Question

A firebox is at 650 K, and the ambient temperature is 250 K. The efficiency of a Carnot engine doing 146 J of work as it transports energy between these constant-temperature baths is 61.5%. The Carnot engine must take in energy 146 J/0.62 = 237.3 J from the hot reservoir and must put out 91.3 J of energy by heat into the environment. To follow Carnot's reasoning, suppose some other heat engine S could have efficiency 70.0%.(a) Find the energy input and exhaust energy output of engine S as it does 146 J of work

Answer 1

Qce = 91.25 J

Step-by-step explanation:

Given

W = 146 J , Tc = 250 k , Th = 650 K

Qhe = W / [ 1 - ( Tc / Th ) ]

Tc / Th = 250 / 650 = 0.384615384

Qhe = 146 J / 0.615384615

Qhe = 237.25 J

So replacing to find the energy exhaust energy

Qce = Qhe - W

Qce = 237.25 J - 146 J

Qce = 91.25 J