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A Carnot heat engine uses a hot reservoir consisting of a large amount of boiling water and a cold reservoir consisting of a large tub of ice and water. In 5 minutes of operation of the engine, the heat rejected by the engine melts a mass of ice equal to 3.30×10−2 kg . Throughout this problem use Lf=3.34×105J/kg for the heat of fusion for water.A) During this time, how much work W is performed by the engine?

Answer :

Answer:

[tex]W=4037.36\ J[/tex]

Explanation:

Given:

mass of ice melted, [tex]m=3.3\times10^{-2}\ kg[/tex]

time taken by the ice to melt, [tex]t=5\ min=300\ s[/tex]

latent heat of the ice, [tex]L=3.34\times 10^5\ J[/tex]

Now the heat rejected by the Carnot engine:

[tex]Q_R=m.L[/tex]

[tex]Q_R=0.033\times 3.34\times 10^5[/tex]

[tex]Q_R=11022\ J[/tex]

Since we have boiling water as hot reservoir so:

[tex]T_H=373\ K[/tex]

The cold reservoir is ice, so:

[tex]T_L=273\ K[/tex]

Now the efficiency:

[tex]\eta=1-\frac{T_L}{T_H}[/tex]

[tex]\eta=1-\frac{273}{373}[/tex]

[tex]\eta=26.81\%[/tex]

Now form the law of energy conservation:

Heat supplied:

[tex]Q_S-W=Q_R[/tex]

where:

[tex]Q_S=[/tex] heat supplied to the engine

[tex]Q_S-\eta\times Q_S=Q_R[/tex]

[tex]Q_S(1-\eta)=Q_R[/tex]

[tex]Q_S=\frac{11022}{1-0.2681}[/tex]

[tex]Q_S=15059.36\ J[/tex]

Now the work done:

[tex]W=Q_S-Q_R[/tex]

[tex]W=15059.36-11022[/tex]

[tex]W=4037.36\ J[/tex]

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