What is the final centigrade temperature of this liquid mixture when thermal equilibrium is reached ?

Question

115.0 ml of milk at 20.0oC are added to 459.5 ml of coffee at 93.6oC . What is the final centigrade temperature of this liquid mixture when thermal equilibrium is reached ?


Assume coffee has the same properties as pure water.


The average density of milk is 1032 kg/m3


The specific heat of milk is 1.97 J/g-oC

Answer 1

First we need the mass of the milk: 1032 kg/m3 = 1.032 g/ml and m = 1.032 g/ml x 115.0 ml = 118.68 g, mass of coffee is 459.5 g.
We assume no heat is lost during the mixture process, which means that the heat, given off by coffee, is the same as the heat, gained by milk. We make symbols: 1 for milk, 2 for coffee
Q1 = Q2
m1 x c1 x ΔT1 = m2 x c2 x ΔT2
We write temperature differences so that they're both positive (a common rule with "mixing" problems):
ΔT1 = Tf − T1, ΔT2 = T2 − Tf (Tf = final temperature)
We solve for Tf (a little bit of algebra):
Tf = (m1 x c1 x T1 + m2 x c2 x T2) / (m1 x c1 + m2 x c2) = 85.6 oC
(the temperatures can be put into the equation in centigrade temperatures)

Answer 2

heat lost = heat gained
coffee is water, mostly
msdeltaT for coffee = msdeltaT for milk
459.5(4.18)(93.6-Tfinal) = 115.0 x 1.032(1.97)(Tfinal - 20)

Multiply out, bring Tfinal to one side, numbers to other and solve for Tfinal.