Help! Finding Final Temp of Mixing Water Together!

1. The problem statement, all variables and given/known data

What would be the final temp if you mixed 1 liter of 20°C water with 4 liters of 40°C water?

This question is very straightforward, and is, of course, assuming total conservation of temp.
(given mass m1 of water at a temperature T1 and mass m2 of water at T2, mix them together)
2. Relevant equations

My book mentions q=mc∆t, but using that equation, I get calories as the units. And this equation I barely know how to use.

3. The attempt at a solution

My guess is the answer is 38.75 °C, but I figured that by reasoning only, as I have no clue what equation to use. I wondered if it might be a proportions prob, but haven't gotten a logical answer that way. I know the answer has to be over 20°C but under 40°C.

I am very very confused, so a clear answer with an good explanation would be the best! Thanks very much!
 

mgb_phys

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My book mentions q=mc∆t, but using that equation, I get calories as the units. And this equation I barely know how to use.
The answer depends on the units of 'c'

My guess is the answer is 38.75 °C,
Not a very good guess

I know the answer has to be over 20°C but under 40°C.
A good start

Using the equation above we know that energy q=mc∆t,
Imagine the final temperature is T, we have a change in temperature of (T-20) for the cold water and (40-T) for the hot water - note the way around for these since one is going up and one is going down - we end up with 5litres at T.

We don't need to know 'c' and we don't need to worry about the mass of 1litre because that is all going to cancel, and we don't care about the actual value of q.

q = 1 c (T-20) + 4 c (40-T) = 5 c T

All you have to do is rearrange to find T.
 
Thank you very much for replying! :smile: Please forgive me, as I understand only a speck of physics.

This makes sence:
The answer depends on the units of 'c'
What do you mean by this?
All you have to do is rearrange to find T.
Is it a matter of using the distributive property in algebra?

q = 1 c (T-20) + 4 c (40-T) = 5 c T
q = 1cT -20c + 80c -4cT = 5cT
q = 60c - 3ct = 5ct
q = 60c = 8ct
q = 7.5 = t

In one of my attempts to figure out the answer, I came up with 8, but assuming 8 is correct, what do I do with it? It's 8 what? 8 bananas? 8 oranges? Is something multiplied by a factor of 8? Or 1/8?

Is this even remotely in the right direction? (Knowing my luck, probably not. :redface: )

I am very sorry for my ignorance! :rolleyes: Thank you for your help! :smile:
 
Last edited:

mgb_phys

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What do you mean by this?
C is the conversion between mass and temperature to energy, it's normally in J/kg/k but you could have it in calories/g/k if you like.

Is it a matter of using the distributive property in algebra?
Sorry I wasn't very clear.

The heat lost by the hot water in cooling from 40 -> T equals the heat gained by the cold water going from 20->T.

So: 1 c (T-20) = 4c(40-T)
Cancels the 'c' and rearrange
 
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