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

=

the focus of lesson 1 is newton's first law of motion - sometimes referred to as the law of inertia. newton's first law of motion is often stated as. an object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.

explanation: