I know homework questions are not welcome on this forum, but I have a take-home problem that involves calculating the heat absorbed by a calorimeter
and I'm having trouble figuring out if the final answer should be positive or negative.
The problem:
2-50mL portions of water, one at a temperature around 60C and another at a temperature around 20C are mixed together in a calorimeter. The final
temperature of the resulting water should ideally be the average of the two starting temperatures, but the temperature is about 1C lower than
expected, calculate the amount of heat absorbed by the calorimeter.
I'll be very up front; I am 100% sure that the answer is 416.3J, the math is rather easy and I'm sure that I didn't make any mistakes. I gave a vague
description of the problem without using exact numbers because I know how to do the math.
I think the answer is +416.3J because heat is being absorbed by the calorimeter (endothermic). I showed the answer to the teacher so he could tell me
if I was on the right track, and he said that I should check my work over. I checked it over and found that nothing was wrong with my math, but that
perhaps the presentation of my answer was incorrect.
Is he trying to throw me off, or is the answer -416.3J?
Thanks for any help.watson.fawkes - 23-11-2011 at 20:43
I think the answer is +416.3J because heat is being absorbed by the calorimeter (endothermic).
Sign
conventions are like standards; there are so many of them. In the present case, "heat absorbed" is positive from the point of view of the calorimeter
instrument, and negative from the point of view of its contents. The trick here is to figure out exactly which question is being asked. From the
sounds of it, it sounds like the former.francis - 24-11-2011 at 09:47
It should be positive. But, what is the calorimeter constant you're using, ie the heat capacity of the calorimeter?
It should be positive. But, what is the calorimeter constant you're using, ie the heat capacity of the calorimeter?
That's the thing, the problem does not say anything about the calorimeter except that it exists and it's made of styrofoam :\Ozone - 25-11-2011 at 07:24
Styrofoam?* Look up coffee-cup calorimeter and constant pressure calorimetry. The volume is not fixed (as in bomb calorimetry) against constant
(atmospheric) pressure. See:http://en.wikipedia.org/wiki/Calorimeter
*typical set up for rxn calorimetry in introductory undergraduate chemistry labs.
Cheers,
O3Ozone - 25-11-2011 at 07:24
Styrofoam?* Look up coffee-cup calorimeter and constant pressure calorimetry. The volume is not fixed (as in bomb calorimetry) against constant
(atmospheric) pressure. See:http://en.wikipedia.org/wiki/Calorimeter
*typical set up for rxn calorimetry in introductory undergraduate chemistry labs.
It should be positive. But, what is the calorimeter constant you're using, ie the heat capacity of the calorimeter?
I don't see where this is relevant. It seems clear that 416.3J are gained by the calorimeter.
No calorimeter details are given so why start making assumptions about it. bbartlog - 25-11-2011 at 16:22
Since the question asks for an amount absorbed, and heat was absorbed, I think it makes little sense to look for a negative quantity. At best your
teacher is being a pedant, at worst he's wrong (which is what I think).francis - 25-11-2011 at 22:05
Yep, I was thinking of when we did a similar lab a while ago, we worked out the calorimeter constant from the heat absorbed by the calorimeter, then
used the Ccal to find enthalpies of neutralisation.
Nah you shouldn't need the calorimeter constant, as long as you have your calculations right.
So for the hot water, using q = m*C*deltaT
(C is the heat capacity of water, deltaT is the difference in temperature between the final solution and the hot or cold water)
q = 50g * (4.184 J/g/degC) * (60 deg - temp final solution)
(call this q hot)
For the cold water, same deal:
q = 50g * (4.184J/g/degC) * (Temp final solution - 20deg C)
Call this q cold
The heat absorbed by the calorimeter (Q cal) is the difference between the two (qcold and qhot), ie
qcold + q hot = Q cal
-------
From the questions, I guess Tfinal = 39 deg C (avg of the two, minus 1 deg).
I worked out the Qcal as the magnitude of heat gained by the calorimeter as 418J, using the approximate numbers you gave.
Maybe your teacher is talking about your working specifically:
In your calculations, does the heat lost by the hot water have a negative sign?
Does the heat gained by the cold, have a positive?
[Edited on 26-11-2011 by francis]
[Edited on 26-11-2011 by francis]White Yeti - 26-11-2011 at 10:20
Interesting approach, but there's an error in your first equation set up. DeltaT is final minus initial, not the opposite.
Other than that, I'm surprised that my final answer (using your approach) is negative O.O
The strange thing is that there are two approaches to find the answer to this problem, both of which are legitimate and both of which yield the same
answer that is either positive or negative.
Here's my approach that yielded a positive answer. I'm not saying one is better than the other, both are legitimate:
I used the difference between the expected average temperature of the water and the actual temperature. In my case, the exact ∆T is -0.995C. I
then plugged this ∆T into the q = mC∆T equation and used 100g for the mass of water.
I called this value "qwater", and it's obviously negative.
Since the heat lost by the water must equal the "heat gained by the calorimeter":
qwater=-qcal
The value for qwater was negative, but the value for qcal ended up positive...
This problem is really getting on my nerves, it's due on Monday -.-
