Question

A Carnot machine operates with 25% efficiency, whose heat rejection reservoir temperature is 300K. Determine the temperature at which the machine absorbs heat.

Answer 1

Answer:

The temperature at which observed heat is 400 K

Explanation:

Given data:

rejection reservoir temperature at exit $T_{L}$ is 300 k

the efficiency of a engine is  η = 25%

we know that efficiency of Carnot is given as

$\eta = (1-\frac{T_{L}}{T_{H}})*100$

Putting all value to obtained  temperature at which observed heat

$0.25 = (1-\frac{300}{T_{H}})$

$T_{H}$ = 400 K