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

Find the equilibrium temperature when equal masses of the first and third are mixed.
Answer in units of โ—ฆC

Respuesta :

Answer:

14.8ยฐC

Explanation:

When the first two are mixed:

m Cโ‚ (Tโ‚ โˆ’ T) + m Cโ‚‚ (Tโ‚‚ โˆ’ T) = 0

Cโ‚ (Tโ‚ โˆ’ T) + Cโ‚‚ (Tโ‚‚ โˆ’ T) = 0

Cโ‚ (9 โˆ’ 15) + Cโ‚‚ (24 โˆ’ 15) = 0

-6 Cโ‚ + 9 Cโ‚‚ = 0

Cโ‚ = 1.5 Cโ‚‚

When the second and third are mixed:

m Cโ‚‚ (Tโ‚‚ โˆ’ T) + m Cโ‚ƒ (Tโ‚ƒ โˆ’ T) = 0

Cโ‚‚ (Tโ‚‚ โˆ’ T) + Cโ‚ƒ (Tโ‚ƒ โˆ’ T) = 0

Cโ‚‚ (24 โˆ’ 26.2) + Cโ‚ƒ (30 โˆ’ 26.2) = 0

-2.2 Cโ‚‚ + 3.8 Cโ‚ƒ = 0

Cโ‚‚ = 1.73 Cโ‚ƒ

Substituting:

Cโ‚ = 1.5 (1.73 Cโ‚ƒ)

Cโ‚ = 2.59 Cโ‚ƒ

When the first and third are mixed:

m Cโ‚ (Tโ‚ โˆ’ T) + m Cโ‚ƒ (Tโ‚ƒ โˆ’ T) = 0

Cโ‚ (Tโ‚ โˆ’ T) + Cโ‚ƒ (Tโ‚ƒ โˆ’ T) = 0

(2.59 Cโ‚ƒ) (9 โˆ’ T) + Cโ‚ƒ (30 โˆ’ T) = 0

(2.59) (9 โˆ’ T) + 30 โˆ’ T = 0

23.3 โˆ’ 2.59T + 30 โˆ’ T = 0

3.59T = 53.3

T = 14.8ยฐC