Question

How much heat energy, in kilojoules, is required to convert 72.0g of ice at -18 degrees celsius to water at 25 degrees celsius?

Answer 1

This is calculated in 5 stages.

1) To heat the ice from -18°C to 0°C (dT = 18°C)

Q = mcdT

         Q = heat ,

m = mass of water = 72 g

        c = specific heat of ice = 2.087 J/g/°C

        dT = temperature difference = 18°C

Q = 72 g x 2.087 J/g/°C x 18 °C = 2704.7 J

Q = 2704.7 J

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

Q = m x Enthalpy of fusion

    = 72 g x 333.6 J/g

    = 24019.2

Q = 24019.2

3) To heat the water from 0°C to 25°C

Q = mcdT where c = specific heat of water , dT = 25°C

   = 72 g x 4.184 J/g/°C x 25°C

= 7531.2 J

Q = 7531.2 J

Total heat required = 2704.7 J + 24019.2 J + 7531.2 J

                             = 34255.1 J

                            = 34.2 kJ

Therefore, 34.2 kJ kJ heat energy is required to heat convert 72.0g of ice at -18 degrees celsius to water at 25 degrees celsius.