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प्रश्न
The efficiency of a Carnot cycle is 75%. If the temperature of the hot reservoir is 727°C, calculate the temperature of the cold reservoir.
उत्तर
Data: η = 75% = 0.75, TH = (273 + 727) K = 1000 K
η = `1-"T"_"C"/"T"_"H"`
∴ `"T"_"C"/"T"_"H"` = 1 − η
∴ TC = TH (1 − η) = 1000 K (1 − 0.75)
= 250 K = (250 − 273)°C
= − 23°C
This is the temperature of the cold reservoir.
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