Question #193530

Suppose that a heat engine is connected to two energy reservoirs, one a pool of molten aluminum

(660°C) and the other a block of solid mercury (-38.9°C). The engine runs by freezing 1.00 g of

aluminum and melting 15.0 g of mercury during each cycle. The heat of fusion of aluminum is 3.97x105

J/kg; the heat of fusion of mercury is 1.18 x104 J/kg. What is the efficiency of this engine?


1
Expert's answer
2021-05-17T11:34:27-0400

The heat to melt 15.0g of Hg is Qc=mLf=∣Q c ​ ∣=mL f ​ = (15×103kg)(1.18×104J/kg)=177J(15×10 ^{−3} kg)(1.18×10 ^4 J/kg)=177J


The energy absorbed to freeze 1.00g of aluminum is

Qh=mLf=(103kg)(3.97×105J/kg)=397J|Q _h ​ ∣=mL_ f ​ =(10 ^{−3} kg)(3.97×10 ^5 J/kg)=397J

and the work output is

Weng=QhQc=220JW eng ​ =∣Q h ​ ∣−∣Q c ​ ∣=220J


e=WengQh​​=220J397J=0.554e={ W _{eng}\over ∣Q _ h ​ ∣} ​ ​ = {220J \over 397J ​ } =0.554 ,

or

The theoretical (Carnot) efficiency is


ThTcTh​​=933K243.1K933K=0.749=74.9%{T _ h ​ −T _c\over T_h} ​ ​ ={ 933K−243.1K \over 933K} ​ =0.749=74.9\%




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