Answer:
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.
<u>Mix 1</u>:
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)
<u>Mix 2</u>:
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)
<u>Mix 3</u>:
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
