How much heat must be spent to turn 2 kg of ice at -15 into steam at 120 and then cool to 100?
| July 1, 2021 | education
Q1 = 2 * 2100 * (0 +15) = 63,000 J – the amount of heat for heating ice
Q2 = 2 * 330,000 = 660,000 J – the amount of heat for melting ice
Q3 = 2 * 4200 * (100 – 0) = 840 000 J – the amount of heat for heating water
Q4 = 2 * 2 300 000 = 4 600 000 J – the amount of heat for evaporation of water
Q5 = 2 * 4225 * (120 – 100) = 169,000 J – the amount of heat for heating steam
Q6 = 2 * 4225 * (100 – 120) = – 169 000 J – the amount of heat for steam cooling
Q = Q1 + Q2 + Q3 + Q4 + Q5 + Q6
Q = 63,000 + 660,000 + 840,000 + 4,600,000 + 169,000 – 169,000 = 6,163,000 J = 6 MJ
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