What amount of heat is needed to obtain 10 kg of 100-degree steam from water taken at 20 degrees.

Initial data: m (mass of water that was turned into steam) = 10 kg; tpap (temperature of the obtained steam) = 100 ºС; t0 (initial water temperature) = 20 ºС.

Constants: by condition C (specific heat) = 4.2 kJ / (kg * ºС) = 4.2 * 10 ^ 3 J / (kg * ºС); L (specific heat of vaporization) = 2300 kJ / kg = 2300 * 10 ^ 3 J / kg.

Required amount of heat: Q = Q1 + Q2 = C * m * (t steam – t0) + L * m = 4.2 * 10 ^ 3 * 10 * (100 – 20) + 2300 * 10 ^ 3 * 10 = 26,360,000 J = 26.36 MJ.

Answer: You need 26.36 MJ of heat.



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