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Calculating mixed water temperature


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#1 kybert

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Posted 13 February 2006 - 12:52 AM

Hi.

Is there a formula that can calculate the resultant temperature of water when two different amounts with different temperatures are mixed?

e.g.

100ml @ 20C added to 50ml @ 60C = 150ml @ ??? C



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#2 insane_alien

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Posted 13 February 2006 - 01:37 AM

calculate the heat content of both with a reference temp(eg. 0 c) then add them and apply that heat to the 150ml.
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#3 ecoli

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Posted 13 February 2006 - 01:43 AM

just a weighted average of the two, no?
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#4 kybert

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Posted 13 February 2006 - 10:00 AM

Can someone give me a worked example?
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#5 swansont

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Posted 13 February 2006 - 11:39 AM

Can someone give me a worked example?


That would kinda defeat the idea of you learning the concept.

The thermal content added is the mass * specific heat capacity * temperature change

Q = m c \Delta T

Can you construct an equation from that? (Note that you don't need to know the actual heat capacity if all the materials are the same, since it will cancel out)
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#6 kybert

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Posted 13 February 2006 - 11:55 AM

OK, now you have really lost me. What has thermal content and mass got to do with it? I want to mix water as a liquid form, i.e. milli-litres not grams.

And i dont know temperature change because thats what im trying to find out!!
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#7 5614

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Posted 13 February 2006 - 10:44 PM

You can use E=mcΔT in the way which insane_alien was saying. Work out the energy (also denoted as Q in swansont's post) relative to 0 degrees C.

You know that 1 litre weights 1kg.

You should be given c, the specific heat capacity of water.

@ecoli: Yeah I do that as a safety check method. Doesn't work if you are mixing two different liquids as each has a different specific heat capacity.
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#8 kybert

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Posted 14 February 2006 - 09:18 AM

What is the specific heat capacity of water? Is that a constant?

Im not at school, and im not a scienctist...!

I have 2 jars of water, and just wondered if it was possible.
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#9 swansont

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Posted 14 February 2006 - 11:46 AM

What you are doing is dividing the excess heat content of the 60 C jar over an amount that is three times as much. The temperature should rise a third of the way from the low to the high temperature.

for water, 1 ml = 1g, and the excess temperature is 60 - 20 = 40 C. The specific heat is the same, so it will cancel out.

So 50g*40C = 150g*\Delta T

\Delta T = 13.3 C ; i.e. you will raise the temperature 13.3 C above the original 20 C, for a final value of T = 33.3 C
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#10 kybert

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Posted 15 February 2006 - 11:43 AM

Hi,


Thanks for that. Is this correct:


((Va * Ta) + (Vb * Tb)) / (Va+Vb) = Tf


Va = volume in tank A whose temperature is Ta
Vb = volume in tank B whose temperature is Tb
Tf = the temperature of the water after mixing.


PS. How do i make the formula into an image like others have done?


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#11 swansont

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Posted 15 February 2006 - 11:57 AM

For the same liquid, and with no phase change, yes.
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#12 kybert

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Posted 15 February 2006 - 11:59 AM

What is a phase change?
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#13 Klaynos

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Posted 15 February 2006 - 01:50 PM

What is a phase change?


A quick description is when a liquid changes to a gas or solid, or the other way round. There are many other examples though, these are just the most common and easily known.
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#14 swansont

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Posted 15 February 2006 - 03:53 PM

The specific examples of what Klaynos said would be boiling or freezing the water. When you boil you add energy (called the latent heat of vaporization), but the water temperature remains at 100 C. And when you freeze it you remove energy (latent heat of fusion) but the water temperature remains at 0 C.
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#15 jbrahy

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Posted 14 August 2012 - 06:52 PM

((Va * Ta) + (Vb * Tb)) / (Va+Vb) = Tf
Va = volume in A whose scoville rating is Ta
Vb = volume in B whose scoville rating is Tb
Tf = the scoville rating of the chile powder after mixing.



ok, I'm mixing chile powders. Would this same formula work?







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#16 Greg H.

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Posted 14 August 2012 - 07:00 PM

ok, I'm mixing chile powders. Would this same formula work?








Why would you expect it to? More importantly why would you resurrect a 6 year old thread to ask a nonsense question about a completely different topic? Isn't it easy enough to start your own nonsense thread?

Edited by Greg H., 14 August 2012 - 07:00 PM.

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#17 etawel

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Posted 24 December 2012 - 07:45 AM

for the mixing tab of water: the temperature hot water is 60 °C; the temperature of cold water is 20 °C. after mixing hot and cold water by the tab, the temperature of mixture watar was 40 °C and the flow rate was 0.061 L/s.
I would like to calculate the percent of hot water is used from the mixture which was 4 Letters
what is equation to find the volume of hot water ?
could you please help me.
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#18 knyazik

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Posted 22 October 2014 - 07:08 AM

When calculating make sure you include the conservation of energy.  So if one of the waters is much warmer then the other then the amount of heat that one of them loses will be the amount that the other gains.  If there is a phase change (for instance going from solid to liquid) first thing to check is that the amount of heat necessary to change all the ice into water is not more then the amount of energy that the other water has above 0 C.  If it is, you can average the rest of the energy and make the same argument about conservation of energy.  If not then only part of the ice will turn into water, and it will be more of a question of given a huge amount of ice and water, when you mix them, how much ice will be left, because a portion will turn into water.  Plug in all the necessary constants like specific heat to make this quantitative.


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