# Question Video: Determining the Characteristics of a Resistive Circuit from the Effect of Its Dissipation

A resistor of an unknown resistance is placed in an insulated container filled with 0.75 kg of water. A voltage source is connected in series with the resistor producing a current of 1.2 A through the resistor that is maintained for 10.0 minutes, producing a water temperature increase of 10°C. Assume a value of 4180 J/kg ⋅ °C for the specific heat capacity of water. What is the resistance of the resistor? What is the voltage supplied by the power supply?

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### 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.