A mixture of water and ice with a total weight of 1 kg is placed in hot water with a mass of 3 kg and a temperature of 60 ° C. If, as a result, a temperature of 30 C is set in the vessel, from the mixture there was ice with a mass of … g? The heat loss should be neglected, the specific heat of water is 4.2 kJ / kg * C, the specific heat of melting of ice is 330 kJ / kg.
Data: m1 (hot water mass) = 3 kg; t1 (hot water temperature) = 60 ºС; m2 (mass of a mixture of water and ice) = 1 kg; tр (equilibrium temperature) = 30 ºС.
Constants: by condition C (specific heat capacity of water) = 4.2 kJ / (kg * ºС) = 4200 J / (kg * ºС); λ (specific heat of melting of ice) = 330 kJ / kg = 330 * 10 ^ 3 J / kg.
1) Heat for heating ice: Q = C * m1 * (t1 – tp) + C * m2 * (tp – 0) = 4200 * 3 * (60 – 30) – 4200 * 1 * (30 – 0) = 252000 J.
2) The mass of ice in the mixture: m = Q / λ = 252000 / (330 * 10 ^ 3) = 0.764 kg.
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