A piece of steel heated to 600 degrees is thrown into a vessel containing 2.8 kg of water at 20 degrees. the water heats up to 100 degrees, and 30 g of it evaporates. determine the mass of the piece of steel.
Initial data: m1 (mass of water) = 2.8 kg; m2 (mass of evaporated water) = 30 g = 0.03 kg; t0 (initial water temperature) = 20 ºС; t1 (end water temperature. steel) = 100 ºС; t2 (initial temperature of a piece of steel) = 600 ºС.
Constants: water C1 = 4200 J / (kg * K); L = 2.3 * 10 ^ 6 J / kg; steel C2 = 500 J / (kg * K).
Q1 + Q2 = Q3: C1 * m1 * (t1 – t0) + L * m2 = C2 * m3 * (t2 – t1).
m3 = (C1 * m1 * (t1 – t0) + L * m2) / (C2 * (t2 – t1)) = (4200 * 2.8 * (100 – 20) + 2.3 * 10 ^ 6 * 0 , 03) / (500 * (600 – 100)) ≈ 4 kg.
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