Mixed 2 kg of water at a temperature of 15 degrees and 3 kg of water at 40 degrees.

Mixed 2 kg of water at a temperature of 15 degrees and 3 kg of water at 40 degrees. Determine the temperature of the resulting mixture.

Q = С * m * ∆t, С – specific heat capacity of water, m – mass of water, ∆t – temperature change (∆t = tн – tк, tн – initial, tк – final).
Qty. heat given off by hotter water:
Q1 = С * m1 * (t1 – t3), m1 = 3 kg, t1 = 40 ° С.
The amount of heat obtained with colder water:
Q2 = С * m2 * (t3 – t2), m2 = 2 kg, t2 = 15 ° С.
Q1 = Q2.
С * m1 * (t1 – t3) = С * m2 * (t3 – t2).
m1 * (t1 – t3) = m2 * (t3 – t2).
3 * (40 – t3) = 2 * (t3 – 15).
120 – 3t3 = 2t3 – 30.
5t3 = 150.
t3 = 30 ° C.