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. 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:0.061

Step-by-step explanation:

Given

`T_C=300 k`

Temperature of soup
`T_H=340 K`

heat capacity of soup
`c_v=33 J/K`

Here Temperature of soup is constantly decreasing

suppose T is the temperature of soup at any instant

efficiency is given by

`\eta =(dW)/(Q)=1-(T_C)/(T)`

`dW=Q(1-(T_C)/(T))`

`dW=c_v(1-(T_C)/(T))dT`

integrating From
`T_H` to
`T_C`

`\int dW=\int_(T_C)^(T_H)c_v(1-(T_C)/(T))dT`

`W=\int_(T_C)^(T_H)33\cdot (1-(300)/(T))dT`

`W=c_v\left [ T-T_C\ln T\right ]_(T_H)^(T_C)`

`W=c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln (T_C)/(T_H)\right )\right ]`

Now heat lost by soup is given by

`Q=c_v(T_C-T_H)`

Fraction of the total heat that is lost by the soup can be turned is given by

`=(W)/(Q)`

`=(c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln (T_C)/(T_H)\right )\right ])/(c_v(T_C-T_H))`

`=(T_C-T_H-T_C\ln ((T_C)/(T_H)))/(T_C-T_H)`

`=(300-340-300\ln ((300)/(340)))/(300-340)`

`=(-40+37.548)/(-40)`

`=0.061`