### Video Transcript

A student investigated how much
energy from the Sun was incident on Earth’s surface at his location. He placed an ice cube that had a
temperature of zero degrees Celsius in an insulated tray under direct sunlight and
measured the time it took for the ice cube to completely melt. The apparatus he used is shown in
Figure one.

Taking a look at this figure, we
see the apparatus consists of a transparent lid — that is one that lets through
direct sunlight — and insulated tray, on top of which rests our ice cube. The idea here is that our ice cube
which starts out at zero degrees Celsius will have energy from the Sun incident on
it through the transparent lid. This energy contained by the
insulated tray will cause the ice cube to melt and the time that takes is what is
measured.

So that’s our setup. And here’s the first question
regarding this scenario.

Write down the equation that
relates the power of a process to the energy transferred by that process and the
amount of time that the process occurs over.

Let’s clear a little bit of space
on screen to write these three variables. These three quantities are power,
energy, and time. And we want to write down the
mathematical relationship that connects all three of these. To do that, here’s what we can
recall. Considering the first of these
three variables, power, we remember that the unit of power is watts. And one watt is equal to a joule
per second, joule being the unit of energy and second being the SI standard unit for
time.

We now have a relationship that
connects our three terms. Here’s what it looks like written
out: we can say that the power in any particular process is equal to the energy
transferred by that process divided by the time over which the process occurs. And if we write this expression in
shorthand, we can say that 𝑃, power, is equal to 𝐸, energy, divided by 𝑡,
time.

Either one of these expressions is
a valid way of writing the equation that connects the power in a process with the
energy transferred and the time passed.

Next, let’s apply this relationship
to our particular experimental scenario.

The ice had completely melted once
2672 joules of energy had been transferred to it. The time taken for the ice to
completely melt was 15 minutes. Calculate the mean power supplied
by the sun to the ice cube.

What’s taking place here is that
energy from the sunlight — specifically 2672 joules of it — is transferred to the
ice cube, which causes it to melt completely over a time of 15 minutes. Based on this information, we wanna
calculate the mean power supplied by the sun to the ice cube, recalling that power
connects energy and time.

To start off, let’s recall the
equation from our previous section. We saw before that power in units
of watts is equal to energy in units of joules divided by time in units of
seconds. That all looks good. We notice though that the energy
we’ve been given is in units of joules, but the time is in units of minutes not
seconds.

This tells us that when we apply
this equation for power to solve for mean power of the Sun, we want to do units
conversion for our time value. And let’s call this mean power 𝑃
sub avg. And that’s equal to our energy 2672
joules divided by our time 15 minutes.

But notice that our answer is to be
in units of watts and that watts specifically are in units of joules per second. That means that before we calculate
𝑃 sub avg, we want to do a bit of work on our denominator to convert it into a time
in units of seconds.

Here is one way we can do that: we
recall that 60 seconds is equal to one minute. This tells us that if we multiply
our given time 15 minutes by the conversion ratio 60 seconds per minute, we’ll get a
time value in seconds like we want. Indeed, notice from looking at this
denominator that as we multiply these two values together, our units of minutes will
cancel out and we’ll be left with the units of seconds.

So the time for this whole melting
process to take place in the units that we want — seconds — is 15 times 60. And now, we’re ready to calculate
this overall fraction and solve for 𝑃 sub avg. When we do, we find the value of
2.97 watts. We report our answer to three
significant figures in the absence of specific guidance in the problem statement
about the precision of our answer.

So that’s it then. The mean or average power supplied
by the Sun to the ice cube over this 15-minute interval is 2.97 watts.

As a next step, let’s consider just
how massive our ice cube which melts completely is.

The specific latent heat of fusion
for water is 334 kilojoules per kilogram. Calculate the mass of ice the
student used in his investigation. Use the correct equation from the
Physics Equation Sheet.

Before we go ahead to calculate
this mass, let’s recall what has been going on with this ice cube. The ice cube we know started out at
zero degrees Celsius, so in other words, right on the borderline between being a
solid and being a liquid. Then, as energy from the Sun’s rays
is incident on the ice cube, the cube starts to melt. But importantly, its temperature
doesn’t go up.

What we mean is that all of the
energy from the Sun’s rays is used to change the state of the ice from solid to
liquid and none of the energy is used to raise its temperature. Even after it’s melted completely,
its temperature is still zero degrees Celsius.

The amount of energy needed to
change a substance from a solid to a liquid is called the specific latent heat of
fusion for that substance. Of course, in this case, it’s
water. It’s not the most intuitive thing
in the world. But in fact, energy is required per
unit mass simply to change the state of a substance from — in this case — solid to
liquid.

It’s this particular amount of
energy per unit mass for water that we’re going to use to solve for the mass of this
ice cube that has melted. To do that, let’s recall a bit
about the energy transferred by the Sun to the ice cube.

From earlier, we saw that that
energy is 2672 joules. And as we mentioned, this energy is
devoted entirely to melting the ice without raising its temperature. Our specific latent heat of fusion
for water, which we’ll call 𝐿, tells us just how much energy is needed per unit
mass to change the state of water from solid to liquid.

We see then that energy, mass, and
specific latent heat of fusion are all connected. And just by looking at the units of
the two terms we’ve written down, we can see how they are. 𝐿, the specific latent heat of
fusion, is an energy per mass. So we can say 𝐿 is equal to 𝐸
divided by 𝑚.

And since the question wants us to
solve for a mass value, let’s rearrange this equation to solve for 𝑚. We find that 𝑚, the mass of our
ice cube, is equal to 𝐸, the energy from the Sun, divided by 𝐿, the specific
latent heat of fusion for water.

Since we’re given both 𝐸 and 𝐿 in
the problem statement, we seem to be in the home stretch, except for one issue. And as you might guess, it’s a
units’ inconsistency issue. See that our energy is in units of
joules, while our specific latent heat of fusion is in units of kilojoules per
kilogram.

To make these units consistent,
let’s convert our energy into units of kilojoules from units of joules. Recalling that one kilojoule is
equal to 1000 joules. That must mean that 2672 joules is
equal to 2.672 kilojoules. And now, we can change the
numerator in our fraction to reflect this.

Notice now what happens to the
units in this fraction. The units of kilojoules cancel out
and the units of kilograms move all the way up to the numerator. We see then that we’re indeed
calculating a mass value. And when we solve for this
fraction, we find a result of 0.008 kilograms. This is the mass of ice that will
be possible to melt after applying a given amount of energy to it.

Next, as a last step, let’s spend
some time thinking about the design of this experiment.

The student’s results can only be
used as an estimate of the mean power at his location. Give one reason why.

Let’s look back over at Figure one
and recall again how this experiment works. We know the idea is that light will
come from the Sun, pass through the transparent lid, and be communicated to the ice
cube, causing it to melt. But here’s the thing: not all of
the energy that passes through the transparent lid will make it to the ice cube. Some of that energy will go into
the insulated tray. And this tray that was indeed
insulated isn’t a perfect insulator. It will transfer some energy to its
surroundings.

So that’s one reason that the
student’s results are estimate results. We know that there is some energy
lost involved through the insulated tray. And that’s not the only reason we
can think of that the student’s results can be used as an estimate of the mean
power.

Think, for example, of the sunlight
that comes in over the time interval of the cube melting, which we know is 15
minutes. We know that sunlight because the
Earth and the Sun are constantly in motion relative to one another changes over
time. It doesn’t stay the same. For example, after 15 minutes of
time, the angle at which the sunlight hits the transparent lid will be
different.

This difference will mean that the
instantaneous power on the ice cube from the sun will change over the 15-minute time
period. That then is another reason why the
calculated results are an estimate for the mean power. And in fact, it’s not just the
angle of solar radiation that will change over this time, there is another factor
that changes too.

Imagine you go outside in the
middle of the day on a bright sunny day in summer time. What does the sunlight feel
like? It’s very intense, right? Now, let’s say that we fast forward
time ahead a number of hours until sunset, what’s the quality of sunlight then? We know that the angle of the rays
of light is different at sunset and noon, but also the intensity of the light is
different.

If we draw a sketch of the Earth
and then Earth’s atmosphere surrounding the planet, if we’re standing at this
particular location on Earth, we see that at the noon hour, sunlight is shining
directly down on us and therefore passes through the smallest sliver of Earth’s
atmosphere, while on the other hand at sunset the Sun has to pass through a
significant section of atmosphere to reach us. And that atmosphere is constantly
damping the intensity of the light.

So we can say this about sun
intensity over time. We can write that the intensity of
the solar radiation may have varied during the investigation. And really, that’s putting it
lightly; it’s almost inevitable that the intensity will have varied some.

Since we were only asked to give
one reason why the student’s results should be treated as estimates, we wouldn’t
need to write all three of these answers we’ve discovered down. That said, any one of these three
would be sufficient to satisfy the question.