### Video Transcript

A woman with a mass of 66.0
kilograms does some vigorous exercise, after which she has a body temperature of
39.0 degrees Celsius. Determine the rate at which she
must cool her body to reduce its temperature to 37.0 degrees Celsius in 3.00 times
10 to the power of three seconds if it continues to be internally heated at a rate
of 110 watts. Use a value of 4180 joules per
kilogram degrees Celsius for the specific heat capacity of the woman’s body.

Okay, so in this question, we’ve
got a woman with a mass of 66.0 kilograms who’s doing some vigorous exercise which
results in the heating up of her body. After her exercise, she has a body
temperature of 39.0 degrees Celsius. What we’ve been asked to do is to
determine the rate at which she must cool her body to reduce this temperature to
37.0 degrees Celsius. And we’ve been given the amount of
time in which she has to do it, 3.00 times 10 to the power of three seconds.

However, there’s a little bit of a
twist. We’ve been told that the body
continues to be internally heated at a rate of 110 watts. This means her body is still
continuing to heat up. And so we’ll have to account for
this when we do our calculations as well. Finally, we’ve been told that the
woman’s body has a specific heat capacity of 4180 joules per kilogram degrees
Celsius. So let’s write down all the
important information in a little corner.

Firstly, the mass of the woman
which we’ll call 𝑚 is 66.0 kilograms. Secondly, the initial temperature
of her body just after she’s finished exercising which we’ll call 𝑇 sub 𝑖 for
initial is equal to 39.0 degrees Celsius. Next, we also know that we want her
final body temperature to be 37.0 degrees Celsius in a time which we’ll call
lowercase 𝑡 of 3.00 times 10 to the power of three seconds.

Next, we also know that her body is
internally being heated at a rate of 110 watts. And we’re calling this 𝑃 sub int
because it’s internal heating and watts is a unit of power. So we’ve been told how much power
her body is expending in order to keep heating itself up. Finally, we also know that her
specific heat capacity is 4180 joules per kilogram degrees Celsius.

What we need to do is to find out
𝑃 sub cool. This is the rate at which her body
must cool in order to reduce her temperature from 39 degrees Celsius to 37 degrees
Celsius in the given time. And because we want to find out the
rate of cooling, this is the same as power. Let’s see why that is.

Well, the reason for this is that
we can recall that power is defined as the rate of energy transfer — in other words,
energy divided by time. And this energy can be various
things depending on the situation. In this case, the energy that we’re
talking about is heat exchange by the woman’s body resulting in either the heating
of her body or the cooling of her body.

Therefore, if we want to calculate
in our final answer the rate at which her body cools, this is the same as the heat
exchange by her body divided by the time taken. And hence, this is a type of
power. So our final answer is going to be
𝑃 sub cool. Okay, so now that that sorted,
let’s get on with solving the problem. So we’re trying to find out how
quickly this woman’s body must give out heat so that her body can cool from 39.0
degrees Celsius to 37.0 degrees Celsius in 3.00 times 10 to the power of three
seconds. The other important thing to note
is that she’s been internally heated at a rate of 110 watts.

And we can also say that 𝑃 sub int
is equal to 𝑄 sub int divided by 𝑡, where 𝑄 sub int is the amount of energy
transferred by her internal heating system — whatever that may be; maybe it’s a
muscles — to her body that’s resulting in the heating up of her body. And of course, 𝑡 is the amount of
time over which this energy is transferred.

So if we rearrange the equation by
multiplying both sides by 𝑡, we find that the energy transferred to her body that’s
resulting in her body heating up is equal to the time for which the body is heating
up multiplied by the power. And this basically is equal to 𝑄
sub int.

Now, if we say on her body diagram
that she needs to give out an energy 𝑄 sub cool in order for her body to cool down,
then we can also say that her body is effectively gaining an energy 𝑄 sub int. Now, of course, this is internal
heating. So it’s basically her muscles that
are heating up her body. But it is the same as if you has
been heated from an external source.

Either way, the result is that the
body is gaining this heat 𝑄 sub int. And thus, the body is being heated
up by this heat. So she needs to give out heat 𝑄
sub cool. And this heat must be much larger
than 𝑄 sub int so that her body can overall cool down.

So once again to clarify, her body
has been heated internally. She has to release this amount of
energy plus a lot more if her body is to cool down properly. So first of all, let’s find out
what the net energy transfer of her body will be if it is to cool down from 39.0
degrees Celsius to 37.0 degrees Celsius in 3.00 times 10 to the power of three
seconds.

To do this, we can recall that the
net energy transfer of an object 𝑄 is given by the mass of that object 𝑚
multiplied by the specific heat capacity of that object multiplied by the change in
temperature of that object Δ𝑇.

Now, we have all the quantities we
need to be able to work out the net heat transfer of her body. So we say that her body’s net heat
transfer is equal to her mass 66.0 kilograms multiplied by her specific heat
capacity 4180 joules per kilogram degrees Celsius multiplied by the change in
temperature of her body which is her final temperature 𝑇 sub f minus her initial
temperature 𝑇 sub i. In other words, Δ𝑇 is equal to 𝑇
sub f minus 𝑇 sub i.

Now, when we evaluate the
right-hand side of this equation here, we find that the heat transfer from her body
is negative 551760 joules. Now, this is the net heat
transfer. And it’s negative because the body
is losing heat. The reason the body must lose heat
overall is because it has to reduce its temperature.

Now, we can find a relationship
between the net energy transfer 𝑄 sub cool and 𝑄 sub int, we can say that the net
energy transfer 𝑄 is equal to the energy gained by her body, which is 𝑄 sub int,
minus the total energy lost by her body, which is 𝑄 sub cool. At this point, we already know what
the net energy transfer is and we’ve got an expression here for 𝑄 sub int.

So first of all, let’s find out
what 𝑄 sub int is. Remember 𝑄 sub int is the internal
heating energy, which is going towards heating up her body. And in a time of 3.00 times 10 to
the power of three seconds, this energy happens to be that time multiplied by 110
watts because remember 110 watts is the power with which her body is being heated up
internally. So we can evaluate this to find
that 𝑄 sub int is equal to 3.30 times 10 to the power of five joules, which means
we officially know the values of 𝑄 and 𝑄 sub int.

So let’s go about finding what 𝑄
sub cool is. Let’s first rearrange the
equation. We find that 𝑄 sub cool is equal
to 𝑄 sub int minus 𝑄. In other words, this ends up being
3.30 times 10 to the power of five, which is 𝑄 sub int, minus 𝑄. But remember 𝑄 is already
negative. So we get a positive sign here. And the value is 551760 and the
unit is going to be in joules. So when we evaluate this, we find
that 𝑄 sub cool is equal to 881760 joules.

So let’s recap exactly what this
means. So the woman’s body has been heated
up by some internal process. And so in a time of 3.00 times 10
to the power of three seconds, an energy 𝑄 sub int is transferred to her body. However, in the same amount of
time, her body is also losing an energy 𝑄 sub cool.

Now, the value of 𝑄 sub cool, the
amount of energy she’s losing, is larger than the value of 𝑄 sub int, the amount of
energy she’s gaining. And so overall, she’s losing
energy. Specifically, her body is losing
551760 joules overall. And this is resulting in a change
in temperature in her body from 39 degrees Celsius to 37 degrees Celsius in this
amount of time.

So what the question actually wants
us to work out is the rate at which she must lose energy, in other words this energy
which she is losing divided by the time over which she loses it. And remember we’ve called this
quantity 𝑃 sub cool. So we say that 𝑃 sub cool is equal
to 𝑄 sub cool, the amount of energy that her body is losing, divided by the time
𝑡.

And remember this is not the
overall energy transfer of her body. We’re now specifically looking at
how much energy her body has to lose in order to be able to compensate for 𝑄 sub
int and also result in a drop in temperature of her body. So anyway, our final answer is
going to be 𝑄 sub cool divided by 𝑡. And at this point, we know what 𝑄
sub cool is. We’ve just worked it out and we
already have been given 𝑡 in the question. Although wouldn’t it be so much
nicer if we were giving tea to drink in the question? However, at this point, we must
make deal with the time 𝑡.

So anyway, 𝑄 sub cool happens to
be 881760 joules and we divide this by 3.00 times 10 to the power of three seconds
cause that’s the amount of time over which her body loses the energy. So we can evaluate this fraction,
remembering of course that joules divided by seconds gives us a unit of watts. And hence, we find that 𝑃 sub cool
is equal to 293.92 watts.

However, all of the quantities that
were given to us in the question were given to three significant figures. So we gave our answer to three
significant figures as well. So here’s the third significant
figure. The next one will tell us whether
the third one rounds up or down. Now, this value is a nine which is
greater than five. So our third significant figure
rounds up. It becomes a four.

And hence, our final answer is that
the rate at which the woman must cool her body is 294 watts.