Question

A carnot engine has efficiency of 50% when its sink is at 227^(@)C. What is the temperature of source

Answer 1

Answer:

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Answer 2

Answer: The temperature of the source is 727 °C

Explanation:

The efficiency of a carnot engine is given by the formula :

η =  $1 - \frac{T_L}{T_H}$

η  ⇒ Efficiency

$T_L$ ⇒ Temperature of sink in Kelvin

$T_H$ ⇒ Temperature of source in Kelvin

Given,

Temperature of sink = 227 °C

Temperature of sink = 227 + 273 Kelvin

Temperature of sink = 500 Kelvin

∴  $\frac{50}{100} = 1 - \frac{500}{T_H}$

⇒  $\frac{500}{T_H} = 1 - \frac{50}{100}$

⇒  $\frac{500}{T_H} = \frac{50}{100}$

⇒  $T_H = 1000$

The temperature of source is 1000 Kelvin

Temperature of source in °C = 1000 - 273

Temperature of source in °C = 727

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