Question

Assuming the domestic refrigerator as reversible engine working between melting point of ice and the room temperature of 27°C, calculate the energy in Joule that must be supplied to freeze one kg of water. (Given melting point of ice = 0°C L = 80 cal g^-1)

Answer 1

Given :

T₁ = 27 + 273 = 300 K

T₂ = 0 + 273 = 273 K

m = 1 kg = 1000 g; L = 80 cal g^-1

Heat to be removed,

Q₂ = mL

= 1000 × 80 cal

= 8 × 10⁴ cal

From the relation,

\(\frac{Q_1}{Q_2}=\frac{T_1}{T_2},\)

\(Q_1=\frac{T_1}{T_2}\times Q_2\) 

\(=\frac{300}{273}\times 8\times10^4\) 

= 87912.1 cal

Energy required to be supplied,

W = Q₁ − Q₂

= (87912.1 – 80,000) cal

= 7912.1 cal 

= 7912.1 × 4.2 J

= 33230.8 J