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Current time:0:00Total duration:11:23

Let's do another states of
matter phase change problem. And we'll deal with
water again. But this one hopefully will
stretch our neurons a little bit further. So let's say I have 500
grams of water. Of liquid water. At 60 degrees Celsius. Now my goal is to get it to
zero degrees Celsius. And the way I'm going to do it
is, I'm going to put ice into this 500 grams of water. And my ice machine at home makes
ice that comes out of the machine at minus 10
degrees Celsius ice. And my question is, exactly
how much ice do I need? So how much, or how
many grams of ice? And I'm going to take the ice
out of the freezer and just plop it into my liquid. How much do I need to bring this
liquid, this 500 grams of liquid water, down
to zero degrees? So the idea, if we just
imagine a cup here. Let me draw a cup. This is a cup. I have some 60 degree
water in there. I'm going to plunk a big
chunk of ice in there. And what's going to happen is
that heat from the water is going to go into the ice. So the ice is going to absorb
heat from the water. So in order for water to go
from 60 degrees to zero degrees, I have to extract
heat out of it. And we're about to figure
out just how much heat. And so we have to say, whatever
was extracted out of the water, essentially has to
be contained by the ice. And the ice can't get
above zero degrees. Essentially, that amount of
ice has to absorb all that heat to go from minus
10 to zero. And then also, that energy will
be used to melt it a bit. But if we don't have enough ice,
then the ice is going to go beyond that and then
warm up even more. So let's see how we do this. So how much energy do we have
to take out of the 500 grams of liquid water? Well, it's the same amount of
energy that it would take to put into zero degrees
liquid water and get it to 60 degrees. So we're talking about
a 50 degree change. So the energy or the heat out
of the water is going to be the specific heat of water,
4.178 joules per grams Kelvin. And I have to multiply that
times the number of grams of water I have to cool down, I
have to take the heat out of. And we know that's 500 grams.
And then I multiply that times the temperature differential
that we care about. And just a side note, I use
this specific heat because we're dealing with
liquid water. Liquid water going
from 60 to zero. So the final thing, I have to
multiply it by the change in temperature. The change in temperature
is 60 degrees. Times 60 degrees. There's a little button on the
side of my pen, when I press it by accident sometimes it
does that weird thing. So let's see what this is. So this is 4.178 times 500 times
change of 60 degrees. It could be a change of 60
degrees Kelvin or a change of 60 degrees Celsius. It doesn't matter. The actual difference is the
same, whether we're doing Kelvin or Celsius. And it's 125,340 joules. So this is the amount of heat
that you have to take out of 60 degree water in order to get
it down to zero degrees. Or the amount of heat you have
to add to zero degree water to get it to 60 degrees. So essentially, our ice has
to absorb this much energy without going above
zero degrees. So how much energy can
the ice absorb? Well let's set a variable. The question is how much ice. So let's set our variable,
maybe we'll call it I. Let's do x. x is always
the unknown variable. So we're going to have
x grams of ice. OK, and it starts at
minus 10 degrees. So when this x grams of ice
warms from minus 10 degrees to zero degrees Celsius, how much
energy will it be absorbing? So to go from minus 10 degrees
Celsius to zero degrees Celsius, the heat that's
absorbed by the ice is equal to-- is equal to the specific
heat of ice, ice water, 2.05 joules per gram Kelvin, times
the amount of ice. That's what we're solving for. So times x. Times the change
in temperature. So this is a 10 degree change
in Celsius degrees, which is also a 10 degree change
in Kelvin degrees. We can just do 10 degrees. I could write Kelvin here, just
because at least when I wrote the specific heat units,
I have a Kelvin in the denominator. It could have been a Celsius,
but just to make them cancel out. This is, of course, x grams.
So the grams cancel out. So that heat absorbed to go from
minus 10 degree ice to zero degree ice is 2.05
times 10 is 20.5. So it's 20.5 times x joules. This is to go from minus 10
degrees to zero degrees. Now, once we're at zero degrees,
the ice can even absorb more energy before
increasing in temperature as it melts. Remember, when I drew that
phase change diagram. The ice gains some energy
and then it levels out as it melts. As the the bonds, the hydrogen
bonds start sliding past each other, and the crystalline
structure breaks down. So this is the amount
of energy the ice can also absorb. Let me do it in a
different color. Zero degree ice to zero--
I did it again-- to zero degree water. Well the heat absorbed now
is going to be the heat of fusion of ice. Or the melting heat,
either one. That's 333 joules per gram. It's equal to 333.55 joules per
gram times the number of grams we have. Once
again, that's x grams. They cancel out. So the ice will absorb 333.55
joules as it goes from zero degree ice to zero
degree water. Or 333.55x joules. Let me put the x there,
that's key. So the total amount of heat
that the ice can absorb without going above
zero degrees. Because once it's at zero degree
water, as you put more heat into it, it's going to
start getting warmer again. If the ice gets above zero
degrees, there's no way it's going to bring the water
down to zero degrees. The water can't get above
zero degrees. So how much total heat
can our ice absorb? So heat absorbed is equal to the
heat it can absorb when it goes from minus 10 to
zero degrees ice. And that's 20.5x. Where x is the number of grams
of ice we have. Plus the amount of heat we can absorb as
we go from zero degree ice to zero degree water. And that's 333.55x. And of course, all of
this is joules. So this is the total amount of
heat that the ice can absorb without going above
zero degrees. Now, how much real energy
does it have to absorb? Well it has to absorb all of
this 125,340 joules of energy out of the water. Because that's the amount of
energy we have to extract from the water to bring it down
to zero degrees. So the amount of energy
the ice absorbs has to be this 125,340. So that has to be equal
to 125,340 joules. We can do a little bit
of algebra here. Add these two things. 20.5x plus 333.55x is 354.05x. Is that right? Yeah, 330 plus 30 is 350. Then you have a 3 with
a 0.5 there. 354.05x and that is equal to the
amount of energy we take out of the water. You divide both sides. So x is equal to 125,340
divided by 354.05. I'll take out the calculator
for this. The calculator tells me 125,340,
the amount of energy that has to be absorbed by the
ice, divided by 354.05 is equal to 354 grams. Roughly;
there's a little bit extra. So actually, just to be careful
maybe I'll take 355 grams of ice. Because I definitely want
my water to be chilled. So our answer is x is equal
to 354.02 grams of ice. So this is interesting. I had 500 grams of liquid. And you know, intuitively I
said, well if I have to bring that down to zero degrees, I'd
have to have a ton of ice. But it turns out, I only need,
what was the exact number? So the liquid is 500 grams.
About, roughly half a pound, if you want to get a sense for
how much 500 grams is. A kilogram is 2.2 pounds. So this ice is 354. So you actually have to have
less ice than water. Which is interesting, because
it seems like the ice isn't making that big of a
temperature change. The ice is only going from
minus 10 to zero degrees. While the water is going all
the way from 60 degrees to zero degrees. So you're like, how does
that work out? And the reason is because so
much energy can be absorbed by the ice in the form of potential
energy as it melts. So if you talk about all the
energy that the ice is asborbing, most of it is
due to the heat of fusion right there. You can actually have zero
degree ice and it can still absorb this huge amount of
energy just by melting it. Without even changing
its temperature. Anyway, hopefully you found
that reasonably useful.