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# mixing ratio

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1. Hi,
how would I go about working this problem out:

3 units of water that are 5oC mix with 2 units of water that are 10oC, what is the resultant temperature of the mixture?

thanks
2. (Original post by mchammer)
Hi,
how would I go about working this problem out:

3 units of water that are 5oC mix with 2 units of water that are 10oC, what is the resultant temperature of the mixture?

thanks
(3 x 5) + (2 x 10) = 15 + 20 = 35, so 35 is the 'total temperature'

There are 5 units, so for the mean temperature, you do 35/5 = 7oc.
3. (Original post by thegodofgod)
(3 x 5) + (2 x 10) = 15 + 20 = 35, so 35 is the 'total temperature'

There are 5 units, so for the mean temperature, you do 35/5 = 7oc.
That would be incorrect.

(Original post by mchammer)
Hi,
how would I go about working this problem out:

3 units of water that are 5oC mix with 2 units of water that are 10oC, what is the resultant temperature of the mixture?

thanks
I'm assuming the specific heat capacity of water = 4.2 J/g degC

The temperature of the 3 unit setup is higher than that of the 2 unit setup. Hence, when mixed together, thermal energy will flow from the 3 units to the 2 units until both are at equal temperatures.

3*4.2*(50-T) = 2*4.2*(T-10)

T = final temperature.
4. I would probably use kelvin for this. I'm not sure if it makes a difference, but 50°C is not 5 times as hot as 10°C and therefore doesn't contribute 5x the energy.
5. (Original post by Repressor)
I would probably use kelvin for this. I'm not sure if it makes a difference, but 50°C is not 5 times as hot as 10°C and therefore doesn't contribute 5x the energy.
It wouldn't make a difference since a change in temp. of 1 K = a change of temp. of 1 deg C

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