Reducing Negative Heat Loss in the Lab: Qfinal vs. Qinitial

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SUMMARY

The forum discussion centers on the calculation of heat loss in a calorimetry experiment involving hot and cold water. Participants clarify that the formula used, Qlost = Qinitial - Qfinal, should yield a positive Qlost, indicating energy loss. The confusion arises from the misconception that Qfinal should exceed Qinitial; however, due to energy loss in a non-ideal calorimeter, Qfinal is indeed smaller than Qinitial. The discussion emphasizes the importance of proper temperature measurement and the necessity of using consistent units, such as converting to Kelvin.

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nmsurobert
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<< Moderator note: This thread is missing the homework template due to originally being posted in another forum. >>

hello everyone, this is my first post and I am looking for some help.
in my lab we did an experiment to calculate specific heat values for various metals using a calorimeter.
the first thing we did was calculate heat lost by adding hot water to some cold water then plugging numbers into formulas and calculating Qlost = Qintial - Qfinal
apparently our Qlost should've been a positive value and we calculated a negative value.
i don't understand why this is wrong. I would think that Qfinal should be larger than Qintial because we arent we adding energy to the system? It doesn't make sense to me why Qfinal should be smaller than Qinitial.
any input would be much appreciated!
 
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How do you calculate Qinital and Qfinal?
 
nmsurobert said:
hello everyone, this is my first post and I am looking for some help.
in my lab we did an experiment to calculate specific heat values for various metals using a calorimeter.
the first thing we did was calculate heat lost by adding hot water to some cold water then plugging numbers into formulas and calculating Qlost = Qintial - Qfinal
apparently our Qlost should've been a positive value and we calculated a negative value.
i don't understand why this is wrong. I would think that Qfinal should be larger than Qintial because we arent we adding energy to the system? It doesn't make sense to me why Qfinal should be smaller than Qinitial.
any input would be much appreciated!
Your measurement is in hot water , in cold water, or in the metal ? And why you adding hot water to cold water and not directly to the metal ?
 
Maybe I should've left out the part about the metal. That was the end goal but first thing to do was calculate heat loss. we had some cold water in the calorimeter then added the the hot water. so we were calculating heat lost with only the cold and the hot water.
These are the two formulas we're using. Letters without subscripts are for hot water, w subscripts is for cold water, c subscript is for the cup.
m = mass, c = specific heat, T = temp.
948B483B-233D-4186-9A56-302E16529E37.png
 
The equations look fine. Show us your working.

It doesn't make sense to me why Qfinal should be smaller than Qinitial.

If everything was perfectly insulated Qfinal would equal Qinitial. That's because all you are doing is mixing two lots of water together. No energy can escape the system if the insulation is perfect.

However the calorimeter doesn't have perfect insulation so some energy is lost during the mixing process. This means that Qfinal is smaller than Qinitial.
 
Last edited:
here is my work.
F6C28DAE-B2C9-467C-86E9-8E0131863F6A.jpg
 
Earlier you wrote..

Letters without subscripts are for hot water, w subscripts is for cold water, c subscript is for the cup.

Which means:

M = mass of the hot water
Mw = mass of cold water

Your equation for Qinitial has the term MCT1 so T1 must be the temperature of the hot water.

So why do you have both T1 and TW = 21.8C in the second line of your working? Was the hot and cold water at the same temperature initially?

I believe you should also be working in Kelvin.

error.jpg
 
Last edited:
ah good catch. that still leaves me with -107 cal though. but atleast its not -4k haha. i think it should be in kelvin is my heat scale is in joules but I am using calories so i don't have to convert masses to kg and temps to kelvin.
 
ill convert everything to joules though if you think itll make a difference. i figured though because all of units matched up then the temp scale didnt matter.
 
  • #10
It's fine to use calories and grams because you have the specific heat capacity of the materials (eg the water) in those units.

However if you use Celsius the absolute amount of heat in something at a temperature of -10C would be negative. That's not possible.
 
  • #11
so i should convert everything to kelvin then? ill do that in a bit then ill get to you.
 
  • #12
I haven't checked if these are the only errors.

What was the hot water temperature?
 
  • #13
80 Celsius
 
  • #14
so converting temps over to kelvin still leaves a negative heat lost. which now that i think about it is what shouldve been expected all i was doing was making the numbers bigger lol.

so we obviously messed up somewhere. but i don't understand what it means to have a negative heat lost. why should the value be positive?
 
Last edited:
  • #15
I also got a negative answer.

Do you understand why the answer should be zero if the calorimeter has perfect (ideal) insulation?

A positive answer would imply that the system had gained energy from somewhere (Qfinal > QInitial). That's not possible because the first equation adds up all the energy before they are mixed. Mixing them in the calorimeter cannot add energy (unless the calorimeter has a heater that was accidentally switched on?).

One possibility is that the mixture wasn't stirred properly? So the 45C final temperature was biased towards the hot water temperature ?
 
  • #16
Might be constructive to work out what the final temperature should be if Qfinal = QInitial.
 
  • #17
CWatters said:
I also got a negative answer.

Do you understand why the answer should be zero if the calorimeter has perfect (ideal) insulation?

A positive answer would imply that the system had gained energy from somewhere (Qfinal > QInitial). That's not possible because the first equation adds up all the energy before they are mixed. Mixing them in the calorimeter cannot add energy (unless the calorimeter has a heater that was accidentally switched on?).

One possibility is that the mixture wasn't stirred properly? So the 45C final temperature was biased towards the hot water temperature ?

ahhh now that makes sense. i figured my math was right but i was having trouble understanding the concept. thank you making it clear. out of 4 groups, all 4 groups got negative numbers. the instructor was going to try the experiment this weekend and see what she got. our lab equipment is not the most up-to-date lol
 
  • #18
Will be interesting to see what she gets. Certainly sounds like there was a procedural error somewhere.
 
  • #19
thanks again for all your help. i have a great instructor but I've been in her office all semester, thought id bug the internet for a change lol.
 

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