Determine the amount of heat needed to turn 50 g of ice at -6 ° C to water at 0 ° C.

Problem data: m (ice mass) = 50 g (0.05 kg); t0 (temperature at which the ice was located) = -6 ºС; t (temperature of the obtained water) = 0 ºС (at this temperature, ice melts).

Reference values: Cl (ice heat capacity) = 2100 J / (kg * ºС); λ (heat of melting of ice) = 34 * 10 ^ 4 J / kg.

1) Heating of ice from -6 ºС to 0 ºС: Q1 = Cl * m * (t – t0) = 2100 * 0.05 * (0 – (-6)) = 630 J.

2) Melting: Q2 = λ * m = 34 * 10 ^ 4 * 0.05 = 17000 J.

3) Required heat: Q = 630 + 17000 = 17630 J.