Question

Calculate the energy in the form of heat (in kJ) required to change 75.0 g of liquid water at 27.0 °C to ice at –20.0 °C. Assume that no energy in the form of heat is transferred to the environment. (Heat of fusion = 333 J/g; heat of vaporization = 2256 J/g; specific heat capacities: ice = 2.06 J/g·K, liquid water = 4.184 J/g·K)

Answer 1

Explanation:

The given data is as follows.

          mass, m = 75 g

      $T_{1} = 0^{o}C$

      $T_{2} = 27^{o}C$

      Specific heat of water = 4.18

First, we will calculate the heat required for water is as follows.

            q = $m C \times (T_{1} - T_{2})$

               = $75 g \times 4.18 J/g^{o}C \times (0 - 27)^{o}C$

               = 8464.5 J/mol

               = 8.46 kJ ......... (1)

Also, it is given that $T_{3} = -20^{o}C$ = (20 + 273) K = 293 K and specific heat of ice is 2.108 kJ/kg K.

Now, we will calculate the heat of fusion as follows.

        q = $mC \times (T_{3} - T_{1})$

           = $0.075 kg \times 2.108 kJ/kg K \times (-293 - 0) K$

           = -46.32 kJ ......... (2)

Now, adding both equations (1) and (2) as follows.

               8.46 kJ - 46.32 kJ

             = -37.86 kJ

Therefore, we can conclude that energy in the form of heat (in kJ) required to change 75.0 g of liquid water at $27.0^{o}C$ to ice at $-20.0^{o}C$ is -37.86 kJ.