Calculating mixed water temperature

[kybert]

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



Joe

[insane_alien]

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

[ecoli]

just a weighted average of the two, no?

[kybert]

Can someone give me a worked example?

[swansont]

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)

[kybert]

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!!

[5614]

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.

[kybert]

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.

[swansont]

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

[kybert]

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?


Jeo

[swansont]

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

[kybert]

What is a phase change?

[Klaynos]

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.

[swansont]

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.