The efficiency of carnot heat engine is 25%. When the temperature of the source alone is raised by 100K, the efficiency becomes 50%. Find the temperature of the source and the sink.

Explanation

Given efficiency `n_1 = 25/100 = 0.25 `
` T_1^1 = (T_1 + 100 ), n_2 = 50/100 = 0.50 `
` T_1 = ? , T_2 = ? `
wkt `n = 1 - T_2/T_1`
i.e., `0.25 = 1 - T_2/T_1 = (T_1 - T_2)/(T_1)`
i.e., ` 0.25 T_1 = T_1 - T_2`
` therefore T_2 = T_1 - 0.25 T_1 = 0.75 T_1`
Also `0.50 = (1 - T_2)/(T_1 - 100)`
i.e., ` 0.50 = (T_1 + 100 - T_2)/(T_1 + 100)`
` 0.50 T_1 + 50 = T_1 + 100 - (0.75 T_1)`
` 0.50 T_1 + 50 = 0.25 T_1 + 100`
` 0.25T_1 = 50`
` T_1 = (50)/(0.25) = 5000/25 = 200 K`
` therefore T_1 = 200 K " and " T_1 = 3/4 xx 200 = 175 K`