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

In summary, a group of students conducted an experiment to calculate specific heat values of various metals using a calorimeter. They first calculated the heat lost by adding hot water to cold water, but their resulting value was negative instead of positive. They were confused as to why this was wrong, thinking that the final heat should be larger than the initial heat since energy was being added to the system. However, this is not possible as the first equation adds up all the energy before mixing, and mixing cannot add energy unless there is an external source. It was suggested that the mixture may not have been stirred properly, resulting in a biased final temperature. The students eventually understood the concept and found out that all groups got negative values, prompting the instructor to
  • #1
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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  • #2
How do you calculate Qinital and Qfinal?
 
  • #3
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 ?
 
  • #4
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
 
  • #5
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.
 
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  • #6
here is my work.
F6C28DAE-B2C9-467C-86E9-8E0131863F6A.jpg
 
  • #7
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
 
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  • #8
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.
 
  • #9
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?
 
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  • #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.
 

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

1. How can I reduce negative heat loss in my lab?

To reduce negative heat loss in the lab, there are a few steps you can take. Firstly, make sure that all windows and doors are properly sealed to prevent cold air from entering the lab. You can also insulate any exposed pipes or surfaces that may contribute to heat loss. Additionally, using a heating system with a programmable thermostat can help regulate the temperature and prevent excessive heat loss.

2. What is the difference between Qfinal and Qinitial in terms of heat loss?

Qfinal and Qinitial refer to the final and initial heat quantities, respectively. Qfinal represents the amount of heat remaining in a system after heat loss has occurred, while Qinitial represents the initial amount of heat in the system before any heat loss has occurred.

3. How does negative heat loss impact experiments in the lab?

Negative heat loss can significantly impact experiments in the lab as it can alter the temperature and conditions required for certain reactions or processes to occur. It can also affect the accuracy and reliability of data collected during experiments, leading to inaccurate results.

4. What are the main causes of negative heat loss in a lab?

The main causes of negative heat loss in a lab include poor insulation, air leaks, and inefficient heating systems. Other factors such as open windows or doors, lack of insulation on equipment, and temperature fluctuations can also contribute to heat loss.

5. How can I measure the amount of negative heat loss in my lab?

To measure the amount of negative heat loss in your lab, you can use a heat flow meter or a thermographic camera. These tools can help identify areas of heat loss and provide a quantitative measurement of the amount of heat being lost. You can also track changes in energy consumption over time to monitor changes in heat loss.

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