009

Three liquids are at temperatures of 14 ◦C, 25◦C, and 33◦C, respectively. Equal masses of the first two liquids are mixed, and the equilibrium temperature is 16◦C. Equal masses of the second and third are then mixed, and the equilibrium temperature is 27.9 ◦C. Find the equilibrium temperature when equal masses of the first and third are mixed.

Answer in units of ◦C.

16.1°C

Explanation:

When the first two are mixed:

m C₁ (T₁ − T) + m C₂ (T₂ − T) = 0

C₁ (T₁ − T) + C₂ (T₂ − T) = 0

C₁ (14 − 16) + C₂ (25 − 16) = 0

-2 C₁ + 9 C₂ = 0

C₁ = 4.5 C₂

When the second and third are mixed:

m C₂ (T₂ − T) + m C₃ (T₃ − T) = 0

C₂ (T₂ − T) + C₃ (T₃ − T) = 0

C₂ (25 − 27.9) + C₃ (33 − 27.9) = 0

-2.9 C₂ + 5.1 C₃ = 0

C₂ = 1.76 C₃

Substituting:

C₁ = 4.5 (1.76 C₃)

C₁ = 7.91 C₃

When the first and third are mixed:

m C₁ (T₁ − T) + m C₃ (T₃ − T) = 0

C₁ (T₁ − T) + C₃ (T₃ − T) = 0

(7.91 C₃) (14 − T) + C₃ (33 − T) = 0

(7.91) (14 − T) + 33 − T = 0

111 − 7.91T + 33 − T = 0

8.91T = 144

T = 16.1°C

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