WEBVTT
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This is College Physics
Answers with Shaun Dychko.
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A pool with 80,000 liters of water is heated
up by one and a half Celsius degrees.
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The question is, how much heat was
transferred into the water of the pool?
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So, that heat is given by the mass of water
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times the specific heat capacity of the
water times the change of temperature.
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Now, we're not told of the mass of
the water as we're given the volume,
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so we need to use the density
formula to figure out that mass.
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So, the density is mass per volume.
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And then, multiply both sides by *V*
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and we get a formula for *M* after
we switch the sides around.
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So, mass of the water is the
water's density times its volume.
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So, we substitute that, and I show
that in red, in place of letter *M*
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and rewrite this formula
for the heat transfer.
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And so, the density of water is one
thousand kilograms per cubic meter
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times 80,000 liters, and since
our density is per cubic meter,
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we need to express her
volume also in cubic meters.
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And so, we multiply this 80,000 liters by
one cubic meter for every 1000 liters.
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And then, we multiply by
the specific heat of water,
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4186 Joules per kilogram
per every Celcius degree,
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and then multiply by the
change in temperature,
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which is an increase of one
and a half Celsius degrees.
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This gives an energy transfer of
5.02 times ten to the 8 Joules.