AAMC practice test 2

Discussion in 'MCAT Study Question Q&A' started by cougarY, May 14, 2008.

  1. cougarY

    cougarY MD aware

    Joined:
    Nov 15, 2006
    Messages:
    70
    Likes Received:
    0
    Status:
    Attending Physician
    I know the first two AAMC practice tests are not very close to the new ones but I still had on a question for anyone out there that has the test. On question 82 on the physical sciences part it takes about how 1.19 W is dissipated and it asked how much the temperature should change between measurements (every 100 sec). I thought this was an easy application of q=mc,deltaT but I always get the wrong answer (D) when the answer in the booklet is C. Can anyone show me what I'm missing?
     
  2. Thread continues after this sponsor message. SDN Members do not see this ad.

  3. cougarY

    cougarY MD aware

    Joined:
    Nov 15, 2006
    Messages:
    70
    Likes Received:
    0
    Status:
    Attending Physician
    I know the first two AAMC practice tests are not very close to the new ones but I still had on a question for anyone out there that has the test. On question 82 on the physical sciences part it takes about how 1.19 W is dissipated and it asked how much the temperature should change between measurements (every 100 sec). I thought this was an easy application of q=mc,deltaT but I always get the wrong answer (D) when the answer in the booklet is C. Can anyone show me what I'm missing?
     
  4. stupidkid30

    Banned

    Joined:
    Mar 23, 2008
    Messages:
    40
    Likes Received:
    0
    Status:
    Pre-Medical
    Would you be kind enough to send me a copy of this thing, [email protected] I would truly apprecite it.:D
     
  5. cwfergus

    Joined:
    May 12, 2008
    Messages:
    164
    Likes Received:
    1
    Status:
    Pre-Medical
    Think you can be a little more explicit with the question?
     
  6. tncekm

    tncekm MS-1

    Joined:
    Jul 18, 2006
    Messages:
    3,622
    Likes Received:
    4
    Status:
    Medical Student
    Okay, I don't have this one, but it has to be a simple q=mcΔT problem, you must just be making a simple mistake.

    Power = 1.19W. Energy transferred = Power x time = Pt

    Energy transferred = q = mcΔT = Pt

    ΔT=1.19*t/mc

    If you did that already, and if you're plugging in the correct mass value (check the units of the c value given and of the mass value, make sure they're the same), I don't know what to tell you.
     
  7. stupidkid30

    Banned

    Joined:
    Mar 23, 2008
    Messages:
    40
    Likes Received:
    0
    Status:
    Pre-Medical
    Hey thanks for the copy, you are such a nice person I appreciate soooo much, It sucks that I couldn't have this before I took my test. Nevertheless, I am going to save it in case I need to retake.
     
  8. jamesrick80

    Joined:
    Jan 27, 2008
    Messages:
    144
    Likes Received:
    0
    Status:
    Pre-Medical
    If I wasn't on vacation I would post the old AAMC paper exams that I have, but I needed a break after that tough May 10th MCAT.
     
  9. Vihsadas

    Vihsadas No summer
    Moderator Emeritus Lifetime Donor Classifieds Approved

    Joined:
    Oct 17, 2007
    Messages:
    5,472
    Likes Received:
    52
    Status:
    Medical Student
    What information are you given in the question?
     
  10. BerkReviewTeach

    BerkReviewTeach Company Rep & Bad Singer
    Exhibitor

    Joined:
    May 25, 2007
    Messages:
    3,821
    Likes Received:
    570
    It starts with Q = mCdeltaT(oil) + CdeltaT(system)
    Note: the CdeltaT of system is because the passage gives the C in terms of J/deltaC

    Q = P x t = 1.19 J/s x 100 s = 119 J

    119 = (500)(2.93)deltaT + 36.8deltaT = 1465deltaT + 36.8deltaT = 1501.8deltaT

    deltaT = 119/1501.8

    Given that 120/1500 = 0.08, the best answer is slightly less than that. Choice C, 0.079, is the best answer.

    The problem you are likely having is that you forgot to consider the heating of the stirrer, cup, thermometer, and resistor.
     
  11. cougarY

    cougarY MD aware

    Joined:
    Nov 15, 2006
    Messages:
    70
    Likes Received:
    0
    Status:
    Attending Physician
    Sorry I just figured you might have a copy. The question reads: If a steady power of 1.19 W is dissipated in the resistor when the potential difference across it is 11.9 V, how much should the temperature change between measurements?
    a. 0.00079 degrees C
    b. 0.00081 degrees C
    c. 0.079 degrees C
    d. 0.081 degrees C

    The answer given in the back is C but I keep getting D.

    In the passage you are given the information about a 10 ohm resistor immersed in a 500 g oil bath contained in a double-walled calorimeter that is thermally isolated from the environment. The oil has a specific heat of 2.93 J/g C at an initial temperature of 20 C. The temperature is recorded every 100 seconds. The heat capacity of the thermometer is 12.6 J/C and that of the stirrer, inner cup, and resistor combined is 24.2 J/C. The electrical power is supplied by a 12 V storage battery. A voltmeter connected in parallel with the resistor has a Resistance of 1 x 10^6 and an ammeter connected in series after the resistor has a Resistance of 1 ohm.

    Using Q=mc(delta)T and using 1.19 W = 500g x 2.93 J/gC x delta T x (1/100 seconds) I always get .081. There must be something I'm missing. Do you have to include the heat capacities of the thermometer or of the other things? Thanks.
     
  12. cougarY

    cougarY MD aware

    Joined:
    Nov 15, 2006
    Messages:
    70
    Likes Received:
    0
    Status:
    Attending Physician
    Hey thanks, I must have been typing as you posted your response. I haven't seen anything like this one the CBT AAMC tests, is this a good type question for the CBT they use now? I know this test is from a 1991 MCAT. And how did you know to include the include the heat capacity of the system (stirrer, inner cup, etc)? Or is that just a given that you have to include everything in this type of question?
     
  13. cougarY

    cougarY MD aware

    Joined:
    Nov 15, 2006
    Messages:
    70
    Likes Received:
    0
    Status:
    Attending Physician
  14. tncekm

    tncekm MS-1

    Joined:
    Jul 18, 2006
    Messages:
    3,622
    Likes Received:
    4
    Status:
    Medical Student
    That looks like a difficult one, and I definitely learned something from that--having said that, the wonderful explanation by BerkReviewTeach is too many calcs for the MCAT, so I figured I'd post the most simple way to answer it.

    Assuming we knew to include "everything else" into the system, rather than just the oil, the quickest way to have reached the answer would have been to just realize that if the oil alone would change, or 0.8 C, then including everything else should causethe change in temp to be slightly smaller.
     
  15. saundersR

    Joined:
    Jul 18, 2014
    Messages:
    1
    Likes Received:
    0
    Status:
    Post Doc
    Does anyone you have a pdf of any of the following AAMC MCAT practice test: 1, 2, 7-10? Might you be wiling to share them?
     

Share This Page