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8614smith
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Homework Statement
Four identical blocks of copper, of mass 20Kg each, are 'perfectly' isolated from the environment. One is kept at a temperature of 573K and the other three at 288K. The heat capacity of copper is 375 J/K/Kg. Assume a carnot process is being used to exploit the temperature differences between the se 4 blocks of copper. How much work can maximally be extracted?
(Hint: the temperature of the copper blocks change continuously from initial to final temperature. Remember that a carnot process is reversible, the total entropy change is zero)
Homework Equations
1. [tex]T_{f}=T_{1}^{\frac{m_{1}}{m_{1}+m_{2}}}*T_{2}^{\frac{m_{2}}{m_{1}+m_{2}}}[/tex]
The Attempt at a Solution
"Treating it as 1 block of 573K at 20Kg and 1 block of 288K at 60Kg and using T1 = 573, T2 = 288 and m1 = 20, m2 = 60 i get;
[tex]T_{mix}=359.25K[/tex]
[tex]T_{f}=573^{\frac{20}{80}}*288^{\frac{60}{80}}[/tex]
[tex]T_{f}=342.04K[/tex]
now using [tex]\Delta{E}=mc\Delta{T}[/tex]
[tex]\Delta{E}=80*375*(359.25-342.04)=516300J[/tex]
[tex]\Delta{S}=\frac{\Delta{E}}{T}-\frac{\Delta{W}}{T}[/tex]
As there's no entropy change the delta S terms is zero which leaves,
[tex]\frac{\Delta{E}}{T}=\frac{\Delta{W}}{T}[/tex]
the T's cancel and your left with extractable work = 516300J"
Does this seem correct? I am not sure about the 'Tmix' part and the last part where delta E is converted to delta W.