Water weighing 0.05 g at a temperature of 80 degrees was mixed with water weighing 0.15 g at a temperature

Water weighing 0.05 g at a temperature of 80 degrees was mixed with water weighing 0.15 g at a temperature of 15 degrees. Determine the temperature of the mixture.

m1 = 0.05kg,
m2 = 0.15kg,
t1 = 80 ° С,
t2 = 15 ° C;
Find: t -?
Let’s write down the heat balance equation:
Q1 = Q2,
where
Q1 = m1 * c * (t1 – t)
and
Q2 = m2 * c * (t – t2);
Here c is the specific heat capacity of water;
We get:
m1 * (t1 – t) = m2 * (t – t2);
Substitute numerical values:
80 – t = 3 * t – 45:
4 * t = 125;
t = 31.25 ° C.