In 1 liter of water at a temperature of 18 degrees, 300 g of molten tin with a temperature of 232
| May 9, 2021 | education
In 1 liter of water at a temperature of 18 degrees, 300 g of molten tin with a temperature of 232 degrees was poured. How many degrees was the water heated?
Initial data: m1 (water) = 1 kg; t1 (water temperature before contact with tin) = 18 ºС; m2 (tin) = 300 g = 0.3 kg.
Constants: C1 = 4200 J / (kg * K); C2 = 230 J / (kg * K); λ = 59 * 10 ^ 3 J / kg; t2 = 232 ºС.
Q1 (water heating) = Q2 (tin crystallization) + Q3 (tin cooling).
C1 * m1 * (t3 – t1) = λ * m2 + C2 * m2 * (t2 – t3).
4200 * 1 * (t3 – 18) = 59 * 10 ^ 3 * 0.3 + 230 * 0.3 * (232 – t3).
4200t3 – 75600 = 17700 + 16008 – 69t3.
4269t3 = 109308.
t3 = 25.6 ºС; Δt = t3 – t2 = 25.6 – 18 = 7.6 ºС.
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