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A Carnot heat engine receives 600 kJ of heat per cycle from a source at 627°C and rejects heat to a sink at 27°C. The amount of heat rejected to the sink per cycle (rounded off to the nearest integer) in kJ is
1. 26
2. 200
3. 400
4. 574

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Correct Answer - Option 2 : 200

Concept:

The efficiency of the heat engine

\(η_{HE} = 1-\frac{Q_L}{Q_H}\)

Where QL is heat rejected to the sink, QH is heat received from the source

Carnot heat engine efficiency

\(η = 1-\frac{T_L}{T_H}\)

Where TL is sink temperature, TH is Source temperature

Calculation:

Given:

QH  = 600 kJ, TH = 627 °C = 627 + 273  = 900 K, TL = 27 °C + 273 = 300 K

Since the above engine is working on the carnot cycle, Therefore,

Carnot efficiency = Heat engine efficiency

η = ηHE

\(1-\frac{T_L}{T_H} = 1-\frac{Q_L}{Q_H}\)

\(\frac{300}{900} = \frac{Q_L}{600}\)

∴ QL = 200 kJ.

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