2 kg of water, the temperature of which is 6C. Added 3 kg of boiling water. What is the temperature of the water?

Task data: m1 (mass of cold water) = 2 kg; t1 (initial temperature of cold water) = 6 ºС; m2 (mass of boiling water) = 3 kg.

Constants: t2 (temperature of boiling water) = 100 ºС.

The steady-state temperature after mixing cold and boiling water is determined from the formula: C * m1 * (t – t1) = C * m2 * (t2 – t).

m1 * (t – t1) = m2 * (t2 – t).

2 * (t – 6) = 3 * (100 – t).

2t – 12 = 300 – 3t.

2t + 3t = 300 + 12.

5t = 312 and t = 312/5 = 62.4 ºС.

Answer: The temperature of the obtained water will be equal to 62.4 ºС.



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