What amount of heat will be required to turn into water and ice with a mass of 2 kg taken at 0s and when the resulting water is heated up to 30s
Given: m (ice mass) = 2 kg; t0 (initial ice temperature) = 0 ºС (melting point); t (temperature that the water must reach) = 30 ºС.
Constants: λ (specific heat of melting of ice) = 3.4 * 10 ^ 5 J / kg; Sv (specific heat capacity of water) = 4200 J / (kg * ºС).
1) The transition of ice to water: Q1 = λ * m = 3.4 * 10 ^ 5 * 2 = 6.8 * 10 ^ 5 J.
2) Heating of the obtained water: Q2 = C * m * (t – t0) = 4200 * 2 * (30 – 0) = 252000 J.
3) The required amount of heat: Q = 6.8 * 10 ^ 5 + 252000 = 932000 J (932 kJ).
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