Question

A steam engine (assume a Carnot engine ) has an efficiency of 50.8%. If the waste heat has a temperature of 72.4 ◦C, what is the temperature of the boiler? Answer in units of ◦C.

Answer 1

Answer:

The temperature of the boiler is approximately 147.1 °C

Explanation:

A Carnot engine is an ideal engine that has the highest efficiency among all the engines because the second law of thermodynamics.That efficiency $\eta$  is:

$\eta=1-\frac{T_{c}}{T_{h}}$

with $T_{h}$ the temperature of the hot reservoir (the boiler temperature) and $T_{c}$ the temperature of the cold reservoir (the steam temperature). Solving for $T_{h}$:

$T_{h}=\frac{-T_{c}}{\eta-1}=\frac{-72.4}{0.508-1}\approx147.1\,C$

Answer 2

Answer:

429.033 °C

Explanation:

We use the equation $n=\frac{Th-Tc}{Th}$ to find the efficiency. So we'll plug in our values.

$0.508=\frac{Th-345.4}{Th}$ ,  345.4 is the waste heat converted to Kelvin

Solve for Th and you get Th=702.033. Convert back to Celcius to get:

Th=429.033 °C