A Carnot engine whose efficiency is 40%, receives heat at 500 K. If the efficiency is to be 50%, the source temperature for the same exhaust temperature is

Options

A. 900 K
B. 600 K
C. 700 K
D. 800 K

Answer

B

Explanation

Efficiency of Carnot engine, `eta = 1 - (T_(2))/(T_(1))`
where `T_(1)` and `T_(2)` are the temperature of source and sink respectively.
`:. (T_(2))/(T_(1)) = 1 - eta = 1 - (40)/(100) = (60)/(100) = (3)/(5) (because eta = 40%)`
or `T_(2) = (3)/(2)T_(1) = (3)/(5) xx 500 K = 300 K` ...(i)
Let `T._(1)` be the temperature of the source for the same sink temperature.
`:. (T_(2))/(T._(1)) = 1 - eta. = 1 - (50)/(100) = (1)/(2) " " (because eta. = 50%)`
or `T._(1) = 2T_(2) = 2 xx 300 K = 600 K` (Using (i))