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A Carnot cycle has an efficiency of `40%` . Its low temperature resrvoir is at `7^(@)C` . What is the temperature of source ? |
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Answer» Correct Answer - `193.66^(@)C` `eta= (T_(2) - T_(1))/(T_(2))` `0.4 = (T_(2)- 280)/(T_(2))` `T_(2)= 466.67 K = 193.66 ^(@)` |
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