pankaj Sanodiya

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