Water weighing 2 kg was heated from 50 degrees. From to a boil and turned into steam.
| April 26, 2021 | education
Water weighing 2 kg was heated from 50 degrees. From to a boil and turned into steam. Calculate the amount of heat received by the water.
m = 2 kg.
t1 = 50 ° C.
t2 = 100 ° C.
C = 4200 J / kg * ° C.
λ = 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 boiling point t2, Q2 is the amount of heat required to evaporate water at boiling point.
Q1 = C * m * (t2 – t1).
Q1 = 4200 J / kg * ° C * 2 kg * (100 ° C – 50 ° C) = 420,000 J.
Q2 = λ * m.
Q2 = 2.3 * 10: 6 J / kg * 2 kg = 4600000 J.
Q = 420,000 J + 4,600,000 J = 5020000 J.
Answer: for the transformation of water into steam, Q = 5020000 J of thermal energy is needed.
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