Answer:
33.85°C
Explanation:
From the question,
Heat lost by the hotter water = heat gained by the colder water
cm'(t₂-t₃) = cm(t₃-t₁)................. Equation 1
Where c = specific heat capacity of water, m' = mass of hot water, m = mass of cold water, t₁ = Initial temperature of cold water, t₂ = Initial temperature of hot water, t₃ = final temperature of the mixture.
But since the density of water is constant, and mass varies directly as volume, We can replace the mass of water with the volume of water. i.e,
cv'(t₂-t₃) = cv(t₃-t₁)................. Equation 2
Where v' and v are the volume of hot water and cold water respectively
make t₃ the subject of the equation
t₃ = (v't₂+vt₁)/(v'+v)............ Equation 3
Given: v' = 5.0 L, v = 60 L, t₁ = 30°C, t₂ = 80°C
Substitute these values into equation 3
t₃ = (5×80+60×30)/(60+5)
t₃ = 2200/68
t₃ = 33.85°C
