Question

Three liquids are at temperatures of 8◦C, 21◦C, and 31◦C, respectively. Equal masses of the first two liquids are mixed, and the equi- librium temperature is 13◦C. Equal masses of the second and third are then mixed, and the equilibrium temperature is 23.2◦C.

Find the equilibrium temperature when equal masses of the first and third are mixed.

Answer in units of ◦C.

Answer 1

Heat absorbed/released due to temperature increase is given by

Q = m*C*dT

m = mass of liquid, dT = temperature change

C = specific heat capacity of liquid

Suppose Heat capacity of 1st liquid = C1, 2nd liquid = C2 & third liquid = C3

then when equal masses of 1st and 2nd are mixed, and equilibrium temperature is 13 C, then

Heat absorbed by 1st liquid = Heat released by 2nd liquid

m*C1*dT1 = m*C2*dT2

C1*(13 - 8) = C2*(21 - 13)

C1*5 = C2*8

C1 = 1.6*C2

Now when equal masses of 2nd and 3rd are mixed, and equilibrium temperature is 23.2 C, then

Heat absorbed by 2nd liquid = Heat released by 3rd liquid

m*C2*dT2 = m*C3*dT3

C2*(23.2 - 21) = C3*(31 - 23.2)

C2*2.2 = C3*7.8

C2 = 3.5455*C3

Now when equal masses of 1st and 3rd are mixed, and suppose equilibrium temperature is T C, then

Heat absorbed by 1st liquid = Heat released by 3rd liquid

m*C1*dT1 = m*C3*dT3

C1*(T - 8) = C3*(31 - T)

Now from above two relations C1 = 1.6*C2 & C2 = 3.5455*C3

gives us

C1 = 1.6*3.5455*C3 = 5.6728*C3

Using this value in above relation

C1*(T - 8) = C3*(31 - T)

5.6728*C3*(T - 8) = C3*(31 - T)

5.6728*T - 5.6728*8 = 31 - T

T = (31 + 5.6728*8)/(1 + 5.6728)

T = 11.45 C = equilibrium temperature

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