The temperatures `T_(1) and T(2)` of two heat reservoirs in an ideal carnot engine are `1500^(@)C and 500^(@)C`. Which of these (a) increasing `T_(1) by 100^(@)C` or (b) decreasing `T_(2) by 100^(@)C` would result in greater improvement of the efficiency of the engine?

Explanation

`T_(1) = 1500^(@)C = 1500+ 273 = 1773 K` and `T_(2)= 500^(@)@C = 500 + 273 = 773K`.
The efficieny of a carnot's engine `eta = 1 - (T_(2))/(T_(1)) , 1- (773)/(1773)` When the temperature of the source is increased by `100^(@)C`, keeping `T_(2)` unchanged , the new temperature of the source is `T'_(1) = 1500+100 = 1600^(@)C= 1873K`.
The efficiency becomes `eta' = 1 - (T_(2))/(T_(1)), = 1 - (773)/(1873) = 0.59`
On the other hand, if the temperature of the sink is decreased by `100^(@)C` , keeping `T_(1)` unchanged , the new temperature of the sink is `T'_(2) = 500-100 = 400^(@)C = 673K`. The efficiency now becomes
` eta'' = 1 - (T'_(2))/(T_(1)) = 1 - (673)/(1773) = 0.62`
Since `eta''` is greater than `eta'`, decreasing the temperature of the sink by `100^(@)C` results in a greater efficiency than increasing the temperature of the source by `100^(@)C`.