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Equilibrium Temperature?

  1. Apr 22, 2005 #1
    How would you find the equilibrium temperature in a problem like this? **I don't need the answer, just some advice on getting there**

    45g of water at 4 degrees C is added to 150g of water at 75 degrees C. What is the mixture's equilibrium temperature?
  2. jcsd
  3. Apr 22, 2005 #2


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    [tex]q = m*C_p *\Delta T[/tex]
    Both objects will have the same heat transfer.............
  4. Apr 22, 2005 #3
    I don't get it...

    m = total mass
    c = forgot it's name of water
    delta T = unknown
    q = unknown

    You have two unknowns...
  5. Apr 22, 2005 #4


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    heat lost by one volume of water at its temperature will be gained by the other. Thus
    q1 (gained)=-q2(lost), you'll need to set the q equal to each other and solve for final temperature, Tf, which will be the same for the two equations.
  6. Apr 23, 2005 #5
    So how do I solve for q? Either way I look at it, I've got two unknowns. Do I need to use solving for simultaneous equations or something?
  7. Apr 23, 2005 #6
    Its Easy......

    hey tokimasa


    we know m,M and we can cancel "c" because both c are equal(we are talking about mixing the same material here-water)...then we will get an equation based on T...and we can find T from it,which is the temperature of the mixture.......

    now i will tell u the idea behind this......we are talking about mixing the same substance...that is water......so the property of heat transfer and temperature changes can be easily related.....when u bring cold water towards hotter water...the hot loses some energy...ie Q...this Q is gained by cold water...then only the condition becomes equillibrium isnt it???????????here one is loosing and the other is gaining...thats y the sign difference occurs
    Last edited: Apr 23, 2005
  8. Apr 23, 2005 #7
    I see where you are coming from now. But where do you get 227 and 348 from?
  9. Apr 23, 2005 #8
    He's converting to kelvin, but he did it wrong.

    The heat gained frmo one equals the heat lost from the other

    [tex] q_1 = -q_2 [/tex]

    [tex] m*c*(T_{celsius}+273) = -m*c*(T_{celsius} + 273) [/tex]

    Where the left side is one of the liquids, and the right side is the other. C is the same for both, and m is the mass of each quantity.
  10. Apr 24, 2005 #9
    y is it wrong man????????we have to incorporate temperature change as well......and i did it Q1=-Q2....ok...dont simply say its wrong
  11. Apr 24, 2005 #10
    But it is wrong.. T - 277? T - 348? Whats that?
  12. Apr 24, 2005 #11
    now what!

    in the problem given by tokimasa........Tf...the final equilibrium temperature is what we have to find......for both the part of liquids change of temperature takes place..........for the cooler water temperature increases from 277K(given by tokimasa) to Tf....so...Q=m*c*(Tf-277)

    for the cooler water temperature decreases from 348K(also given) to Tf....
    so Q=M*c*(Tf-348)
    remember......delta T is always final-initial...so Tf-277 and Tf-348 likewise....now do u understand????????????????
  13. Apr 25, 2005 #12
    First of all, is it really necessary to change it to Kelvin? Is the answer meant to be in Kelvin?

    It is quite simple. You find m*c*DT for both of the "waters". You know the mass of both, the specific heat capacity (c) is 4190, and you know the initial temperature of both systems. You want to find the final temperature.

    m1*c*DT1 + m2*c*DT2 = 0

    so you get 0.045*4190*(Tf-4) + 0.15*4190*(Tf-75) = 0

    Then you just solve it algebraically.

    Cehck that your answer is between 4 degrees and 75 degrees.
    Last edited: Apr 25, 2005
  14. Apr 25, 2005 #13
    Yes, kelvin is necessary. Other than that this is all correct..
    I just noticed I forgot the deltas in my equations above.
  15. Apr 25, 2005 #14
    celsius and kelvin depends upon the unit of specific heat constant...that is "c"...if u r taking c on the basis of celcius...use celcius...if c is in SI use kelvins.....
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