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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)

    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


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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


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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


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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


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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
    I hope everything is clear now.
    mgb_phys solved the problem on the basis of
    My solution was on the basis of

    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


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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


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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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