Formula: final temperature when mixing water with water

Sanne

Back to basics - sorry, my class doesn't have a textbook or the like.

Can one of you nice people please tell me: the formula for the final temperature when mixing two different amounts of water of different temperatures?

If it's complicated you're of course welcome to explain ;)

Hootenanny

You could always derive it for yourself. Start by considering the equation of the specific heat capacity.

Eskild

To find the temperature of water mixed with water in a calorimeter you can use this equation.

Temperature[final] = (Mass[1]*Heat specific[1]*Temperature[1]+Mass[2]*Heat specific[2]*Temperature[2])/(Mass[1]*Heat specific[1]+Mass[2]*Heat specific[2])

EXAMPLE:

1: 1 kilo gram of water at 20 degrees Celsius.
2: 0.3 kilo gram of water at 9 degrees Celsius
The heat specific of water is 4182.

(1*4182*20+0.3*4182*9c)/(1*4182+0.3*4182) = 17.461 Degrees Celsius.

-Eskild.

the_ugly1

because your adding the two different masses of water together the equation:
q1+q2=0 where q1 is the first amount of water at t1 and q2 is the second amount of water at t2. We need to state that the final temp is tf. the value is 0 because the system neither gains nor looses heat.

If you expand the equation you get:
m1C1(tf-t1)+m2C2(tf-t2)=0

expanding further we get:
[(m1*C1*tf)-(m1*C1*t1)+(m2*C2*tf)-(m2*C2*t2)]=0
rearranging:
(m1*C1*tf)+(m2*C2*tf)=(m1*C1*t1)+(m2*C2*t2)

now solve for the item you need.

deex91

hootenanny youre such a dick, answer the question and just give the equation, your comment is so helpful

gneill

Responding with vitriol to a post that was made in 2006 (and which was perfectly in line with forum policy) is not an auspicious beginning to your posting career, deex91!