A jet of centigrade water vapor is directed onto a piece of ice with a mass of 10 kg and a temperature of 0.

| September 11, 2021 | education

A jet of centigrade water vapor is directed onto a piece of ice with a mass of 10 kg and a temperature of 0. What temperature will be established after the ice melts, if the mass of consumed steam is 2 kg.

To find the steady-state temperature after ice melting, we use the equality: Q1 + Q2 = Q3 + Q4 and λ * m1 + Cw * m1 * tr = L * m2 + Cw * m2 * (t – tp).

Constants and variables: λ – beats. heat of melting of ice (λ = 3.4 * 10 ^ 5 J / kg); m1 is the mass of melted ice (m1 = 10 kg); Cv – beats heat capacity of water (Cw = 4200 J / (kg * K)); L – beats. heat of condensation of water vapor (L = 2.3 * 10 ^ 6 J / kg); m2 – mass of consumed steam (m2 = 2 kg); t – condensation temperature (t = 100 ºС).

Calculation: 3.4 * 10 ^ 5 * 10 + 4200 * 10 * tr = 2.3 * 10 ^ 6 * 2 + 4200 * 2 * (100 – tr).

3.4 * 10 ^ 6 + 42000 * tr = 5.44 * 10 ^ 6 + 840,000 – 8400 * tr.

50400tr = 2.04 * 10 ^ 6.

tр = 40.48 ºС.



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