Formula: final temperature when mixing water with water

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SUMMARY

The final temperature when mixing two different amounts of water at varying temperatures can be calculated using the formula: 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]). In this discussion, an example is provided where 1 kg of water at 20 degrees Celsius is mixed with 0.3 kg of water at 9 degrees Celsius, resulting in a final temperature of approximately 17.461 degrees Celsius. The specific heat capacity of water is noted as 4182 J/kg°C, which is essential for these calculations. Additionally, the discussion elaborates on the heat exchange principle, emphasizing that the system neither gains nor loses heat during the mixing process.

PREREQUISITES
  • Understanding of specific heat capacity
  • Basic knowledge of thermal equilibrium
  • Familiarity with algebraic manipulation of equations
  • Concept of mass and temperature in thermodynamics
NEXT STEPS
  • Study the principles of calorimetry
  • Learn about the conservation of energy in thermal systems
  • Explore the concept of thermal equilibrium in different materials
  • Investigate the effects of varying specific heat capacities in mixtures
USEFUL FOR

Students in physics or chemistry, educators teaching thermodynamics, and anyone interested in understanding heat transfer and thermal mixing processes.

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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