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Can anyone help me with calorimeters and change in temperature?

  1. Dec 18, 2008 #1
    1. The problem statement, all variables and given/known data
    A 0.10 kg piece of copper at an initial temperature of 95(degrees)C is dropped into 0.20 kg of
    water contained in a 0.28 kg aluminum calorimeter. The water and calorimeter are initially at 15(degrees)C. What is the final temperature of the system when it reaches equilibrium? (Cp of Copper= 387j/kg X (degrees)C, Cp of Aluminum= 899j/kg X (degrees)C, Cp of water= 4186j/kg X (degrees)C)

    2. Relevant equations
    I am homeschooled and unfortunately, when I have problems figuring the physics questions out, such as this one, I have a very hard time understanding what exactly is being done (its not like I can ask the teacher for a better explanation). I try to use common sense but some times it doesn't work out. I have tried looking for examples on the internet and come here only as a last resort. Any help here would be appreciated.


    3. The attempt at a solution
    Mass of Copper= 0.10 kg Initial Temp.= 95(degrees)C Cp= 387j/kg X 95= 36,765
    Mass of Water= 0.20 kg Initial Temp.= 15(degrees)C Cp= 4186j/kg X 15= 62,790
    Mass of AL.= 0.28 kg Initial Temp.= 15(degrees)C Cp= 899j/kg X 15= 13,485

    Formula I used: [Cp,w X Mw(Tf-Tw)= Cp,c X Mc(Tf-Tc)= Cp,al - Mal(Tf-Tw)]

    The first 1/3 works out like this:
    62,790 X .2(Tf-15)
    62790 X .2Tf -3
    (62,787 X .2Tf)=

    Second part:
    36,765 X .1(Tf-95)
    36,765 X .1Tf -9.5
    (36,755.5 X .1Tf)=

    Final Part:
    13,485 X .28(Tf -15)
    13,485 X .28Tf -4.2
    (13,480.8 X .28Tf)=

    62,787 X .2Tf= 36,755.5 X .1Tf= 13,480.8 X .28Tf

    From there it was simply a matter of division, subtraction and isolating the Final Temp. or Tf.
    I worked it out to:
    (39,512.3= 14Tf)
    Tf=2822.31

    This didn't make sense so I figured that since the answer was in Tf I should then divide it by the original temp., so I added all 3 (95 + 15 + 15) and divided (2822/125).

    The answer I came up with is 22.6(degrees)C

    Am I even close?
     
  2. jcsd
  3. Dec 18, 2008 #2

    mgb_phys

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    Call the final temperature T
    0.1kg Cu at 95degC cools to T so energy lost is 0.1kg * 387J/kg/K * (95-T) K
    0.20 kg H20 at 15degC warms to T so energy gained = 0.2kg * 4186J/kg/K * (T-15) K
    and 0.28kg of Al also warms to T = 0.28kg * 899J/kg/K * (T-15) K

    Note the way the temperature terms are different because one is rising to T the temperature difference is positive, the one cooling the temperature difference is negative.

    As you said the energy lost by the Cu must equal the energy gained by the Al+H2O
    0.1 * 387 * (95-T) = 0.2 * 4186 * (T-15) + 0.28 * 899 * (T-15)
    3676.5-38.7T = 837.2T - 12558 + 251.7 T - 3775.8
    3676.5 + 12558 + 3775.8 = 837.2T + 251.7 T + 38.7T
    20 010 = 1127.6 T, so T = 17.7 deg C
     
  4. Dec 18, 2008 #3
    Close, but I got a different answer.
    For simplicity I worked in degrees Kelvin. I just added 273° to each Centigrade value.
    So 95°C becomes 368°K, and 15°C becomes 288°K.
    Then I summed all the heat energy before putting in the copper.
    (0.1*368*387)+(0.2*288*4186)+(0.28*288*899)=327850,56 joules
    I know that at equilibrium this will equal
    ((0.1*387)+(0.2*4186)+(0.28*899))*FinalTemp°K = 1127,62*FinalTemp°K
    And when I've equated these two expressions and solved for FinalTemp I find it is 290,74°K or 17,74°C.
    Since you are clearly on your in this I would feel very bad if I've made an error or mislead you so I hope someone else will have a look at this and check both out workings.
     
  5. Dec 18, 2008 #4

    mgb_phys

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    This is exactly the same method.
    There is no need to use Kelvin since we are only concerned with changes in temperature the 1kelvin = 1deg C.
    17.74 is a rather optimistic number of decimal places given that you are only told the mass to 2sig figures ( i shoudl have really put 18 but I didn't want to confuse the OP if some answer software didn't accept it)
     
  6. Dec 18, 2008 #5
    mgb phy
    My answer was directed to seanmcgowan. You and I both posted about the same moment. I was not commenting on your post!
    I would say my method is simpler and makes just as much physical sense.
    You are right about the optimistic number of decimal places.
     
  7. Dec 18, 2008 #6

    mgb_phys

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    Fair enough, I do think your explanation of the total energy is simpler to follow.
    The OP had managed to do the maths - just made a little slip in the signs.
     
    Last edited: Dec 18, 2008
  8. Dec 18, 2008 #7
    Thanks. so everything was right, i just made a simple mistake in the conversion from the question to writing down the formula?
     
  9. Dec 18, 2008 #8

    mgb_phys

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    You got the direction of the temperature changes wrong, it has to be different between the thing heating up and the thing cooling down.
    So it's either a typo in the equation or a fundamental misunderstanding - depending if you understood this.
     
  10. Dec 18, 2008 #9
    I misunderstood the textbook. It is hard for me to just read and understand material without somone to explain things further. thank you.
     
  11. Dec 19, 2008 #10
    sean
    I hope everything is clear now.
    mgb_phys solved the problem on the basis of
    HEAT LOST = HEAT GAINED
    My solution was on the basis of
    TOTAL HEAT AT BEGINNING = TOTAL HEAT AT EQUILIBRIUM

    If the copper you added into the calorimeter had been previously cooled in liquid nitrogen to -196°C then that would have introduced another small complexity. Would you be confident about dealing with that?
     
  12. Dec 24, 2008 #11
    I don't quite understand. are you replacing the water with nitrogen? or just changing the temperature of the copper? if it were just the temperature change, wouldn't i use the same formula? or would the fact that everything would freeze be problematic?
     
  13. Dec 24, 2008 #12

    mgb_phys

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    The only extra complexity is that some energy is used to boil the nitrogen, the energy needed to convert a mass of liquid into gas is called the latent heat of vaporisation.
     
  14. Dec 24, 2008 #13
    Ok, so here is another problem that a friend had, that i helped him with using carid's method and i think i got it right (just wan to be sure im not messing something up here)

    the Question: Find the final eqilibrium temperature when 10.0 g of milk at 10.0 C is added to 1.60 x 10^2 g of coffee with a temperature of 90.0 C. Assume the specific heats of coffee and milk are the same for water (Cp,w = 4.19j/ g*C).

    The answer should be 85.29 C, or 85.3 C (when i use sig. figures)

    om almost positive this is right, just wanna make sure i fully understand what ou guys were saying.
     
  15. Dec 24, 2008 #14

    mgb_phys

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    Since it's all water you can ignore the specific heat - the c cancels.
    So using total energy
    10g * c * (10+273)K + 160g* c *(90+273)K = 170g * c *(T+273)K
    Rearange and solve for T (the final temperature in Celcius)
    Note that you have to use absolute temperatures because you are multipying.

    Or you can use the temperature differences. Again c cancels but no need to use absolute temperature because you are only taking the difference.
    10g *c * (10-T) = 160g * c *(T-90)
     
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