# Video: Rate of Heat Transfer

A woman with a mass of 66.0 kg does some vigorous exercise, after which she has a body temperature of 39.0°C. Determine the rate at which she must cool her body to reduce its temperature to 37.0°C in 3.00 × 10³ s if it continues to be internally heated at a rate of 110 W. Use a value of 4180 J/kg ⋅ °C for the specific heat capacity of the woman’s body.

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### Video Transcript

A woman with a mass of 66.0 kilograms does some vigorous exercise, after which she has a body temperature of 39.0 degrees Celsius. Determine the rate at which she must cool her body to reduce its temperature to 37.0 degrees Celsius in 3.00 times 10 to the power of three seconds if it continues to be internally heated at a rate of 110 watts. Use a value of 4180 joules per kilogram degrees Celsius for the specific heat capacity of the woman’s body.

Okay, so in this question, we’ve got a woman with a mass of 66.0 kilograms who’s doing some vigorous exercise which results in the heating up of her body. After her exercise, she has a body temperature of 39.0 degrees Celsius. What we’ve been asked to do is to determine the rate at which she must cool her body to reduce this temperature to 37.0 degrees Celsius. And we’ve been given the amount of time in which she has to do it, 3.00 times 10 to the power of three seconds.

However, there’s a little bit of a twist. We’ve been told that the body continues to be internally heated at a rate of 110 watts. This means her body is still continuing to heat up. And so we’ll have to account for this when we do our calculations as well. Finally, we’ve been told that the woman’s body has a specific heat capacity of 4180 joules per kilogram degrees Celsius. So let’s write down all the important information in a little corner.

Firstly, the mass of the woman which we’ll call 𝑚 is 66.0 kilograms. Secondly, the initial temperature of her body just after she’s finished exercising which we’ll call 𝑇 sub 𝑖 for initial is equal to 39.0 degrees Celsius. Next, we also know that we want her final body temperature to be 37.0 degrees Celsius in a time which we’ll call lowercase 𝑡 of 3.00 times 10 to the power of three seconds.

Next, we also know that her body is internally being heated at a rate of 110 watts. And we’re calling this 𝑃 sub int because it’s internal heating and watts is a unit of power. So we’ve been told how much power her body is expending in order to keep heating itself up. Finally, we also know that her specific heat capacity is 4180 joules per kilogram degrees Celsius.

What we need to do is to find out 𝑃 sub cool. This is the rate at which her body must cool in order to reduce her temperature from 39 degrees Celsius to 37 degrees Celsius in the given time. And because we want to find out the rate of cooling, this is the same as power. Let’s see why that is.

Well, the reason for this is that we can recall that power is defined as the rate of energy transfer — in other words, energy divided by time. And this energy can be various things depending on the situation. In this case, the energy that we’re talking about is heat exchange by the woman’s body resulting in either the heating of her body or the cooling of her body.

Therefore, if we want to calculate in our final answer the rate at which her body cools, this is the same as the heat exchange by her body divided by the time taken. And hence, this is a type of power. So our final answer is going to be 𝑃 sub cool. Okay, so now that that sorted, let’s get on with solving the problem. So we’re trying to find out how quickly this woman’s body must give out heat so that her body can cool from 39.0 degrees Celsius to 37.0 degrees Celsius in 3.00 times 10 to the power of three seconds. The other important thing to note is that she’s been internally heated at a rate of 110 watts.

And we can also say that 𝑃 sub int is equal to 𝑄 sub int divided by 𝑡, where 𝑄 sub int is the amount of energy transferred by her internal heating system — whatever that may be; maybe it’s a muscles — to her body that’s resulting in the heating up of her body. And of course, 𝑡 is the amount of time over which this energy is transferred.

So if we rearrange the equation by multiplying both sides by 𝑡, we find that the energy transferred to her body that’s resulting in her body heating up is equal to the time for which the body is heating up multiplied by the power. And this basically is equal to 𝑄 sub int.

Now, if we say on her body diagram that she needs to give out an energy 𝑄 sub cool in order for her body to cool down, then we can also say that her body is effectively gaining an energy 𝑄 sub int. Now, of course, this is internal heating. So it’s basically her muscles that are heating up her body. But it is the same as if you has been heated from an external source.

Either way, the result is that the body is gaining this heat 𝑄 sub int. And thus, the body is being heated up by this heat. So she needs to give out heat 𝑄 sub cool. And this heat must be much larger than 𝑄 sub int so that her body can overall cool down.

So once again to clarify, her body has been heated internally. She has to release this amount of energy plus a lot more if her body is to cool down properly. So first of all, let’s find out what the net energy transfer of her body will be if it is to cool down from 39.0 degrees Celsius to 37.0 degrees Celsius in 3.00 times 10 to the power of three seconds.

To do this, we can recall that the net energy transfer of an object 𝑄 is given by the mass of that object 𝑚 multiplied by the specific heat capacity of that object multiplied by the change in temperature of that object Δ𝑇.

Now, we have all the quantities we need to be able to work out the net heat transfer of her body. So we say that her body’s net heat transfer is equal to her mass 66.0 kilograms multiplied by her specific heat capacity 4180 joules per kilogram degrees Celsius multiplied by the change in temperature of her body which is her final temperature 𝑇 sub f minus her initial temperature 𝑇 sub i. In other words, Δ𝑇 is equal to 𝑇 sub f minus 𝑇 sub i.

Now, when we evaluate the right-hand side of this equation here, we find that the heat transfer from her body is negative 551760 joules. Now, this is the net heat transfer. And it’s negative because the body is losing heat. The reason the body must lose heat overall is because it has to reduce its temperature.

Now, we can find a relationship between the net energy transfer 𝑄 sub cool and 𝑄 sub int, we can say that the net energy transfer 𝑄 is equal to the energy gained by her body, which is 𝑄 sub int, minus the total energy lost by her body, which is 𝑄 sub cool. At this point, we already know what the net energy transfer is and we’ve got an expression here for 𝑄 sub int.

So first of all, let’s find out what 𝑄 sub int is. Remember 𝑄 sub int is the internal heating energy, which is going towards heating up her body. And in a time of 3.00 times 10 to the power of three seconds, this energy happens to be that time multiplied by 110 watts because remember 110 watts is the power with which her body is being heated up internally. So we can evaluate this to find that 𝑄 sub int is equal to 3.30 times 10 to the power of five joules, which means we officially know the values of 𝑄 and 𝑄 sub int.

So let’s go about finding what 𝑄 sub cool is. Let’s first rearrange the equation. We find that 𝑄 sub cool is equal to 𝑄 sub int minus 𝑄. In other words, this ends up being 3.30 times 10 to the power of five, which is 𝑄 sub int, minus 𝑄. But remember 𝑄 is already negative. So we get a positive sign here. And the value is 551760 and the unit is going to be in joules. So when we evaluate this, we find that 𝑄 sub cool is equal to 881760 joules.

So let’s recap exactly what this means. So the woman’s body has been heated up by some internal process. And so in a time of 3.00 times 10 to the power of three seconds, an energy 𝑄 sub int is transferred to her body. However, in the same amount of time, her body is also losing an energy 𝑄 sub cool.

Now, the value of 𝑄 sub cool, the amount of energy she’s losing, is larger than the value of 𝑄 sub int, the amount of energy she’s gaining. And so overall, she’s losing energy. Specifically, her body is losing 551760 joules overall. And this is resulting in a change in temperature in her body from 39 degrees Celsius to 37 degrees Celsius in this amount of time.

So what the question actually wants us to work out is the rate at which she must lose energy, in other words this energy which she is losing divided by the time over which she loses it. And remember we’ve called this quantity 𝑃 sub cool. So we say that 𝑃 sub cool is equal to 𝑄 sub cool, the amount of energy that her body is losing, divided by the time 𝑡.

And remember this is not the overall energy transfer of her body. We’re now specifically looking at how much energy her body has to lose in order to be able to compensate for 𝑄 sub int and also result in a drop in temperature of her body. So anyway, our final answer is going to be 𝑄 sub cool divided by 𝑡. And at this point, we know what 𝑄 sub cool is. We’ve just worked it out and we already have been given 𝑡 in the question. Although wouldn’t it be so much nicer if we were giving tea to drink in the question? However, at this point, we must make deal with the time 𝑡.

So anyway, 𝑄 sub cool happens to be 881760 joules and we divide this by 3.00 times 10 to the power of three seconds cause that’s the amount of time over which her body loses the energy. So we can evaluate this fraction, remembering of course that joules divided by seconds gives us a unit of watts. And hence, we find that 𝑃 sub cool is equal to 293.92 watts.

However, all of the quantities that were given to us in the question were given to three significant figures. So we gave our answer to three significant figures as well. So here’s the third significant figure. The next one will tell us whether the third one rounds up or down. Now, this value is a nine which is greater than five. So our third significant figure rounds up. It becomes a four.

And hence, our final answer is that the rate at which the woman must cool her body is 294 watts.