Three liquids are at temperatures of 9 β—¦C, 24β—¦C, and 30β—¦C, respectively. Equal masses of the first two liquids are mixed, and the equilibrium temperature is 15β—¦C. Equal masses of the second and third are then mixed, and the equilibrium temperature is 26.2 β—¦C.

Find the equilibrium temperature when equal masses of the first and third are mixed.
Answer in units of β—¦C

Respuesta :

Answer:

14.8Β°C

Explanation:

When the first two are mixed:

m C₁ (T₁ βˆ’ T) + m Cβ‚‚ (Tβ‚‚ βˆ’ T) = 0

C₁ (T₁ βˆ’ T) + Cβ‚‚ (Tβ‚‚ βˆ’ T) = 0

C₁ (9 βˆ’ 15) + Cβ‚‚ (24 βˆ’ 15) = 0

-6 C₁ + 9 Cβ‚‚ = 0

C₁ = 1.5 Cβ‚‚

When the second and third are mixed:

m Cβ‚‚ (Tβ‚‚ βˆ’ T) + m C₃ (T₃ βˆ’ T) = 0

Cβ‚‚ (Tβ‚‚ βˆ’ T) + C₃ (T₃ βˆ’ T) = 0

Cβ‚‚ (24 βˆ’ 26.2) + C₃ (30 βˆ’ 26.2) = 0

-2.2 Cβ‚‚ + 3.8 C₃ = 0

Cβ‚‚ = 1.73 C₃

Substituting:

C₁ = 1.5 (1.73 C₃)

C₁ = 2.59 C₃

When the first and third are mixed:

m C₁ (T₁ βˆ’ T) + m C₃ (T₃ βˆ’ T) = 0

C₁ (T₁ βˆ’ T) + C₃ (T₃ βˆ’ T) = 0

(2.59 C₃) (9 βˆ’ T) + C₃ (30 βˆ’ T) = 0

(2.59) (9 βˆ’ T) + 30 βˆ’ T = 0

23.3 βˆ’ 2.59T + 30 βˆ’ T = 0

3.59T = 53.3

T = 14.8Β°C