Question

Calculate the amount of heat that must be absorbed by 50.0 grams of ice at -12.0oC to convert it to water at 20.0oC. (specific heat capacity of ice = 2.09 J/g.K; specific heat capacity of liquid water = 4.18 J/g.K; specific heat capacity of water vapor = 1.84 J/g.K; heat of fusion of ice = 334 J/g

Answer 1

This is calculated in 3 stages.

1) To heat the ice from -12°C to 0°C (dT = 12°C =12K)

Q = mcdT

Q = heat ,

m = mass of water = 50 g

        c = specific heat capacity of ice = 2.09 J/g/°C

        dT = temperature difference = 12°C = 12 K

Q = 50g x 2.09 J/g/K x 12 K = 1254 J

Q = 1254 J

2) To melt the ice at 0°C to water at 0°C.

Q = m x heat of fusion of ice

    = 50 g x 334 J/g

    = 16700 J

Q = 16700 J

3) To heat the water from 0°C to 20°C

Q = mcdT where c = specific heat of capacity liquid water , dT = 20°C = 20 K

   = 50 g x 4.18 J/g/K x 20°C

= 4180 J

Q = 4180 J

Total heat to absobed = 1254 J + 16700 J + 4180 J

                                = 22134 J

Therefore,   22134 J heat must be absorbed by 50.0 grams of ice at -12.0^oC to convert it to water at 20.0^oC.