A Carnot engine takes 5000 kcal of heat from a reservoir at 727 ° C and gives heat to a sink at 127 ° C . The work done by the engine is

Step-by-step solution

Given, Q Heat = 5000   kcal = 5 × 10 6   cal T 2 Sink = 127 ° C = 127 + 273 K = 400   K T 1 Source = 727 ° C = 727 + 273 K = 1000   K The efficiency of the Carnot cycle is given by, η = 1 - T sink T source = W Q ⇒ 1 - 400 1000 = 6 10 = W 5000 ⇒ W = 3000   kcal = 3 × × 4 . 2 × 10 6   J = 12 . 6 × 10 6   J .