How much heat is needed to make steam from 3 kg ice taken at a temperature of -18?
| February 7, 2021 | education
Q = Q1 + Q2 + Q3 + Q4.
Q1 = C1 * m * (tк1 – tн1), С1 – beats. warm ice (C1 = 2100 J / (kg * K)), m – mass (m = 3 kg), tк1 = 0 ºС, tн1 = -18 ºС.
Q2 = λ * m, λ – beats. heat of fusion (λ = 34 * 10 ^ 4 J / kg).
Q3 = C2 * m * (tк2 – tн2), C2 – beats. heat capacity of water (С2 = 4200 J / (K * kg)), tк2 = 100 ºС, tн2 = 0 ºС.
Q4 = L * m, L – beats. heat steam (L = 2.3 * 10 ^ 6 J / kg).
Q = 2100 * 3 * (0 – (-18)) + 34 * 10 ^ 4 * 3 + 4200 * 3 * (100 – 0) + 2.3 * 10 ^ 6 * 3 = 9293400 J = 9293.4 kJ …
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