Question

11. A Carnot engine has engine has efficiency of 80%. If its sink is at 127°C, then find the temperature of source.

Answer 1

Answer is 1727°C or 2000Kelvin

Explanation- just see the uploaded photo

Answer 2

$T_H=2000\ K$ is the temperature of the source.

Explanation:

Given:

• Carnot efficiency of an engine, $\eta=0.8$

• temperature of sink, $T_L=127+273=400\ K$

We know that the Carnot efficiency is given as:

$\eta=1-\frac{T_L}{T_H}$

where:

$0.8=1-\frac{400}{T_H}$

$T_H=2000\ K$ is the temperature of the source.

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TOPIC: Carnot efficiency of a heat engine

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