### Video Transcript

A resistor of an unknown resistance
is placed in an insulated container filled with 0.75 kilograms of water. A voltage source is connected in
series with the resistor producing a current of 1.2 amps through the resistor that
is maintained for 10.0 minutes, producing a water temperature increase of 10 degrees
Celsius. Assume a value of 4180 joules per
kilogram degrees Celsius for the specific heat capacity of water. What is the resistance of the
resistor? What is the voltage supplied by the
power supply?

Okay, this is an interesting
exercise. To get started on it, let’s record
the information we’re given off to the side. And then, we’ll clear some space on
screen to make a sketch of our setup. Right, so we’ve condensed our
information given in the problem statement over to this right hand side of our
screen.

The mass of water, 𝑚 sub 𝑤, we’re
given as well as the current 𝐼, the time that the current ran through the resistor,
the change in temperature, we’ve called 𝛥𝑇, and the specific heat capacity of
water, 𝐶 sub 𝑤. Knowing all that, we want to solve
for the resistance value 𝑅 of the unknown resistor as well as the voltage supplied
to it.

If we make a sketch of this
scenario, we have an unknown resistor, that’s what we’ve called 𝑅, and an unknown
power supply, we’ve called 𝑉. The power supply sends current
through this circuit which heats up the resistor 𝑅, which then heats up the water
that’s in this insulated container. We’re told that, after 10 minutes
of heating, that water temperature has increased by 10 degrees Celsius. And this rate of heating it turns
out will help us solve for the resistance value of our resistor.

We know that, in this process, some
amount of heat energy is being transmitted to the water over this amount of time,
10.0 minutes. If we think in terms of these
values, energy per unit time, what is that? Well, energy per unit time is
power. And that’s helpful to us because
there’s another relationship for power we can recall. The power in an electrical circuit
is equal to the current in that circuit squared times the resistance of the
circuit.

Remember that we know the current
that runs through this loop, which heats up the water in our insulated
container. That means if we can also solve for
the power supplied to the water, then we’ll have enough information to solve for the
resistance of our resistor. We can put it this way. The resistance of the resistor is
equal to the power supply to heat the water divided by the current in this circuit
squared. And knowing that power is equal to
energy per unit time, we can write that resistance as the energy supplied to the
water divided by 𝐼 squared times 𝑡.

Looking at the information we’ve
been given, we know the current 𝐼 as well as the time 𝑡 that the water is being
heated. But what about the energy supplied
to it, 𝐸? We can figure out that energy by
working with a specific heat capacity of water.

Take a look for a second at the
units in this specific heat capacity, joules per kilogram degrees Celsius. What this value tells us is just
how much energy in joules is required to raise the temperature of one kilogram of
water by one degree Celsius.

Focusing again on the units of this
expression, we realize that if we were to multiply this specific heat capacity by
the mass of the water heated and then multiplied that by the change in temperature
the water experienced, then look what happens when we plug in for those two values,
the units of kilograms. The units of mass cancel out as do
the units degrees Celsius. This leaves us with a value in
units of joules. That is, it’s an energy.

Since this is an expression for
energy, that means we can take this value and then divide it by 𝐼 squared times 𝑡,
where 𝐼 is 1.2 amperes, as given in our problem statement. And 𝑡 we convert to a value in
seconds. 10.0 minutes is 10.0 times 60
seconds. And when we calculate this, what
we’ve solved for according to our equation on the top left is the value of our
resistor 𝑅.

In our numerator, we have the
energy supplied to heat the water. And in our denominator, we have the
current squared times the time over which the water was heated. When we calculate this value, to
two significant figures, it’s 36 ohms. That’s the resistance value of our
previously unknown resistor.

And now that we know the resistance
of our resistor, the next thing that we want to solve for is the voltage of our
power supply 𝑉. For this, we can use the
relationship known as Ohm’s law that the potential difference or voltage in a
circuit is equal to the current in a circuit times the resistance of the
circuit. Once again, using 1.2 amps for the
current and then 36 ohms for the resistance we’ve just solved for, we find a voltage
of 44 volts.

Knowing this, we now know how to
build a circuit that could heat 0.75 kilograms of water, 10 degrees Celsius over a
time of 10 minutes.