Three liquids are at temperatures of 6 ◦C, 23◦C, and 38◦C, respectively. Equal masses of the first two liquids are mixed, and the equilibrium temperature is 13◦C. Equal masses of the second and third are then mixed, and the equilibrium temperature is 26.8 ◦C. Find the equilibrium temperature.

The equilibrium temperature is T13=3.12 ◦C

Explanation:

Given

The temperature of liquids: T1=6◦C, T2=23◦C, T3=38◦C

The temperature of 1+2 liquids mix: T12 = 13◦C.

The temperature of 2+3 liquids mix: T23=26.8 ◦C.

The temperature of 1+3 liquids mix: T13=?

1. When the first two liquids are mixed:

mC1 (T1-T12) + mC2 (T2-T12) = 0 C1 (6-13) = C2 (23-13) = 0 7C1=10C2 C1=1.42C2

2. When the second and third liquids are mixed:

mC2 (T2-T23) + mC3 (T3-T23) = 0 C2 (23-26.8) = C3 (38-26.8) = 0 3.8C2=12.8C3 C2=3.36C3

3. When the first and third liquids are mixed:

mC1 (T1-T13) + mC3 (T3-T13) = 0 C1 (6-T13) + C3 (38-T13) = 0 C1=1.42C2 C2=3.36C3 C1=1.42C2 (3.36C3) C1=4.77C3 C1 (6-T13) + C3 (38-T13) = 0 4.77C3 (6-T13) + C3 (38-T13) = 0 By solving the equation we get, T13=3.12 ◦C The equilibrium temperature is T13=3.12 ◦C