The efficiency of a heat engine is defined as the ratio of the mechanical work done by the engine in one cycle to the heat absorbed from the high temperature source . `eta = (W)/(Q_(1)) = (Q_(1) - Q_(2))/(Q_(1))` Cornot devised an ideal engine which is based on a reversible cycle of four operations in succession: isothermal expansion , adiabatic expansion. isothermal compression and adiabatic compression.

For carnot cycle `(Q_(1))/(T_(1)) = (Q_(2))/(T_(2))`. Thus `eta = (Q_(1) - Q_(2))/(Q_(1)) = (T_(1) - T_(2))/(T_(1))` According to carnot theorem "No irreversible engine can have efficiency greater than carnot reversible engine working between same hot and cold reservoirs".
A carnot engine whose low temperature reservoir is at `7^(@)C` has an efficiency of `50%`. It is desired to increase the efficiency to `70%`. By how many degrees should the temperature of the high temperature reservoir be increased?

Options

A. `273K`
B. `(1120)/(3)K`
C. `140K`
D. None of these

Answer

B

Explanation

`eta= (T_(1) - T_(2)) /(T_(1)) = 0.5 implies T_(1) = 560K`
`eta' = 0.7 = (T'_(1) - T_(2))/(T'_(1)) = (T'_(1) - 280)/(T'_(1)) implies (2800)/(3)`
`implies T'_(1) - T_(1) = (1120)/(3)`