Question

A particular refrigerator cools by evaporating liquefied dichlorodifluoromethane, CCl2F2. How many kilograms of this liquid must be evaporated to freeze a tray of water at 0°C to ice at 0°C? The mass of the water is 571 g, the heat of fusion of ice is 6.02 kJ/mol, and the heat of vaporization of dichlorodifluoromethane is 17.4 kJ/mol.

Answer 1

Answer:

1.33 kilograms of this liquid must be evaporated to freeze a tray of water at 0°C to ice at 0°C.

Explanation:

Let the mass of liquefied dichlorodifluoromethane be x.

Mass of water to freeze = 571 g

Moles of water =$\frac{571 g}{18 g/mol}=31.7 mol$

Heat of fusion of ice = 6.02 kJ/mol

Heat lost when 1 mole of water freeze's = -6.02kJ/mol

Heat lost when 31.7 moles of water freeze's: Q

$Q=-6.02 kJ/mol\times 31.7 mol=-191. kJ$

Heat required to evaporate x amount of liquefied dichlorodifluoromethane: Q'

Q'= -(Q) = 191. kJ

Moles of liquefied dichlorodifluoromethane =$\frac{x}{121 g/mol}$

Heat of vaporization of dichlorodifluoromethane = 17.4 kJ/mol

$Q'=17.4 kJ/mol\times \frac{x}{121 g/mol}$

$191. kJ=17.4 kJ/mol\times \frac{x}{121 g/mol}$

Solving for x:

x = 1328.2 g = 1.33 kg

1 g = 0.001 kg

1.33 kilograms of this liquid must be evaporated to freeze a tray of water at 0°C to ice at 0°C.