Question

11.

Consider the following specific heat capacities:

H2O (s) = 2.09 J/g·°C

H2O (l) = 4.18 J/g·°C

H2O (g) = 2.03 J/g·°C

The heat of fusion for water is 334 J/g and its heat of vaporization is 2260 J/g. Calculate the amount of heat required to convert 93 g of ice at -36°C completely to liquid water at 35°C.

52 kJ

21 kJ

7 kJ

38 kJ

Answer 1

This is calculated in 3 steps.

1) To heat the ice from -36°C to 0°C (dT = 20°C)

Q = mcdT

         Q = heat ,

m = mass of ice = 93 g

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

        dT = temperature difference = 36°C

Q = 93 g x 2.09 J/g/°C x 36°C = 6997.32 J

Q = 6997.32 J

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

Q = m x Enthalpy of fusion of water

    = 93 g x 334 J/g

    = 31062 J

Q = 31062 J

3) To heat the water from 0°C to boiling at 35°C

Q = mcdT where c = specific heat of water ,dT = 100°C

   = 93 g x 4.18 J/g/°C x 35°C

= 13605.9 J

Q = 13605.9 J

Therefore,

Total amount of heat required = 6997.32 J + 31062 J + 13605.9 J

= 51665 J

= 52 kJ

Hence,

Ans = 52 kJ