A person tries to heat up her bath water by adding 5.0 L of water at 80°C to 60 L of water at 30°C. What is the final temperature of the water? Group of answer choices

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

it would be d. henry moseley

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