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Homework Help: Heat and final temperature question

  1. Jan 30, 2006 #1
    I am stuck on this problem...
    What would be the final temperature when 100 g of 25 degree C water is mixed with 75 g of 40 degree C water?

    I know that mc?T gained = mc? T lost
    so (100g) (1.0 cal/g C) (Tf -25) = (75g) (1.0 cal/g C)(40 degrees - Tf)

    Where do I go from here?
    any push in the right direction would be great.

  2. jcsd
  3. Jan 30, 2006 #2
    100g/75g = 4/3 = (40 - Tf)/(Tf - 25).
    If I push you for such a small thing you will fall down dear kid.
    Generally take this as a formula when same object of differant amounts at differant temperatures are kept in contact with each other and the temperature of equilibrium temperature is asked - M1/M2 = dT1/dT2.

    Anyway do you really need a push to solve the above equation?
  4. Jan 30, 2006 #3
    i guess so

    I guess I need to be hit in the head with a hammer cause I just cant "see" it.
    My teacher gave me the following formula
    (100g) (1.0 cal/g C) (Tf -25) = (75g) (1.0 cal/g C)(40 degrees - Tf)

    and I just can't see first what the heck Tf stands for, where the 1.0 cal came from an how to solve it at all. I get to this point and then falter.
  5. Jan 30, 2006 #4


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    With regards to [itex]T_f[/itex] think about it logically. Go through each parameter in your equation. [itex]T_f[/itex] is the only unresolved variable and therefore must be the thing you are trying to calculate.

    As for the 1 cal/g C, this is the specific heat capacity of water. It is usually given in the question or on a data sheet.
  6. Jan 31, 2006 #5
    It was set up wrong!!

    I figured it out!
    The problem should be set up like so:
    (100g) (1.0 cal/g Co ) (Tf- 25 degrees) + (75g) (1.0 cal/g Co) (40 degrees - Tf) = 0
    Resulting in this:

    100Tf – 2500 + 75Tf – 3000 = 0
    175Tf = 2500 + 3000
    175Tf = 5500
    175Tf/175 = 5500/175
    Tf = 31.4

    I knew something was missing! Yea!!!! I'm not dumb! Thanks to those who tried to help.
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