How much heat needs to be spent to turn 4 kg of ice taken at -30C into steam at 100C?

Q = Q1 + Q2 + Q3 + Q4.

Q1 (ice) = С1 * m * (tк1 – tн1), where С1 = 2100 J / (kg * К), m = 4 kg, tк1 = 0 ºС, tн1 = -30 ºС.

Q2 (ice) = λ * m, where λ = 34 * 104 J / kg.

Q3 (water) = C2 * m * (tк2 – tн2), where С2 = 4200 J / (К * kg), tк2 = 100 ºС, tн2 = 0 ºС.

Q4 (water) = L * m, where L = 2.3 * 106 J / kg.

Q = 2100 * 4 * (0 – (-30)) + 34 * 104 * 4 + 4200 * 4 * (100 – 0) + 2.3 * 106 * 4 = 252000 + 1360000 + 1680000 + 9200000 = 12492000 ≈ 12 , 5 MJ.

Answer: You need to spend 12.5 MJ of heat.




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