120 liters of water were mixed in a bath at a temperature of 10 degrees with 160 liters of water at a temperature of 70

120 liters of water were mixed in a bath at a temperature of 10 degrees with 160 liters of water at a temperature of 70 degrees what is the temperature of the resulting mixture

Given:
V1 = 120L = 0.12m ^ 3,
V2 = 160L = 0.16m ^ 3,
t1 = 10 ° С,
t2 = 70 ° C;
To find:
t -?
From the heat balance equation:
Q1 = Q2,
where
Q1 = m1 * c * (t – t1);
here
m1 = V1 * ρ;
ρ is the density of water;
Q2 = m2 * c * (t2 – t);
m2 = V2 * ρ;
Substitute these values into the equation:
ρ * V1 * c * (t – t1) = ρ * V2 * c * (t2 – t);
V1 * (t – t1) = V2 * (t2 – t);
t * (V1 + V2) = V2 * t2 + V1 * t1;
t = (V2 * t2 + V1 * t1) / (V1 + V2) = (11.2 + 1.2) /0.28;
t = 44.3 ° C.