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

i guess it’s a and b

spectral lines lyman is one set