How much energy is spent on heating water with a mass of 750 g from 20 to 100 and the subsequent formation of steam with a mass of 250 g? The specific heat capacity of water is 4200 J, the specific heat of vaporization of water is 2.3 MJ / kg
mw = 750 g = 0.75 kg.
C = 4200 J / kg * ° C.
t1 = 20 ° C.
t2 = 100 ° C.
mp = 250 g = 0.25 kg.
λ = 2.3 MJ / kg = 2.3 * 10 ^ 6 J / kg.
The required amount of thermal energy is expressed by the formula: Q = Q1 + Q2, where Q1 is the required amount of thermal energy only for heating water from temperature t1 to boiling point t2, Q2 is the amount of thermal energy required to evaporate water at boiling point.
Q1 = C * mw * (t2 – t1).
Q1 = 4200 J / kg * ° C * 0.75 kg * (100 ° C – 20 ° C) = 252000 J.
Q2 = λ * mp.
Q2 = 2.3 * 10 ^ 6 J / kg * 0.25 kg = 9200000 J.
Q = 252000 J + 9200000 J = 9452000 J.
Answer: to convert water into steam, you need Q = 9452000 J of thermal energy.
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