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".
Efficiency of a carnot's cycle change from `(1)/(6)` to `(1)/(3)` when source temperature is raised by `100K`. The temperature of the sink is-

Options

A. `(1000)/(3)K`
B. `(500)/(3)K`
C. `250K`
D. `100K`

Answer

A

Explanation

`eta= 1 - (T_(2))/(T_(1)) = (1)/(6) , eta'= 1 - (T_(2))/(T_(1) + 100) = (1)/(3) implies T_(2) = (1000)/(3)K`