Mix 3 liters of water at 20 ° C and 2 liters of boiling water. Determine the temperature of the mixture.

Initial data: V1 (volume of cold water) = 3 l (m1 = 3 kg); t1 (cold water temperature) = 20 ºС; V2 (boiling water volume) = 2 l (m2 = 2 kg).

Reference data: t2 (boiling water temperature) = 100 ºС.

If there is no heat loss during mixing, then the ratio is correct: C * m1 * (t3 – t1) = C * m2 * (t2 – t3) and m1 * (t3 – t1) = m2 * (t2 – t3)

Let’s substitute the known values into the expression:

3 * (t3 – 20) = 2 * (100 – t3).

3t3 – 60 = 200 – 2t3.

5t3 = 260.

t3 = 260/5 = 52 ºС.

Answer: The temperature of the mixture was 52 ºС.