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

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

14.8°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 (9 − 15) + C₂ (24 − 15) = 0

-6 C₁ + 9 C₂ = 0

C₁ = 1.5 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 (24 − 26.2) + C₃ (30 − 26.2) = 0

-2.2 C₂ + 3.8 C₃ = 0

C₂ = 1.73 C₃

Substituting:

C₁ = 1.5 (1.73 C₃)

C₁ = 2.59 C₃

When the first and third are mixed:

m C₁ (T₁ − T) + m C₃ (T₃ − T) = 0

C₁ (T₁ − T) + C₃ (T₃ − T) = 0

(2.59 C₃) (9 − T) + C₃ (30 − T) = 0

(2.59) (9 − T) + 30 − T = 0

23.3 − 2.59T + 30 − T = 0

3.59T = 53.3

T = 14.8°C

54 degrees incident angle = reflected angle

angle is from normal turned to line of ray