[xoedusk]

Good afternoon,

If I am trying to find out how mixing different amounts of different temperature liquids will affect the final solution temperature, is it just a straight-forward application of mixing concentrations, except I use temperature instead of normal concentrations?

I.e., will mixing equal amounts of 0 C liquid water and 100 C liquid water yield 50 C water?

My hesitation in saying this is true is that temperature is a measure of average kinetic energy. Just wondering if this would mess it up.

[DevaDevil]

you use the specific heat capacity in order to find the temperature of a mixture of compounds.

this is how it works in mixing: Balance in liquids A and B, with final AB:
Energy before = energy after (assume no loss in energy)
{integrate between T_A,initial and T_A,end} n_A^.T_A^.C_p,A + {integrate between T_{B, initial} and T_{B, end}} n_B^.T_B^.C_p,B = 0
n = moles
C_p = specific heat capacity (at constant pressure)
T = temp in Kelvin

solution:

n_A^.C_p,A(T_{A, initial}-T_{A, end}) + n_B^.C_p,B(T_{B, initial}-T_{B, end}) = 0
And as T_{A, end} = T_{B, end} (mixture) this is an equation with only 1 unknown.

with 2 equal volumes (=equal moles) of water (both with the same heat capacity) it becomes very easy:
n_A^.C_p,A(T_{A, initial}-T_{A, end}) = - (n_B^.C_p,B(T_{B, initial}-T_{B, end}) )

or: T_{A, initial} - T_{A, end} = - (T_{B, initial} - T_{B, end})

and since T_{A, end} = T_{B, end} = T_end

this becomes T_{A, initial} + T_{B, initial} = 2* T _end)
and indeed in this case the end temperature will be the average of both.

with different solutions (salts soluted for example), or different compunds (alcohol and water for example) you will need C_p.


Edited for spelling >.>

[xoedusk]

Thank you DevaDevil. I knew this at some point in my schooling - it left my brain very quickly I suppose. I appreciate your post.