If you have 0.306 m3 of water at 25.0°C and add 0.102 m3 of water at 95.0℃, what is the final temperature of the mixture? Use 1000 kg/m3 as the density of water at any temperature. Number O Previous ⓧ Give Up & View Solution. Check Answer 0 Next | E Hint Is the amount of heat gained by the cold water the same as the amount of heat lost by the hot water?

The final temperature of the mixture = 42.5 °C

Explanation:

Heat gained by cold water = heat lost by hot water

cm₁ (t₃-t₁) = cm₂ (t₂-t₃) ... Equation 1

Where c = specific heat capacity of water, m₁ = mass of cold water, m₂ = mass of hot water, t₁ = initial temperature of cold water, t₂ = initial temperature of hot water, t₃ = temperature of the mixture.

m₁t₃ - m₁t₁ = m₂t₂ - m₂t₃

m₁t₃ + m₂t₃ = m₂t₂+m₁t₁

t₃ (m₁+m₂) = m₂t₂+m₁t₁

t₃ = m₂t₂+m₁t₁ / (m₁+m₂) ... Equation 2

Given: t₁ = 25 °C, t₂ = 95 °C,

but mass = density*volume.

m₁ = D*v₁,

where v₁ = volume of cold water, D = Density of water.

v₁ = 0.306 m³, D = 1000 kg/m³

m₁ = 1000*0.306 = 306 kg.

Also

m₂ = D*v₂

Where v₂ = volume of hot water = 0.102 m³, D = 1000

m₂ = 1000*0.102 = 102 kg

Substituting these values into equation 2,

t₃ = 102 (95) + 306 (25) / (306+102)

t₃ = 9690+7650 (408)

t₃ = 17340/408

t₃ = 42.5 °C.

Thus the final temperature of the mixture = 42.5 °C