Formula: final temperature when mixing water with water

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Homework Help Overview

The discussion revolves around determining the final temperature when mixing two different amounts of water at varying temperatures, focusing on the principles of heat transfer and specific heat capacity.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the derivation of the final temperature formula, with some suggesting the use of specific heat capacity equations. Others provide equations and examples to illustrate the concept, while some express frustration over the nature of responses.

Discussion Status

The discussion includes various attempts to clarify the formula for final temperature, with some participants providing equations and examples. There appears to be a mix of constructive contributions and some interpersonal conflict regarding the responses given.

Contextual Notes

Participants note the absence of a textbook for reference, which may contribute to the need for clarification and guidance on the topic.

Sanne
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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 ;)
 
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You could always derive it for yourself. Start by considering the equation of the specific heat capacity.
 
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.
 
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.
 
Last edited:
hootenanny youre such a dick, answer the question and just give the equation, your comment is so helpful
 
deex91 said:
hootenanny youre such a dick, answer the question and just give the equation, your comment is so helpful

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!
 

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