### Video Transcript

A prankster applies 300 volts to a 65.0-microfarad capacitor and then tosses it to an unsuspecting victim. The victim’s finger is burned by the discharge of the capacitor through 0.130 grams of flesh. What is the temperature increase of the flesh? Take the specific heat of tissues to be 3500 joules per kilogram Kelvin.

Alright, in this case, we have a capacitor which is charged to 300 volts of potential difference and then tossed to an unsuspecting victim, who catches the capacitor with their hand and has a finger burned as it discharges. We want to solve for the temperature increase of the victim’s finger, which is burned by the discharge. And we’ll call that temperature increase Δ𝑇.

We know that the source of this temperature increase is all of the energy stored in the capacitor suddenly being discharged into the victim’s hand. How much energy is this? Well, we can recall that if we call 𝑢 the energy stored in a capacitor, it’s equal to one-half the capacitance 𝑐 times the potential difference squared. We’re given what those capacitance and potential values are. So we can calculate the energy that is stored in the capacitor.

We’re also told that the specific heat of tissues in the victim’s hand is 3500 joules per kilogram Kelvin. That means that it would take 3500 joules in order to raise one kilogram of these tissues one Kelvin. This value for specific heat will come in handy when we want to solve for Δ𝑇, the total temperature increase of the victim’s hand.

Here’s one way we can figure that change in temperature out. If we start with this value, the specific heat of tissues, we see from a unit’s perspective that if we multiply this by some mass value, specifically the mass of flesh that was heated up, then those units of mass will cancel out. So let’s do that. Let’s replace 𝑀 with the mass of the flesh that was heated up. Having converted that mass value into units of kilograms, we see now that if we were to multiply these two values together, those units would indeed cancel. But we’re not ready for that quite yet. Let’s keep going.

Remember that our charged capacitor had some amount of energy that it was prepared to discharge. If we multiply our specific heat of tissues by one over that energy, then this energy 𝑢, which is expressed in units of joules, will cancel out that energy unit in our specific heat value. And then finally, if we multiply this expression by our change in temperature of the flesh, Δ𝑇, then that will take care of the third and final set of units, Kelvin.

What we have created here by multiplying these values all together is an expression which is unitless and which overall is equal to one. That is, if we take the specific mass of flesh that was heated up divided by the specific energy of the capacitor and multiply it by the actual change in temperature of the flesh and then multiply all that by our specific heat of tissues, we’ll get a value which is unitless and equal to one.

This is a very good thing because, in this expression, we see the value we want to solve for, Δ𝑇. And what we can do now is to rearrange this whole expression to isolate Δ𝑇 on one side. When we do, we find that Δ𝑇 is equal to the energy stored in the capacitor times one over the mass of the flesh that was heated up multiplied by the inverse of the specific heat of tissues.

Now, all that’s left for us to do is to plug in for the energy stored in the capacitor given as one-half 𝑐 times 𝑣 squared. With all that done, take a look at what happens to our units on the right side of this expression. The units of kilograms in numerator and denominator cancel out. And we know that a farad times a volt squared is equal to the unit of a joule. And since we have one over joules in the rest of our expression, those units cancel out as well.

In the end, we’ll be left with a number in units of Kelvin, that is, a temperature. And that temperature will represent the change in temperature of the flesh of the victim.

So now, all that’s left for us to do is to multiply through this string of numbers. And when we do, to three significant figures, we find a result of 6.43 Kelvin. That’s how much the temperature of the victim’s finger increases thanks to the energy discharge from the capacitor.