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

=

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hi there!

out of the two, (b) precision would be more reasonable.   however, both a and b could be pretty reasonable.

-awesomerepublic   : )