WEBVTT

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This is Giancoli Answers
with Mr. Dychko.

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When ice is placed
within this calorimeter,

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it's going to gain some heat
to increase its temperature

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from its initial temperature
of negative 8.5 degrees Celsius up to 0.

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And then once it gets to 0,

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it's gonna absorb
some more heat energy

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in order to change its phase
from solid into the liquid,

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and so that's gonna be mass of the ice
times the latent heat of fusion.

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And then furthermore,
it'll increase its temperature

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once it becomes water.

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And it'll absorb
this much energy mass

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times specific heat of water

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times the final temperature
minus the initial temperature of 0

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which is what it has after
it's done its phase change

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from ice into water.

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And I've got <i>m</i> ice here.

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Even though it's water in this term,

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it's the same mass
as it is here and here.

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It's the original mass
of the ice cube.

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And all that heat energy gained
by this ice cube, which turns into water,

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is going to equal
the heat energy lost

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by the aluminum
that makes up the calorimeter container

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and the heat energy lost
by the water within the calorimeter.

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And we can factor out
<i>m</i> ice from all 3 terms here.

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And so we have <i>m</i> ice times 8.5

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times the specific heat of ice
plus the latent heat of fusion

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plus 17
times the specific heat of water

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because it ends up
with a temperature of a,

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a final temperature of 17 degrees.

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So, 17 minus 0 is 17 there.

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And the right hand side is unchanged.

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And then divide both sides
by this bracket.

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And you end up mass
of the ice must be the

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0.095 kilograms of aluminum
that makes up the calorimeter

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times 900 joules
per kilogram Celsius degree,

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specific heat of aluminum,

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times the initial temperature
of the aluminum of 20 degrees Celsius

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minus the final temperature of 17

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plus the 0.31 kilograms
of water within the calorimeter

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times 4,186 joules
per kilogram Celsius degree,

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specific heat of water,

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times 20 minus 17,
it has the same temperature change

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as the aluminum did.

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And divide that
by this whole bracket here.

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So, that's 8.5 times 2,100
which is the specific heat of ice

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plus the latent heat
of fusion for water,

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333 times 10
to the 3 joules per kilogram,

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and then plus 17 Celsius degrees

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times 4,186 joules
per kilogram Celsius degree.

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And that's when the ice
is turned into water

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and it's heating up to its
final temperature 17, starting from 0.

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And this works out to 9.8 grams

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must have been
the mass of the ice cube.