Question
Download Solution PDFA Carnot engine rejects 30% of absorbed heat to a sink at 30°C. The temperature of the heat source is
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
The maximum efficiency of the engine is given as, \(\eta=1-\frac{T_L}{T_H}\)
And the actual efficiency of the heat engine is given as, \(\eta=\frac{W}{Q_A}\)
Calculation:
Given:
The heat absorbed by the engine is QA
And heat rejected to sink is QR = 0.30QA
Therefore, the work done by engine is, W = 0.70QA
Sink temperature, TL = 303 K
Therefore by equating both efficiencies, we will get,
\(\frac{W}{{{Q_A}}} = 0.7 = 1 - \frac{{303}}{T}\)
Or \(\frac{{303}}{T} = 0.3\)
Or T = 1010 K = 737°CLast updated on May 28, 2025
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