Mixed 100 g of water having a temperature of 20C, with 300 g of water having a temperature of 60C.
| May 1, 2021 | education
Mixed 100 g of water having a temperature of 20C, with 300 g of water having a temperature of 60C. What is the temperature of the mixture?
Problem data: m1 (mass of cool water) = 100 g (0.1 kg); t1 (cool water temperature) = 20 ºС; m2 (mass of hot water) = 300 g (0.3 kg); t2 (hot water temperature) = 60 ºС.
The steady-state temperature of the resulting mixture is determined from the equality: Cw * m1 * (t – t1) = Cw * m2 * (t2 – t).
m1 * (t – t1) = m2 * (t2 – t).
0.1 * (t – 20) = 0.3 * (60 – t).
0.1t – 2 = 18 – 0.3t.
0.4t = 20 and t = 20 / 0.4 = 50 ºС.
Answer: The temperature of the resulting mixture will be 50 ºС.
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