### Video Transcript

Two students investigated the rate at which water cooled in a glass beaker. They measured the temperature of the water at regular intervals over a 10-minute
period. Figure 1 shows the apparatus that each student used.

Taking a moment to look at Figure 1, we see that it shows us two different
experimental setups, one for student A and one for student B. The setups are identical as far as the glass beaker used and the amount of water in
each beaker. But we do see that there is one important difference between the two setups. Student A is using a digital thermometer whereas student B is using a regular
thermometer. We might call it an analog thermometer. Keeping this difference in mind, let’s move on to our question.

Choose one advantage of using student A’s apparatus. Tick one box. Student A’s measurements were more accurate; Student A’s measurements had a higher
resolution; Student B’s apparatus only measured categorical variables.

Considering these three answer options, let’s look at the last one first, the one
which talks about categorical variables. What are categorical variables after all? This is a variable that describes something in a qualitative rather than quantitative
way. For example, color is a categorical variable. We can describe an object as red or blue or green. And that’s in a qualitative sense.

Now, if that’s what a categorical variable is, then the question is: does student B’s
set up only measure categorical variables? When we look at the up-close view of the thermometer in student B’s setup, we see
that when we read it, we’ll get a quantitative value. That is, we’ll get a number. This tells us that actually student B’s apparatus doesn’t only measure categorical
variables. It gives at least one quantitative variable, the temperature of the water. We won’t choose our third answer option then. So let’s move on to another.

What about the first option, that student A’s measurements were more accurate? We can remember that the accuracy of a measurement or a result goes up as that
measurement gets closer to the true value of some quantity. In other words, to know just how accurate a measurement is, we need to have a
reference point or a standard to measure it against. We need to know the true value of that quantity.

In our case, we don’t know the actual temperature of the water. We don’t have an absolute reference to tell us that. All we have is student A’s measurement of the temperature and student B’s
measurement. Without a reference to compare to these two values, we don’t know which student is
making a more accurate measurement. This tells us that we can’t choose the first option as one advantage of student A’s
apparatus.

This leaves us with the second option which says that student A’s measurements had a
higher resolution. If we look at student B’s setup, we see that on this thermometer, there are
marked-out large hash marks every five degrees Celsius. And then, in between those larger markings, we see one, two, three, four smaller hash
marks which indicate a change of one degree Celsius. This means that student B has a measurement apparatus with a resolution of one degree
Celsius. If the temperature of some object was in between degrees, we would just have to round
our result to the nearest whole degree Celsius.

This is in contrast to student A’s set up which includes a digital thermometer that
we can see reports temperatures to the nearest one one hundredth of a degree. This tells us that student A’s apparatus has 100 times the temperature resolution as
student B’s apparatus. This second option then, that student A’s measurements have a higher resolution, is
accurate. That’s true based on the difference between the type of thermometer used by student A
and student B. Let’s move on now to consider a bit about student B’s experimental process.

Student B measured the temperature of the water every 30 seconds. For the time between each reading, student B took the thermometer out of the water
and placed it on the table. When taking each reading, she took the reading as soon as the thermometer was in the
water. Explain why doing this would cause measurement errors.

Here’s what’s happening here. Student B, as we know, is putting a thermometer in the water and taking a measurement
of the temperature. Right after she takes the measurement, we’re told that she takes the thermometer out
of the water and sets it on the table by the beaker. 30 seconds later, the thermometer goes back in the beaker of water and a reading of
the temperature is taken immediately. The question is: why would this experimental procedure lead to measurement
errors?

To start off, let’s consider the difference between the environments of the water in
the beaker and the air in the room around the beaker. We saw earlier that the water in the beaker had a temperature of about 56 degrees
Celsius. On the other hand, the temperature in the room, if it’s normal indoor air
temperatures, will be around 20 degrees Celsius. In other words, there is a temperature difference between the water and the air
surrounding the beaker. This means that as our thermometer is put into and then taken out of the beaker over
and over, it will constantly be adjusting to the temperature of its environment. When it’s out on the table, it will adjust to the temperature of the air. And when it’s back in the water, it will adjust to that temperature.

Moreover, we know that this thermometer won’t be able to immediately reflect the
temperature of its environment. It takes a few seconds for the thermometer to register the new temperature of what
it’s in. So picture this, after about 30 seconds of the thermometer adjusting to the
temperature of the air, it’s put back into the water and the temperature reading is
taken immediately afterward. That is before the reading on the temperature would actually be able to adjust to its
new environment. This is why this experimental method would lead to errors. And here is what we can write about it.

We can first say that when the thermometer is out of the water, it will cool to the
ambient temperature of the room. And as we said, this is in the neighbourhood of 20 degrees Celsius. Then, when the thermometer is returned to the water, it takes a few seconds to adjust
to the water temperature. So if the measurement of temperature is taken immediately, and we’re told that it is,
then it won’t be accurate. It will be somewhere in between the temperature of the air and the temperature of the
water. This is the reason why this particular experimental procedure would lead to
measurement error. Let’s move on now to consider some of the data resulting from this experiment.

Student A’s results are shown in Figure 2.

Let’s take a look at Figure 2. And we see that it shows the temperature in degrees Celsius of the water in the
beaker versus the time in seconds at which the measurement was made. When we look at the curve of the overall measured results, we see that it has a
downward gradient to it. That makes sense because the water is cooling as it approaches room temperature. Now on to our question.

What was the decrease in temperature between zero and 100 seconds. Tick one box. 16 degrees Celsius, 40 degrees Celsius, 48 degrees Celsius, 42 degrees Celsius.

To figure out the answer to this question, we’ll want to know the temperature of the
water at these two times, at zero seconds and 100 seconds. Going to the horizontal axis of our plot, if we locate these two times on that axis,
the next thing we want to do is solve for the water temperature at a time of zero
seconds and then at a time of 100 seconds. We’ll do this by seeing where these two time values intersect with our data
curve.

To find that intersection point, we trace a vertical line up from these time
values. From zero seconds, we find it intersects the curve there. And for 100 seconds, we marked that point of intersection. To find out what the corresponding temperatures of these two points on our graph are,
we’ll trace lines over horizontally until we run into our vertical axis. We see that in the case of our first data point, at zero seconds, that isn’t
necessary because the intersection point is already on the vertical axis. Our job then is to interpret the temperature of this particular point on the
curve. It seems to be around 93 or 94 degrees Celsius. So let’s choose 93. We’ll say 93 degrees Celsius is the temperature of the water at zero seconds.

Then, for our second data point at 100 seconds, we do need to trace our horizontal
line over to our vertical axis. And we see it intersects that axis here. That looks to be at a temperature of about 45 degrees Celsius. We’ll say that’s the water temperature at 100 seconds.

Going back to our question, we want to calculate the decrease in temperature between
these two time values. To do that, we’ll subtract 45 degrees Celsius from 93 degrees Celsius. This is equal to 48 degrees Celsius which matches up perfectly with one of our answer
options. But this might raise a question. What if we had estimated our temperatures differently? For example, what if instead of 93 degrees Celsius, we had assumed that the
temperature at zero seconds was 94 degrees Celsius?

We can see how that would change our calculated temperature decrease and we wouldn’t
have gotten an answer which is exactly one of our answer options. Notice though as we look over these four answer options that none of them are within
one degree Celsius of another choice. That means that even if we get approximately the correct answer from our approximate
reading of the temperature on a vertical axis, we’ll still be able to home in on the
correct choice. We’ll just choose the result that’s nearest to the one we calculated. The box we’ll tick here is for 48 degrees Celsius. That’s the decrease in water temperature between zero and 100 seconds. Now that we’ve considered some results having to do with temperature, let’s look at a
result that has to do with time.

The experiment was carried out in a room where the temperature was 19 degrees
Celsius. Use Figure 2 to determine how long it took for the water to cool to the same
temperature as the room.

Looking once more at figure two, we see that it shows us a correspondence between the
temperature of the water and the time passed in seconds. We’ll use this correspondence to figure out at what time value the water temperature
was the same as the temperature in the room. To get started, we’ll locate the room temperature, 19 degrees Celsius, on our
vertical axis. And we see it’s located there at this mark we’ve made.

Our next job is to very carefully trace out a horizontal line from this point until
that horizontal line intersects our data curve. When we do this, making sure to stay on the same grid line on our graph, we see the
point of intersection occurs here, which is at a time value of 550 seconds. This tells us that approximately 550 seconds after the water started to cool, it had
reached the same temperature as the room it was in. As the water cooled, we know that it released energy to its surroundings. Let’s consider that energy next.

Calculate the energy transferred to the surroundings as 0.80 kilograms of water cools
from 94 degrees Celsius to 19 degrees Celsius. The specific heat capacity of water is 4180 joules per kilogram degrees Celsius. Use the correct equation from the physics equation sheet.

We’re talking here about energy transferred. And it has to do with the water in our beaker which starts out at 94 degrees Celsius
cooling down to the temperature of the room it’s in, 19 degrees Celsius. As the water cools down, it releases its heat energy to its surroundings. And it’s that energy that we want to solve for. And to do it, we’ll be applying an equation from the physics equation sheet. The equation we’ll use is this. It says that the energy transferred by a change in temperature of some material is
equal to the mass of that material multiplied by its specific heat capacity 𝑐
multiplied by the change in temperature the material goes through.

Now in our case, that material is the water that’s in the glass beaker. And we’re told what the mass of that water is as well as its specific heat capacity
and its change in temperature. Our next step then is to plug in for these three values based on what we’re told in
the problem statement. We’re told that the mass of water in the beaker is 0.80 kilograms, that its specific
heat capacity is 4180 joules per kilogram degrees Celsius, and that the water starts
out at 94 degree Celsius but then cools to 19 degrees Celsius. So this last term is equal to the change in the water’s temperature, Δ𝑡.

Before we calculate our result, notice something interesting about the units in these
three terms. If we look at the units of mass, we see that one factor of kilograms appears in the
numerator and one in the denominator. So when we multiply these numbers together, those factors cancel out. Likewise, the temperature in degrees Celsius cancels out as well. We have one factor upstairs, so to speak, and one factor downstairs. We’ll be left then with a final answer in units of joules which is the unit of
energy. When we multiply through, we find a result of 250800 joules. That’s how much energy is released by the water to its surroundings as the water
cools. And this large number has to do with the fact that water has a large specific heat
capacity. Lastly, as we consider the water cooling down, let’s consider just how far it might
go as it loses heat.

If the students continue to measure the temperature of the water for more than 10
minutes, would they see the temperature of the water fall below 19 degrees
Celsius? Give a reason for your answer.

Here is the idea. We saw that the students measure the temperature of the water up to 10 minutes but
not past that. They stopped making measurements 10 minutes after they started. The question is: if they had gone beyond that, would they have seen the temperature
of the water continues to cool down until it actually gets cooler than the
temperature of the room that it’s in? To figure this out, we can think from an energy balance perspective.

We know that when the temperature of the water was greater than the temperature of
the air in the room, the water would release heat to its surroundings. And thanks to that release of heat energy, the temperature of the water would
decrease. And we saw it cool down until it reached about the room temperature. Let’s say that we’ve waited long enough for that time to come, for the water
temperature to be equal to the room temperature. The question becomes: when the water and the room are at the same temperature, is
there any reason for heat to move from one substance to the other?

The answer is no because the temperature of both of these substances, the water in
the beaker and the air in the room, is the same. There is no thermal imbalance to be balanced out by the transfer of heat energy. Therefore, when the water reaches 19 degrees Celsius, it won’t cool down any further
because it has nothing to give heat energy to. If it tried to give energy to the air, so to speak, the air would just give it right
back. Here then is how we can write our answer.

We can say that no, the temperature of the water will not fall below 19 degrees
Celsius no matter how long the students make measurements because the water cannot
fall below the temperature of its surroundings. Once the water and air have the same temperature, no heat will flow between them. So it was reasonable for the students to stop making measurements at around 10
minutes because the water couldn’t have gotten much colder than they measured it to
get.