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Homework Help: Thermo-Chem, Final Temp of a Mixture

  1. Jun 3, 2012 #1
    1. The problem statement, all variables and given/known data
    A 155-g piece of copper at 188oC is dropped into 250.0 g of water at 23.7oC. (The specific heat of copper is 0.385 J/goC.) Calculate the final temperature of the mixture. (Assume no heat loss to the surroundings.)

    155g Copper Initial Temp 188 C
    250.0g Water Initial Temp 23.7 C
    Spec Heat Copper .385 J/G C
    Spec Heat Water 4.184 J / G C

    Final Temp Water

    C - Celsius
    J - Joules
    G - Grams
    T - Final Temp

    2. Relevant equations

    Heat lost by copper = Heat gained by Water
    my equation setup

    155g × 0.385J / g °C × (Tfinal-188 °C) = 250.0g × 4.184j / g °C × (Tfinal-23.7 °C)

    3. The attempt at a solution

    Well I have several pages of attempts here but I'll take a screenshot of my most recent.
    Oddly enough I took a guess on the system we use (wileyplus) to get a hint as it gives you a hint on attempt 2 and got it correct. But it's bugging me that I can't solve it even after repeated attempts.

    ^ wiley Plus screen saying my 32.5 guess was correct /

    This attempt I dropped the units and just solved for T
    Someone in the class mentioned trying this method but the result is off or I'm completely brain dead this evening and failing at math /

    155g x 0.385 J/g C x (Tfinal-188 C) = 250.0g x 4.184 J/g C x (Tfinal-23.7 C)
    59.675 x (t - 188) = 1046 x (t - 23.7)
    59.675t - 11218.9 = 1046t - 24790.2
    -986.325t = -13571.3
    t = 13.759

    Which would be way to far off from the +-1 sigfig on the 3rd digit allowance that said 32.5 C was correct.

    When I go for the dimensional analysis route I can't isolate Celsius, I always end up with joules as a unit on my final answer.

    155g | .385 J | (T - 188 C) | = 250g | 4.184 J | (T - 23.7 C)
    | g C | | | g C |

    Gives me the same result with Joules left.

    I'm totally lost here, the bad part of my brain keeps saying you guessed the right answer so move on, but I can't leave this alone I have to know how to solve it. Any help would be much appreciated.
    Last edited: Jun 3, 2012
  2. jcsd
  3. Jun 3, 2012 #2
    Hi prwraith!! Welcome to physics forums :smile:

    You're almost there, with a right approach, but this equation has a wrong sign :wink:
  4. Jun 3, 2012 #3
    by wrong sign do you mean the -?
  5. Jun 3, 2012 #4
    Oh, no. I mean the equation should have (188-T) instead. You are only calculating the amount of heat lost, and this being negative(as in your equation) gives no meaning.
  6. Jun 3, 2012 #5
    hmm ok let me run the numbers, I just ran the numbers on switching the sign to + while waiting for a reply and got 36.5 which is wayyy closer. Ty for the help, lets see what this round gives.
  7. Jun 3, 2012 #6
    Yay I got the answer !

    155g x .385 J / g C x (188C - T) = 250g x 4.184 j / g C x (T - 23.7C)
    -59.675 T + 11218.9 = 1046 T - 24790.2
    -1105.675 T = -36009.1
    C = 32.5

    Yay I love you for the help with switching that sign on my equation !
  8. Jun 3, 2012 #7
    Great! :approve:

    But I hope you understood why you had to switch the signs :wink:
  9. Jun 3, 2012 #8
    Well after you point it out it does actually make sense haha. Why in gods name would I be subtracting my highest temperature, something that's clearly going to go down after being plunged into water from the final temperature which I already knew was far far lower.

    This half of the equation would have resulted in a sub-zero Celsius answer. Which clearly isn't going to happen lol!.
  10. Jun 3, 2012 #9
    Sub zero Celsius temperatures are possible, but the main idea is that the amount of heat cannot be negative :wink:

    Good going, though! :biggrin:
  11. Jun 3, 2012 #10
    Just to clarify

    155g × 0.385J / g °C × (188°C - Tfinal) = 250.0g × 4.184j / g °C × (Tfinal-23.7 °C)

    The underlined portion is now correct because we're looking for the amount of heat lost en route from the original temperature to the final temperature.

    And the bold portion is correct because we're not looking for the temperature that was already present we're looking for the net gain by the water correct?

    Sorry to be a bother i'm just trying to really understand it so I never have trouble with this concept in the future.
  12. Jun 3, 2012 #11
    Yep, you got it :smile:
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