If you mix three liquids of the same mass and specific heat capacity, but different temperatures

If you mix three liquids of the same mass and specific heat capacity, but different temperatures of 10, 40, 100 * C, then the temperature of the resulting mixture will be equal to ..

m1 = m2 = m3 = m.

C1 = C2 = C3 = C.

t1 = 10 ° C.

t2 = 40 ° C.

t3 = 100 ° C.

T – ?

Q1 + Q2 = Q3 is the heat balance equation.

Q1 = C1 * m1 * (t – t1).

Q2 = C2 * m2 * (t – t2).

Q3 = C3 * m3 * (t3 – t).

C1 * m1 * (t – t1) + C2 * m2 * (t – t2) = C3 * m3 * (t3 – t).

C1 * m1 * t – C1 * m1 * t1 + C2 * m2 * t – C2 * m2 * t2 = C3 * m3 * t3 – C3 * m3 * t.

Since m1 = m2 = m3 = m, C1 = C2 = C3 = C.

C * m * t – C * m * t1 + C * m * t – C * m * t1 = C * m * t3 – C * m * t.

3 * C * m * t = C * m * t1 + C * m * t2 + C * m * t3.

3 * t = t1 + t2 + t3.

t = (t1 + t2 + t3) / 3.

t = (10 ° C + 40 ° C + 100 ° C) / 3 = 50 ° C.

T = 50 ° C + 273 = 323 K.

Answer: when mixing three liquids, the temperature will be set T = 323 K.