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ss19.2.9 In an ideal thermal engine, 300 J is generated with each kilojoule of energy absorbed...

ss19.2.9

In an ideal thermal engine, 300 J is generated with each kilojoule of energy absorbed by the warmer reservoir. The temperature of the colder reservoir is 280 K. What is the temperature of the warmer reservoir? (Carnot Process)

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Answer #1

Solution -

Efficiency = work / heat input = 300 / 1000 = 0.3

Efficiency = 1 - T2 / T1   

0.3 =   1 - T2 / T1   

T2 / T1 = 0.7

280  / T1 = 0.7

T1 =   280 / 0.7 = 400 K

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