Question

Calculate the least amount of work that must be done to freeze 1g of water at 0°C by means of refrigerator. Temperature of surroundings is 27°C. How much heat is passed into the surroundings?

Answer 1

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$T_1={27}^{o}C=300K$

$T_2=0^{o}C=273K$

$Q_2=mL=1\times 80=80cal$

$leastamountofwork,$

$\frac{Q_2}{WD}=\frac{T_{\frac{2}{}}}{T_1-T_2}$

$WD=\frac{(T_1-T_2)}{T_2}{Q}_2=\frac{(300-273)80}{273}=7.91cal$

$Q_1=Q_2+W=80+7.91=87.91cal$