Three liquids are at temperatures of 4 ◦C, 24◦C, and 29◦C, respectively. Equal masses of the first two liquids are mixed, and the equi - librium temperature is 21◦C. Equal masses of the second and third are then mixed, and the equilibrium temperature is 26.1 ◦C. Find the equilibrium temperature when equal masses of the first and third are mixed. Answer in units of ◦C.

T₁₃ = 24.1°C

Explanation:

Given

m = mass of each liquids (all masses are equal)

C₁ = specific heat of the first liquid

C₂ = specific heat of the second liquid

C₃ = specific heat of the third liquid

T₁ = 4°C (temperature of the first liquid)

T₂ = 24°C (temperature of the second liquid)

T₃ = 29°C (temperature of the third liquid)

Temperature of 1+2 liquids mix: T₁₂ = 21°C

Temperature of 2+3 liquids mix: T₂₃ = 26.1°C

Temperature of 1+3 liquids mix: T₁₃ = ?

We can apply the relation ∑Q = 0

We assume the system is isolated and the process is adiabatic.

Mix 1:

Q₁ + Q₂ = 0

where

Q₁ = m*C₁ * (T₁-T₁₂)

and

Q₂ = m*C₂ * (T₂-T₁₂)

then

m*C₁ * (T₁-T₁₂) + m*C₂ * (T₂-T₁₂) = 0

⇒ C₁ * (T₁-T₁₂) + C₂ * (T₂-T₁₂) = 0

⇒ (4°C-21°C) * C₁ + (24°C-21°C) * C₂ = 0

⇒ - 17°C*C₁ + 3°C*C₂ = 0

⇒ C₁ = (3/17) * C₂ = 0.176*C₂ (i)

Mix 2:

Q₂ + Q₃ = 0

where

Q₂ = m*C₂ * (T₂-T₂₃)

and

Q₃ = m*C₃ * (T₃-T₂₃)

then

m*C₂ * (T₂-T₂₃) + m*C₃ * (T₃-T₂₃) = 0

⇒ C₂ * (T₂-T₂₃) + C₃ * (T₃-T₂₃) = 0

⇒ (24°C-26.1°C) * C₂ + (29°C-26.1°C) * C₃ = 0

⇒ - 2.1°C*C₂ + 2.9°C*C₃ = 0

⇒ C₃ = 0.724*C₂ (ii)

Mix 3:

Q₁ + Q₃ = 0

where

Q₁ = m*C₁ * (T₁-T₁₃)

and

Q₃ = m*C₃ * (T₃-T₁₃)

then

m*C₁ * (T₁-T₁₃) + m*C₃ * (T₃-T₁₃) = 0

⇒ C₁ * (T₁-T₁₃) + C₃ * (T₃-T₁₃) = 0

⇒ (4°C-T₁₃) * C₁ + (29°C-T₁₃) * C₃ = 0

If we use the following relations C₁ = 0.176*C₂ and C₃ = 0.724*C₂ we obtain

(4°C-T₁₃) * 0.176*C₂ + (29°C-T₁₃) * 0.724*C₂ = 0

⇒ (4°C-T₁₃) * 0.176 + (29°C-T₁₃) * 0.724 = 0

⇒ 0.706°C - 0.176*T₁₃ + 21°C - 0.724*T₁₃ = 0

⇒ 0.9*T₁₃ = 21.706°C

⇒ T₁₃ = 24.1°C