What amount of heat must be given to ice taken at a temperature of -10 with a mass of 2 kg

univerkov | April 8, 2021 | education

What amount of heat must be given to ice taken at a temperature of -10 with a mass of 2 kg in order to melt it, and heat the resulting water to a boil, and then evaporate?

Given: m (mass) = 2 kg; tн (initial temp.) = -10 ºС.

Constants: Cl = 2100 J / (kg * K); tmelt = 0 ºС; λ = 34 * 10 ^ 4 J / kg; Cw = 4200 J / (kg * K); tboil = 100 ºС; L = 23 * 10 ^ 5 J / kg.

1) Ice heating: Q1 = Cl * m * (tm – tn) = 2100 * 2 * (0 – (-10)) = 42000 J.

2) Melting ice: Q2 = λ * m = 34 * 10 ^ 4 * 2 = 680,000 J.

3) Heating water: Q3 = Sv * m * (tboil – tmelt) = 4200 * 2 * (100 – 0) = 840000 J.

4) Evaporation: Q4 = L * m = 23 * 10 ^ 5 * 2 = 4600000 J

5) Q = 42000 + 680000 + 840000 + 4600000 = 6.162 MJ.

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