Question

We want to condense 1 kg of water vapour at 100 c to a temperature of -5 c (Ice). How much heat should be removed?

Answer 1

Let us calculate the change of energy. First, the vapour should be transformed into water with temperature 100 C, so (see https://en.wikipedia.org/wiki/Enthalpy_of_vaporization)

$|\Delta Q_1| = L_wm_w = 2260000\,\mathrm{J/kg}\,\cdot1\,\mathrm{kg} = 2260000\,\mathrm{J}.$

Next, we should cool this water to the temperature 0 C, so

$|\Delta Q_2| = c_wm_w\Delta T = 4200\,\mathrm{J/kg/K}\cdot1\,\mathrm{kg}\cdot100\,\mathrm{K} = 420000\,\mathrm{J}.$

Next, we should transform the water into ice, so (see https://en.wikipedia.org/wiki/Enthalpy_of_fusion)

$|\Delta Q_3| = \lambda_i m_w =334000\,\mathrm{J/kg}\cdot1\,\mathrm{kg} = 334000\,\mathrm{J}.$

Then, we should cool this ice, so

$|\Delta Q_4| = c_im_w\Delta T_2 = 2110\,\mathrm{J/kg/K}\cdot1\,\mathrm{kg}\cdot5\,\mathrm{K} = 10550\,\mathrm{J}.$

Therefore, the total amount of heat to be removed is

$|\Delta Q| = |\Delta Q_1| +|\Delta Q_2| +|\Delta Q_3| +|\Delta Q_4| = 2260000 + 420000+334000+10550 = 3024550\,\mathrm{J} \approx 3\,\mathrm{MJ}.$