How much heat needs to be spent to bring 7 kg water taken at a temperature of 0 degrees to a boil and then completely evaporate it?
m = 7 kg. t1 = 0 “C. t2 = 100” C. C = 4200 J / kg * “C. λ = 2.3 * 10 ^ 6 J / kg. Q -? The amount of heat Q for complete evaporation of water will be the sum: Q = Q1 + Q2, where Q1 is the amount of heat for heating water from t1 = 0 “C to t2 = 100” C, Q2 is the amount of heat for evaporation at the boiling point. Q1 = C * m * (t2 – t1). Q1 = 4200 J / kg * “C * 7 kg * (100 “C – 0” C) = 2940000 J. Q2 = λ * m. Q2 = 2.3 * 10 ^ 6 J / kg * 7 kg = 16100000 J. Q = 2940000 J + 16100000 J = 19040000 J = 1904 * 10 ^ 4 J.
Answer: to evaporate water, you need Q = 1904 * 10 ^ 4 J of thermal energy.
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