Question

Say that you are in a large room at temperature TC = 300 K. Someone gives you a pot of hot soup at a temperature of TH = 340 K. You set the bowl up so that as it cools to room temperature the heat first flows through a Carnot Engine. The soup has Cv= (33 J/K). Assume that the volume of the soup does not change.
1. What fraction of the total heat QH that is lost by the soup can be turned into useable work by the engine?

Answer 1

Answer:

Explanation:

Heat energy given out by the soup

= C_v  x ( t₂ - t₁ )

= 33 x ( 340 - 300)

= 1320 J

This heat is given to Carnot engine . Efficiency of engine

= (340 - 300 ) / 340

= 40 / 340

2 / 17

This fraction of total heat given is converted into useable work by the engine.