WEBVTT
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This is College Physics
Answers with Shaun Dychko
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An aluminum bowl with some
soup is placed in a freezer,
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and the soup and bowl have an initial
temperature of 25 degrees Celsius.
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And we’re told that 377 kiloJoules
of energy are taken out of the soup,
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which is 377 times 10 to the 3 Joules.
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And the question is “what is the
final temperature of the soup?”
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So, we have three heat
transfer terms to consider:
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one is the heat transfer to change
the temperature of the soup and bowl
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from 25 degrees Celsius to zero;
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and then once it’s at zero,
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we have the heat transfer to freeze
the water from a liquid into a solid,
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and that’s the heat of fusion;
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and then after that,
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we have a further heat transfer to reduce
the temperature from zero degrees Celsius
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to whatever the final temperature
is that we’re going to calculate.
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So that’s the total energy taken
out, which we’re told is this much.
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So some information that we,
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I need to research here is the
latent heat of fusion for water
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which is 334 times 10 to
the 3 Joules per kilogram.
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The specific heat of aluminum is 900
Joules per kilogram per Celsius degree.
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And the specific heat of the soup, we’re
told to assume is the same as that of water,
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which is 4186 Joules per
kilogram per Celsius degree.
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And the specific heat of ice,
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which will be needed for this
final heat transfer term here,
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where we’re changing the temperature
of the frozen soup and bowl
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from zero to whatever final temperature is.
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The specific heat of ice will be this much.
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Okay, and we’re told that there’s .25
kilograms of aluminum in the bowl,
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and .8 kilograms of soup.
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So, the first step is to substitute
for the heat transfer
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to change temperature from
25 degrees Celsius to zero.
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Now, a small oddity is that I’ve
reversed these temperature terms.
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Normally, you have final minus initial,
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and the final temperature is zero,
and the initial temperature is 25.
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But I’ve reversed them around
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because we want to have the heat
transferred out of the soup.
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This *Q* formula, you know,
*Q* equals *mc delta T*,
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is the amount of heat
transferred into something.
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And when heat is transferred
out, it becomes negative.
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And when it becomes negative,
we can have, you know,
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it’s negative *mc T final* minus *T
initial*, if you write it the proper way.
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But this can be instead written as
positive *mc* times *T initial* minus *T final*.
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And so, you can make this negative *Q*,
you can write it instead like this:
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you have *mc* times the temperatures
with the initial written first,
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and the final written second.
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And so that’s why this is written that way.
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So, we’re making this whole
term positive in other words.
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And then we have the heat of fusion:
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the mass of the soup times the
latent heat of fusion for water.
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And then add to that the
mass of the aluminum
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times the specific heat of aluminum
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plus the mass of the soup times
the specific heat of ice,
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as we go from zero degrees to some final
temperature that we will calculate.
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So, we distribute this
bracket into this one here,
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which conveniently turns to
only two terms because...
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or one if depending on how you look at it
because multiplied by zero makes zero.
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And so we have this term for
the change in temperature
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from 25 degrees Celsius
to zero written here,
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and that’s a minus because I
subtracted it from both sides
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so I’m taking this to the left hand side,
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and then this term also is
going to the left hand side
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or we’re subtracting it from
both sides, you might say.
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And then we’re left with
this, by itself, on one side
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and this bracket here
multiplied by zero makes zero.
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And this bracket is getting
multiplied by negative *T final*.
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So it’s negative *T final*.
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And then I switch the sides
around as well after all that
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in order to have the unknown on the left.
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Okay, so we have negative *T final*
times all this written here.
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And then equals the total energy
transferred from the soup
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minus each of these two terms here.
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Okay.
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And then, we’ll solve for *T final*.
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We’ll divide both sides by this bracket here.
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So that’s divided by negative *m al C
al*, plus *m soup C ice*, both sides.
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Now, this negative makes each of the
terms in the numerator change the sign.
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And so that’s why we have
a positive *ms Lf* here,
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and a positive bracket times *T
i*, and then the *Q t* is negative.
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Then on the bottom we’ve written positives.
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Okay, then we plug in numbers
and get our answer.
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So the final temperature is:
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the quarter kilograms of aluminum
multiplied by specific heat of aluminum;
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plus the .8 kilograms of soup
multiplied by specific heat of water,
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while it’s in the water liquid state;
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times 25 Celsius degrees ---
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change in temperature from 25 to zero;
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plus the heat of fusion,
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which is the mass of the soup multiplied
by the latent heat of fusion for water;
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and then subtract from that the
total energy taken from the soup;
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and then divide by the mass of aluminum
times specific heat of aluminum
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plus the mass of the water
times the specific heat of ice.
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And end up with a final temperature
of negative 10.8 degrees Celsius.
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And there we go.