How much energy is needed to convert 4 kg water taken at a temperature of 50 degrees into steam?
| February 10, 2021 | education
m = 4 kg.
C = 4200 J / kg * ° C.
t1 = 50 ° C.
t2 = 100 ° C.
q = 2.3 * 10 ^ 6 J / kg.
Q -?
The required amount of heat Q will be the sum: Q = Q1 + Q2, where Q1 is the amount of heat required to heat water from temperature t1 to t2, Q2 is the amount of heat required to evaporate water at boiling point.
Q1 = C * m * (t2 – t1).
Q1 = 4200 J / kg * ° C * 4 kg * (100 ° C – 50 ° C) = 840,000 J.
Q2 = q * m.
Q2 = 2.3 * 10 ^ 6 J / kg * 4 kg = 9200000 J.
Q = 840,000 J + 9,200,000 J = 10,040,000 J.
Answer: to evaporate all the water, you need Q = 10040000 J of thermal energy.
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