Question

A carnot’s engine whose temperature of the source is 127ºC takes 200 calories of heat at this temperature and rejects 100 calories of heat to the sink. What is the temperature of the sink? Also calculate the 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%

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