The efficiency of a Carnot engine is 50%. The temperature of the heat sink is `27^(@)C`. Find the temperature of the heat source.

Explanation

Temperature of the heat sink,
`T_(2) = 27 + 273 = 300 K`
Efficiency, `eta = 50% = 50/100 = 1/2`
If the temperature of the heat source is `T_(1)`.
`eta = 1 - (T_2)/(T_1)`
i.e., `T_(1) = (T_2)/(1 - eta) = (300)/(1 - 1/2)`
`= 600 K = (600 - 273)^(@)C = 327^(@)C`.