In 3 liters of water at 60 degrees, water was added at 20 degrees, receiving water at 40 degrees

In 3 liters of water at 60 degrees, water was added at 20 degrees, receiving water at 40 degrees. How much water was added?

Given: V1 (hot water volume) = 3 l (m1 = 3 kg); t1 (hot water temp.) = 60 ºС; t2 (temp. cool water) = 20 ºС; t (temp. equilibrium) = 40 ºС.

1) The mass (volume) of the added water is determined from the equality: C * m1 * (t1 – t) = C * m2 * (t – t2), whence m2 = m1 * (t1 – t) / (t – t2) = 3 * (60 – 40) / (40 – 20) = 3 kg (3 L).

2) Change in temperature of hot water is 20 ºС, cool water – 20 ºС. To change the temperatures by the same value, it is required to mix the same volumes of water and V1 = V2 = 3 liters.