A piece of ice weighing 4 kg at a temperature of -20 ° C was immersed in water having a temperature of 80 ° C.

A piece of ice weighing 4 kg at a temperature of -20 ° C was immersed in water having a temperature of 80 ° C. Water mass 10 kg. What temperature will the water have when all the ice has melted? The specific heat of water is 4200 JDkg * ° C), the specific heat of ice is 2100 JDkg * ° C), and the specific heat of melting of ice is 340 kJ / kg.

Q1 = Sv * m1 * (t1 – tk), Sv = 4200 J / (kg * K), m1 = 10 kg, t1 = 80 ° С.
Q2 = Сl * m2 * (0 – tl) + λ * m2 + Sv * m2 * (tк – 0), Сl = 2100 J / (kg * K) = 2100 J / (kg * K, m2 = 4 kg, tl = – 20 ºС, λ = 340 kJ / kg = 340 * 10³ J / kg.
Q1 = Q2.
Sv * m1 * (t1 – t3) = SL * m2 * (0 – tl) + λ * m2 + Sv * m2 * (t1 – 0).
4200 * 10 * (80 – tk) = 2100 * 4 * 20 + 340 * 10³ * 4 + 4200 * 4 * tk.
3360000 – 42000tk = 168000 + 1360000 + 16800tk.
58800tk = 3360000 – 168000 – 1360000.
58800tk = 1832000.
tк = 31.16 ºС.
Answer: Water will have a temperature equal to 31.16 ºС.

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