How much will be released during the transformation of steam with a mass of m = 1 kg
| May 2, 2021 | education
How much will be released during the transformation of steam with a mass of m = 1 kg, taken at a temperature of t1 = 100 degrees, and ice at a temperature of t2 = 0 degrees.
Q = Q1 + Q2 + Q3.
Q1 = L * m, where L – beats. heat of vaporization of water (L = 2.3 * 10 ^ 6 J / kg), m is the mass of steam (m = 1 kg).
Q2 = C * m * (tn – tc), where C is the specific heat capacity of water (C = 4200 J / (K * kg)), tn is the beginning. temperature (tн = 100 ºС), tк – end. temperature (tк = 0 ºС).
Q3 = λ * m, where λ – beats. heat of melting of ice (λ = 34 * 10 ^ 4 J / kg).
Q = Q1 + Q2 + Q3 = L * m + C * m * (tн – tк) + λ * m = 2,3 * 10 ^ 6 * 1 + 4200 * 1 * (100 – 0) + 34 * 10 ^ 4 * 1 = 3,060,000 J = 3.06 MJ.
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