A Carnot's engine operates with an efficiency of `40%` with its sink at `27^(@)C`. By what amount should the temperature of the source be increased with an aim to increase the efficiency by `10%`

Options

A. `50 K`
B. `150 K`
C. `80 K`
D. `100 K`

Answer

D

Explanation

Let `T_(1)` be the initial temperature of source, then
Using `eta = 1-(T_2)/(T_1)`
We have `(40)/(100)=1-((273+27K))/(T_1)`
or `T_(1) = 500K`
For the efficiency to be 10% more, i.e. `50%` let `T_(1)^(')` be the new temperature of the sink.
then, `50/100 = 1-((273+27K))/(T_(1)^('))`
or `T_(1)^(') = 600K`
The requried increase in the temperature of the source ltbr. `T_(1)^(') - T = 600K-500K = 100K`.