How much energy is required for 400g of ice at 0 degrees to turn into steam at 100 degrees.
| June 23, 2021 | education
m = 400 g = 0.4 kg.
λ = 3.4 * 10 ^ 5 J / kg.
C = 4200 J / kg * ° C.
q = 2.3 * 10 ^ 6 J / kg.
t1 = 0 ° C.
t2 = 100 ° C.
Q -?
First, it is necessary to turn the ice into water, heat the resulting water to the boiling point, and turn the water into steam at the boiling point.
Q = Q1 + Q2 + Q3.
Q1 = λ * m.
Q1 = 3.4 * 10 ^ 5 J / kg * 0.4 kg = 1.36 * 10 ^ 5 J.
Q2 = C * m * (t2 – t1).
Q2 = 4200 J / kg * ° C * 0.4 kg * (100 ° C – 0 ° C) = 1.68 * 10 ^ 5 J.
Q3 = q * m.
Q3 = 2.3 * 10 ^ 6 J / kg * 0.4 kg = 9.2 * 10 ^ 5 J.
Q = 1.36 * 10 ^ 5 J + 1.68 * 10 ^ 5 J + 9.2 * 10 ^ 5 J = 12.24 * 10 ^ 5 J.
Answer: to convert ice into steam, you need Q = 12.24 * 10 ^ 5 J of thermal energy.
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