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Heat Engine Efficiency and Heat Output

A heat engine absorbs \mathrm{3000 J} of heat from a high-temperature reservoir and releases \mathrm{1500 J} of heat to a low-temperature reservoir during each cycle. Calculate the efficiency of the heat engine and the work done by the engine during each cycle.

Option: 1

\mathrm{1500 J}


Option: 2

\mathrm{1600 J}


Option: 3

\mathrm{1800 J}


Option: 4

\mathrm{2000 J}


Answers (1)

best_answer

The efficiency (\eta) of a heat engine is given by the formula:

\mathrm{ \eta=\frac{W}{Q_h}=1-\frac{Q_c}{Q_h} }

Given:

\mathrm{ Q_h=3000 \mathrm{~J} \text { (heat absorbed from high-temperature reservoir) } }

\mathrm{ Q_c=1500 \mathrm{~J} \text { (heat released to low-temperature reservoir) } }

Calculate the efficiency:
\mathrm{ \begin{aligned} & \eta=1-\frac{Q_c}{Q_h} \\ & \eta=1-\frac{1500 \mathrm{~J}}{3000 \mathrm{~J}} \\ & \eta=0.5 \end{aligned} }
The work \mathrm{(W)} done by the engine can be calculated using the work-efficiency relationship:

\mathrm{ W=\eta \cdot Q_h }

Substitute the values:

\mathrm{ \begin{aligned} & W=(0.5) \cdot 3000 \mathrm{~J} \\ & W=1500 \mathrm{~J} \end{aligned} }

Therefore, the correct option is 1.

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