Lost Heat. When q hot + q cold ≠ 0 We have made the assumption that all heat is transferred,...
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Transcript of Lost Heat. When q hot + q cold ≠ 0 We have made the assumption that all heat is transferred,...
![Page 1: Lost Heat. When q hot + q cold ≠ 0 We have made the assumption that all heat is transferred, whether that’s objects or reactions. This is a ridiculous.](https://reader035.fdocuments.in/reader035/viewer/2022062519/5697bfda1a28abf838cb02a8/html5/thumbnails/1.jpg)
Lost Heat
![Page 2: Lost Heat. When q hot + q cold ≠ 0 We have made the assumption that all heat is transferred, whether that’s objects or reactions. This is a ridiculous.](https://reader035.fdocuments.in/reader035/viewer/2022062519/5697bfda1a28abf838cb02a8/html5/thumbnails/2.jpg)
When qhot + qcold ≠ 0
We have made the assumption that all heat is transferred, whether that’s objects or reactions.
This is a ridiculous assumption:
No container perfectlyinsulates.
![Page 3: Lost Heat. When q hot + q cold ≠ 0 We have made the assumption that all heat is transferred, whether that’s objects or reactions. This is a ridiculous.](https://reader035.fdocuments.in/reader035/viewer/2022062519/5697bfda1a28abf838cb02a8/html5/thumbnails/3.jpg)
How much?
Finding out how much heat is lost is actually pretty easy:
qhot + qcold + qlost = 0
Or
qreaction + qwater + qlost = 0
![Page 4: Lost Heat. When q hot + q cold ≠ 0 We have made the assumption that all heat is transferred, whether that’s objects or reactions. This is a ridiculous.](https://reader035.fdocuments.in/reader035/viewer/2022062519/5697bfda1a28abf838cb02a8/html5/thumbnails/4.jpg)
How much?
Now substitute:
qhot + qcold + qlost = 0mcT + mcT + qlost = 0
Or
qreaction + qwater + qlost = 0H*moles + mcT + qlost = 0
![Page 5: Lost Heat. When q hot + q cold ≠ 0 We have made the assumption that all heat is transferred, whether that’s objects or reactions. This is a ridiculous.](https://reader035.fdocuments.in/reader035/viewer/2022062519/5697bfda1a28abf838cb02a8/html5/thumbnails/5.jpg)
How much?
Now substitute:
qhot + qcold + qlost = 0mcT + mcT + qlost = 0
Or
qreaction + qwater + qlost = 0H*moles + mcT + qlost = 0
If you measure everything else, qlost can be found.
![Page 6: Lost Heat. When q hot + q cold ≠ 0 We have made the assumption that all heat is transferred, whether that’s objects or reactions. This is a ridiculous.](https://reader035.fdocuments.in/reader035/viewer/2022062519/5697bfda1a28abf838cb02a8/html5/thumbnails/6.jpg)
For Example
Mix 15g of hot water (77oC) with 22g of cold water (22oC). If the final temperature is 36oC, then…
qhot + qcold + qlost = 0
![Page 7: Lost Heat. When q hot + q cold ≠ 0 We have made the assumption that all heat is transferred, whether that’s objects or reactions. This is a ridiculous.](https://reader035.fdocuments.in/reader035/viewer/2022062519/5697bfda1a28abf838cb02a8/html5/thumbnails/7.jpg)
For Example
Mix 15g of hot water (77oC) with 22g of cold water (22oC). If the final temperature is 38oC, then…
qhot + qcold + qlost = 015*4.184*(-39) + 22*4.184*16 + qlost = 0
![Page 8: Lost Heat. When q hot + q cold ≠ 0 We have made the assumption that all heat is transferred, whether that’s objects or reactions. This is a ridiculous.](https://reader035.fdocuments.in/reader035/viewer/2022062519/5697bfda1a28abf838cb02a8/html5/thumbnails/8.jpg)
For Example
Mix 15g of hot water (77oC) with 22g of cold water (22oC). If the final temperature is 38oC, then…
qhot + qcold + qlost = 015*4.184*(-39) + 22*4.184*16 + qlost = 0
qlost = 975 J
The hot water produced 2448 JThe cold water absorbed 1473 J
![Page 9: Lost Heat. When q hot + q cold ≠ 0 We have made the assumption that all heat is transferred, whether that’s objects or reactions. This is a ridiculous.](https://reader035.fdocuments.in/reader035/viewer/2022062519/5697bfda1a28abf838cb02a8/html5/thumbnails/9.jpg)
Fair Comparisons
Mix 15g of hot water (77oC) with 22g of cold water (22oC). If the final temperature is 38oC, then…
qhot + qcold + qlost = 015*4.184*(-39) + 22*4.184*16 + qlost = 0
qlost = 975 J
Just like with reactions, comparing qlost probably isn’t fair unless we standardize the number. In this case, we’ll do it per degree (assuming the container also started at 22oC):
975 J / 16oC = 61 J/oC = Ccal
![Page 10: Lost Heat. When q hot + q cold ≠ 0 We have made the assumption that all heat is transferred, whether that’s objects or reactions. This is a ridiculous.](https://reader035.fdocuments.in/reader035/viewer/2022062519/5697bfda1a28abf838cb02a8/html5/thumbnails/10.jpg)
What is Ccal?
Just like with reactions, if qlost / T = Ccal
then
qlost = Ccal*T
Ccal is a measure of how well a container insulates–’for every degree that the contents heat up, how much heat is lost?’
(or, ‘for every degree that the contents cool down, how much extra heat the container provide?’)
![Page 11: Lost Heat. When q hot + q cold ≠ 0 We have made the assumption that all heat is transferred, whether that’s objects or reactions. This is a ridiculous.](https://reader035.fdocuments.in/reader035/viewer/2022062519/5697bfda1a28abf838cb02a8/html5/thumbnails/11.jpg)
Why cal?The ‘cal’ is short for ‘calorimeter’, which essentially means “a container that you do a heat transfer experiment in”.
A perfect calorimeter would have Ccal = 0 J/oC
Note: this value covers all the heat lost in your setup to everything—cup, air, thermometer, zombies—so long as you keep your setup reasonably the same.
![Page 12: Lost Heat. When q hot + q cold ≠ 0 We have made the assumption that all heat is transferred, whether that’s objects or reactions. This is a ridiculous.](https://reader035.fdocuments.in/reader035/viewer/2022062519/5697bfda1a28abf838cb02a8/html5/thumbnails/12.jpg)
SummaryHeat lost is just another q
Find it by measuring all the other q values
q = Ccal*T
There are now three choices for what to substitute in for q: object, reaction, or calorimeter.