Heat energy in a mixture to get final temperature

[johns123]

Homework Statement

A glass mug has mass 125g empty. It contains 180g of
coffee. Both are at 70 C. I add 15g of creme at 5 C.
Find final temperature. Assume creme has specific
heat of 2900 J/kg.C .

Homework Equations

Q = m C (delta)T in joules or Kcalories

m = mass in kg

C is specific heat where C-coffee 1 kcal/kg.C
C-glass .2 kcal/kg.C
C-creme 2900 J/kg.C

delta T is *change* in temperature

The Attempt at a Solution

Attempt at solution: First I don't see a "change"
in temperature. But I'm somehow suppose to apply
the above formula for Q-heat energy. There are 3
Qs .. 2 at 70 C .. and 1 at 5 C. I am somehow
suppose to weight these separate energies, calculate
a change in temperature. Then show the final equilibrium
temperature.

I can only guess that I can do something like

Q-coffee = mC-coffee x 70 to get an absolute ?? energy

Q-glass = mC-glass x 70

Q-creme = mC-creme x 5

and somehow weight the absolute thermal energies to
get delta-T .. and then T ??

 

[barryj]

Two hints: First, get your constants into the same units.
Second: Remember that the heat lost by the glass and coffee is the heat gained by the creme.

 

[johns123]

Thanks Barryj .. I got it. I did the heat gained = heat lost Conservation of Energy ( heat )

m1C1(70 - T) + m2C2(70 - T) = m3C3( T - 5 ) with the m3 conversion going back to kcal/kg.C

and I got a very reasonable answer. Amazing! Not one word of how to balance mixtures is anywhere in the
chapter.