Question

Assuming the domiestic refrigerator as reversable 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 temperature of water 0° C, L = 80clag-1 )​

Answer 1

AnSwer :-

T₁ = 27 + 273 = 300K

T₂ = 0 + 273 = 273K

m = 1kg = 100g

L = 80cal g$^{-1}$

Heat to be removed Q₂ = mL

= 1000 × 80 cal

= 8 × 20⁴cal

from the relation

$\sf \dfrac{Q_1}{Q_2} = \dfrac{T_1}{T_2}$

$\sf {Q_1} = \dfrac{T_1}{T_2} \times Q_2 = \dfrac{300}{273} \times 8 \times {10}^{4}$

$\sf = 87912.1cal$

Energy required to be supplied

➠ W = Q₁ - Q₂

➠ W = (87912.1 - 80,000) cal

➠ W = 7912.1 cal

➠ W = 7912.1 × 4.2J

➠ W = 33230.8 J

$\:$

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