How much heat must be imparted to a piece of ice with a mass of 50 g at 0 ℃ in order to melt

How much heat must be imparted to a piece of ice with a mass of 50 g at 0 ℃ in order to melt it and heat the formed water to a boil.

Given: m (mass of a piece of ice) = 50 g (0.05 kg); t0 (temperature of a piece of ice) = 0 ºС.

Constants: tmelt (temperature of the beginning of melting) = 0 ºС; λ (specific heat of melting of ice) = 34 * 104 J / kg; С (specific heat capacity of water) = 4200 J / (kg * ºС); tboil (temperature of the beginning of boiling) = 100 ºС.

1) Melting ice: Q1 = λ * m = 34 * 10 ^ 4 * 0.05 = 17000 J.

2) Heating to boiling: Q2 = C * m * (tboil – tmelt) = 4200 * 0.05 * (100 – 0) = 21000 J.

3) Amount of heat: Q = Q1 + Q2 = 17000 + 21000 = 38000 J (38 kJ).