Question

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 4.80×10−2 kg . Throughout this problem use Lf=3.34×105J/kg for the heat of fusion for water.

Part A

During this time, how much work W is performed by the engine?

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Answer 1

Work-done in Carnot engine is given by:

W = Qc + Qh

Qc/Qh = -Tc/Th

We know that: for a heat engine Qc is negative and Qh is positive, So

Given that Qc = Heat required to melt the ice = -m*Lf

Qc = -4.80*10^-2*3.34*10^5 = -16032 J

Now

Qh = -Qc*(Th/Tc)

Th = Temperature of hot reservoir = 273 + 100 = 373 K

Tc = Temperature of cold reservoir = 273 + 0 = 273 K

Qh = -(-16032)*(373/273)

Qh = 21904.5 J

So, work performed by engine will be:

W = Qc + Qh

W = -16032 + 21904.5

W = 5872.5 J

W = 5.87*10^3 J