Question

the temperature of source of a carnot heat engine is 127°C. It takes 500 Calories of heat from the source and reject 400 calorie to the sink per cycle. Calculate the temperature of the sink and efficiency of the engine.

Answer 1

The temperature of the sink will be 47°C and the efficiency will be 20%

Explanation :

Q₁ = 500

Q₂ = 400

T₁ = 127 °C = 273 + 127 = 400 k

we know that

Q₂/Q₁ = T₂/T₁

=> T₂ = Q₂T₁/Q₁ = 400 x 400 / 500 = 320 k

= 320 - 273 = 47 °C

Hence the temperature of the sink is 47 °C

Since the engine takes 500 cal of heat and reject 400 cal

so energy output = 500 - 400 = 100 cal

we know that. efficiency is given as

η = output / input

= 100/500 = 0.2

= 20%

Hence the efficiency of the engine will be 20%