0.4-L glass of water at 20°C is to be cooled with ice to 5°C. The density of water is 1 kg/L, and the specific heat of water at room temperature is c = 4.18 kJ/kg·°C. The specific heat of ice at about 0°C is c = 2.11 kJ/kg·°C. The melting temperature and the heat of fusion of ice at 1 atm are 0°C and 333.7 kJ/kg. Determine how much ice needs to be added to the water, in grams, if the ice is at 0°C. Also determine how much water would be needed if the cooling is to be done with cold water at 0°C. The melting temperature and the heat of fusion of ice at atmospheric pressure are 0°C and 333.7 kJ/kg, respectively, and the density of water is 1 kg/L.

70 g

Explanation:

ice at zero degree will melt and cool down water from 20 degree to 5 degree

heat gained by ice in melting

= mass x heat of fusion

= m x 333.7 x 10³ J

Heat gained by water at zero to water at 5 degree

= mass x specific heat x rise in temperature

= m x 4.18 x 10³ x 5

Total heat gained

= m x 333.7 x 10³ + m x 4.18 x 10³ x 5

= 354.6 x 10³ m J

mass of water =.4 x 1

=.4 kg

heat lost by hot water at 20 degree to 5 degree

=.4 x 4.18 x 10³ x (20-5)

=25.08 X 10³ J

Heat lost = heat gained

354.6 x 10³ m = 25.08 X 10³

m = 25.08 / 354.6

=.070 kg

70 g

b)

Let the water required be m kg

heat gained by water

= m x 4.18 x 10³ x 15

m x 4.18 x 10³ x 15 = 25.08 X 10³

m = 25.08 / (4.18 x 15)

=.4 kg

= 400 gm

=