The efficiency of Carnot's enegine is `50%`. The temperature of its sink is `7^(@)C`. To increase its efficiency to `70%`. What is the increase in temperature of the source?

Answer

`373.3K`

Explanation

Efficiency in first state `eta = 50% = 1//2`
`T_(2) = 293 +7 = 280K`
Formula `eta = 1 - (T_(2))/(T_(1)) , (1)/(2) = 1 - (280)/(T_(1)) hArr (280)/(T_(1)) = (1)/(2)`
or `T_(1) = 560^(@)K` (temperature of source)
In the second state (i) `(70)/(100) = 1 - (280)/(T_(1))`
`:. T_(1) = (2800)/(3) = 9.33.3K`
`:.` Increase in source temperature `= (933.3 - 560)`
`= 373.3 K`