800 grams of boiling water is poured into water weighing 1 kg, the temperature of which is 10 ° C.

800 grams of boiling water is poured into water weighing 1 kg, the temperature of which is 10 ° C. What will be the final temperature of the mixture? The specific heat of water is 4200 J / kg × ° С.

Legend:
t1 – cold water temperature;
t2 is the boiling water temperature;
t is the final temperature;
c – heat capacity of water;
m1 is the mass of cold water;
m2 is the mass of boiling water.

The amount of heat that cold water will receive:

сm1 (t – t1).

The amount of heat that boiling water will give:

сm2 (t2 – t).

Heat balance equation:

сm1 (t – t1) = сm2 (t2 – t).

m1 (t – t1) = m2 (t2 – t)

m1t – m1t1 = m2t2 – m2t;

t = (m2t2 + m1t1) / (m1 + m2) =
= (0.8 kg * 100 ° C +1 kg * 10 ° C) / (1 kg + 0.8 kg) =
= ((80 + 10) / 1.8) ° С = 50 ° С.

Answer: 50 ° C.