# AAMC practice test 2

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

1. ### cougarY MD aware 10+ Year Member

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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. OP

### cougarY MD aware 10+ Year Member

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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?

3. ### stupidkid30 Removed

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Would you be kind enough to send me a copy of this thing, [email protected] I would truly apprecite it.

4. ### cwfergus 10+ Year Member

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Think you can be a little more explicit with the question?

5. ### tncekm MS-1 2+ Year Member

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Okay, I don't have this one, but it has to be a simple q=mc&#916;T problem, you must just be making a simple mistake.

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

Energy transferred = q = mc&#916;T = Pt

&#916;T=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.

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6. ### stupidkid30 Removed

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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.

7. ### jamesrick80 10+ Year Member

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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.

8. ### Vihsadas No summer Moderator EmeritusLifetime DonorVerified Account 5+ Year Member

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What information are you given in the question?

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9. ### BerkReviewTeach Company Rep & Bad Singer Vendor 10+ Year Member

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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.

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10. OP

### cougarY MD aware 10+ Year Member

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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.

11. OP

### cougarY MD aware 10+ Year Member

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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?

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### cougarY MD aware 10+ Year Member

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13. ### tncekm MS-1 2+ Year Member

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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.

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14. ### saundersR

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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?

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