Lost Heat

Post on 30-Dec-2015

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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 assumption: No container perfectly insulates. How much?. Finding out how much heat is lost is actually pretty easy: - PowerPoint PPT Presentation

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Lost Heat

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.

How much?

Finding out how much heat is lost is actually pretty easy:

qhot + qcold + qlost = 0

Or

qreaction + qwater + qlost = 0

How much?

Now substitute:

qhot + qcold + qlost = 0mcT + mcT + qlost = 0

Or

qreaction + qwater + qlost = 0H*moles + mcT + qlost = 0

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.

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

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

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

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

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?’)

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.

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.