750 cm3 of ice was made in the refrigerator at a temperature of -5 degrees Celsius. how much heat was removed from the water and ice, if the initial temperature of the water was 15 degrees Celsius
Initial data: V (ice volume) = 750 cm ^ 3 = 750 * 10 ^ -6 m ^ 3; t (ice temperature) = -5 ºС; t0 (initial water temperature) = 15 ºС.
Constants: ρ (ice density) = 900 kg / m ^ 3; C1 (specific heat capacity of ice) = 2100 J / (kg * K); C2 (specific heat capacity of water) = 4200 J / (kg * K); λ (specific heat of melting of ice) = 330 * 10 ^ 3 J / kg.
Ice (water) mass: m = ρ * V = 900 * 750 * 10 ^ -6 = 0.675 kg.
Q = Q1 + Q2 + Q3 = C1 * m * (-t) + m * λ + C2 * m * t0 = 2100 * 0.675 * 5 + 330 * 10 ^ 3 * 0.675 + 4200 * 0.675 * 15 = 272 362, 5 J.
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