I'd like to know how I could calculate the balance temperature obtained after mixing two fluids at different temperature.
I think it'd go along these lines, but am not sure:
m1c1T1 = m2c2T2
where m = mass
c = heat capacity of the fluid
T = temperature (in kelvin degrees I guess, if using kg for mass)
I'd like to calculate how much mass and at which temperature should be fluid1 to bring a known mass of fluid2 at a given temperature, say 20°C.
Thank youDraconicAcid - 20-2-2013 at 09:17
Your calculation should work, regardless of units (you can work in any mass unit, and any temperature scale), except that instead of T being
temperature, you need deltaT (change in temperature). A bit of algebra will rearrange that to give T = (m1c1T1 + m2c2T2)/(m1c1 + m2c2).
HOWEVER, if your two fluids are sufficiently different, there will be an enthalpy of solution which will generate/absorb heat when the two fluids are
mixed. ( http://en.wikipedia.org/wiki/Enthalpy_change_of_solution ) The extra gain/loss of heat will render the above equation moot.math - 20-2-2013 at 11:22
I see, thank you.
Do I need deltaT in both T1 and T2?
I still don't get it straight enough.DraconicAcid - 20-2-2013 at 11:35
Yes- the heat given off by one fluid will be absorbed by the other one (and we're ignoring heat given off or absorbed to/from the surroundings).
q1 = q2
In the absence of a chemical reaction or phase change, q = mc(delta)T (the good old Scottish formula), so
m1c1(Tf - T1) = -m2c2(Tf - T2) (the final temp is the same for both fluids).
Multiply out, and rearrange to get Tf = (m1c1T1 + m2c2T2)/(m1c1 + m2c2).
So, for example, if you're mixing two dilute aqueous solutions (same heat capacity, no reaction occurring), say 50 mL of one at 15 oC and the other
one 100 mL at 25 oC, then....
Tf = (m1c1T1 + m2c2T2)/(m1c1 + m2c2)
= (50g x 4.184 J/gK x 15 oC + 100 g x 4.184 J/gK x 25 oC)/(50g x 4.184J/gK + 100 g x 25 oC)
Now, the heat capacities are the same, so they are a common factor of the numerator and denominator, so...
= (50 g x 15oC + 100 g x 25 oC)/(50 g + 100 g)
= 21.7 oC
Of course, if these were dilute solutions of aqueous nitric acid and aqueous sodium hydroxide, the calculation would give the entirely wrong answer,
as it ignores the enthalpy of rxn of the acid and base. If we were to mix water and carbon tetrachloride, the heat capacity wouldn't cancel out,
because we'd have two different ones (and since the two don't actually dissolve in each other, no enthalpy of solution is expected).
If you were to mix, say, ethanol and water, you'd have to worry about the enthalpy of solution. I wonder if the heat capacity of the mixture would
have an effect on it....
