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mixing ratio Watch

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    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
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    (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.
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    (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.
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    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.
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    (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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