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

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