Question

How many grams of water can be cooled from 42 ∘c to 20 ∘c by the evaporation of 51 g of water? (the heat of vaporization of water in this temperature range is 2.4 kj/g. the specific heat of water is 4.18 j/g⋅k.)?

Answer 1

Let us first calculate heat obtained by the evaporation of 51 g of water.

Given, heat of vaporization of water = 2.4 kJ/ g

∴ Heat obtained by evaporation of 51 g of water = 2.4 × 51 = 122.4 kJ

This is the heat energy available that can be used to cool water from 42°C to 20°C.

Specific heat of water is given by,

$C=\frac{Q}{mdt}$

Here,

C is the specific heat of water = 4.18 J/gK

Q is the amount of heat = 122400 J

m is the mass of the water that can be cooled.

dt is the change in temperature= 42°C ₋ 20°C = 22°C ( The numerical value will be the same if Kelvin unit is used.)

Substituting the values we get,

$4.18=\frac{122400}{m*22}$

m = 1331 g

1331 grams of water can be cooled from 42°C to 20°C by evaporation of 51 g of water.