How much heat must be spent to melt and heat ice to a temperature of 20 ° C, if the mass of water is 1 kg?

Given: m (obtained mass of water) = 1 kg; tк (final temperature of the formed water) = 20 ºС; we assume that the ice was taken at the melting temperature (0 ºС).

Constants: λ (specific heat of melting of ice) = 34 * 104 J / kg; C (specific heat of water) = 4200 J / (kg * ºС).

1) Melting ice: Q1 = λ * m = 34 * 10 ^ 4 * 1 = 34 * 10 ^ 4 J.

2) Heating of the obtained water: Q2 = C * m * (tc – 0) = 4200 * 1 * (20 – 0) = 84000 J.

3) The expended heat: Q = Q1 + Q2 = 34 * 10 ^ 4 + 84000 = 424000 J (424 kJ).