What amount of heat must be given to ice taken at a temperature of -10 and weighing 2 kg
| December 16, 2020 | education
What amount of heat must be given to ice taken at a temperature of -10 and weighing 2 kg in order to melt it and heat the resulting water to a boil and then evaporate?
Q1 = m * C * (tpl-t1)
[Q] = J
Q2 = m * “lambda”
Q3 = m * C * (tк-tpl)
Q4 = m * L
Q = Q1 + Q2 + Q3 + Q4
Q1 = 2 * 2100 * 100 = 42000J
Q2 = 2 * 3.4 * 10 ^ 5 = 680,000J
Q3 = 2 * 4200 * (100-0) = 840000J
Q4 = 2 * 2.3 * 10 ^ 6 = 4600000J
Q = 4600000 + 840000 + 680000 + 42000 = 6162000J
Answer: 6162000J or 6.162MJ
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