# Question Video: Calculating a Temperature Increase due to Mechanical Power Dissipation

In a study of healthy young men, doing 20 push-ups in 1 minute required an amount of energy per kg that for a 70.0-kg man corresponds to 8.06 calories (kcal). The average specific heat capacity of the human body is 0.830 kcal/kg ⋅ °C. How much would a healthy, young, 70.0-kg man’s temperature rise if he did not lose any heat while doing 20 push-ups in 1 minute?

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### Video Transcript

In a study of healthy young men, doing 20 push-ups in one minute required an amount of energy per kilograms that for a 70.0-kilogram man corresponds to 8.06 calories, in parentheses kilocalories. The average specific heat capacity of the human body is 0.830 kilocalories per kilogram degrees Celsius. How much would a healthy young 70.0-kilogram man’s temperature rise if he did not lose any heat while doing 20 push-ups in one minute?

Alrighty! Let’s bring this question down into digestible chunks by underlining all the important information that’s been given to us. So we know that there was a study of healthy young men, and they were doing 20 push-ups in one minute. To do this required an amount of energy per kilogram that for a 70-kilogram man corresponds to 8.06 kilocalories. Now as a quick aside, kilocalories is the actual unit of measurement and calories is just a colloquialism.

In other words, the actual unit, kilocalories, is often referred to as just calorie. So often you’ll see on TV that a certain chocolate bar for example has only 200 calories. Well in reality, it’s actually got 200 kilocalories, and a ’kilo’ is a prefix meaning 1000, so that chocolate bar has 200000 calories. However, as we said earlier, this is a colloquialism where we refer to one kilocalorie as just a calorie. So this can get a bit confusing. But remember that whenever we’re doing science, we use the actual unit, and if it’s just an everyday conversation then we can call it calories if we want to.

Generally, though it’s best to be accurate, so stick with kilocalories as much as you can. Anyway, moving on, we’re told that the average specific heat capacity of the human body is 0.830 kilocalories per kilogram degrees Celsius. Now notice here that the proper unit has been used because this time we’re given an actual quantity. We’ve been given a specific heat capacity quantitatively, and it needs to have the correct units. So we’re asked how much would a healthy young 70-kilogram man’s temperature rise if he did not lose any heat while doing 20 push-ups in one minute.

Now this is quite a confusing question. It contains a fair amount of information that’s not really relevant. So we’re gonna have to try our best to sort through it as much as possible. Let’s start with doing 20 push-ups in one minute. The only reason this information is relevant at all is because we’re told the amount of energy it takes to do 20 push-ups in one minute, and we’re trying to calculate the temperature rise in a man’s body when he does 20 push-ups in one minute. So aside from that, the fact that he’s doing 20 push-ups or doing it in one minute is not really relevant. What is relevant is how much energy that takes and how much the temperature rises by when doing that activity.

The healthy young men part is exactly the same. The only reason that’s relevant is because we’ve been given an information from a study talking about healthy young men, and we’re calculating the temperature rise in a healthy young man’s body. This is also true for the fact that the man in question has a mass of 70 kilograms. Again, we’re told that the amount of energy that a 70-kilogram man will expend doing a certain activity. And we’re just trying to calculate the temperature rise by doing that activity for a 70-kilogram man.

However, this time the value that we’ve been given, 70.0 kilograms, is actually relevant to our question. We actually need to know that value in order to calculate the temperature rise. So this is why this quantity is different to the number of push-ups being done or the amount of time it takes, because the temperature rise in the man’s body is directly affected by the mass of the man. We can see this by the specific heat capacity.

The specific heat capacity is the amount of energy it takes to increase the man’s body temperature by one degree Celsius for every kilogram of body mass. So he has a 70-kilogram body mass, which is going to be relevant. Now let’s look at the fact that we’ve been told that doing 20 push-ups in one minute requires a certain amount of energy per kilogram. Well this is a kind of there just to confuse us because we’ve been told that it requires an amount of energy per kilogram. We don’t know what the energy per kilogram is such that a 70-kilogram man will require 8.06 kilocalories.

So again, the energy per kilogram is irrelevant. We know the total energy required for a 70-kilogram man. The energy per kilogram would only be relevant if they’d given that to us, for example a 70- kilogram man, and wanted us to calculate something for a 50-kilogram man for example, basically if they’d given us the energy for kilogram for a specific man and wanted us to calculate the energy expended by another man with a different mass. But that’s not the case.

We’ve been told the energy for a 70-kilogram man, and we’re trying to calculate the temperature change for a 70-kilogram man. Now, we’ve also been given the average specific heat capacity. Well if this is the average specific heat capacity, then we can assume that our average human being, the bloke, is doing 20 push-ups has a body which has that specific heat capacity. And that happens to be the 0.830 kilocalories per kilogram degrees Celsius. And finally, we’re told to assume that the body did not lose any heat while doing 20 push-ups in one minute. So there’s no wastage of energy, and the rest of orange lines are not really that relevant, but I’m gonna underline them in pink anyway just to make it look nice. We do want consistency after all, don’t we?

Anyway, so let’s start labelling quantities. We know the amount of energy it takes for a man to do 20 push-ups in one minute if that man weighs 70 kilograms. We can call this energy capital 𝐸. And this energy is 8.06 kilocalories. We’ve also been told the average specific heat capacity of the human body, which we’ve said is the specific heat capacity of the bloke doing the push-ups. So we can label this as well as 𝑐. We know the bloke’s mass, which we’ll label as 𝑚. And we’re trying to find out the change in his body temperature, Δ𝑇.

Now let’s look at the quantity that we’ve labelled 𝐸, the energy that he expends doing 20 push-ups in one minute. Well he’s expending this energy by doing these push-ups, but that energy is what is going towards increasing his body temperature. In other words, this energy is the heat supplied to his body in order to raise its temperature. So instead of calling it 𝐸, the energy expended by the dude doing the push-ups, we can call it 𝑄, the heat supplied to the body by the muscles in order to increase the body temperature. Now at this point, we can see that there’s a relationship that we need to find between the heat supplied to the body, the specific heat capacity of the body, the mass of the body, and the temperature rise in the body. Have I said body enough times yet?

Anyway! This is the relationship we’re looking for: 𝑄 is equal to 𝑚𝑐Δ𝑇. This comes directly from the definition of specific heat capacity because, remember, the specific heat capacity is defined as the amount of heat that we need to supply to an object in order to induce a one-degree-Celsius temperature change for every kilogram of mass that the object has. And if we rearrange this definition of specific heat capacity, that leads us to this equation above. We could of course start from this equation and work our way forwards. But this is the way the equation is most commonly formulated. And so that maybe the formulation that’s most familiar to us.

Either way, it doesn’t matter which we start with because they’re the same equation. What we’re trying to do is to solve for Δ𝑇, so let’s go about doing that. If we take the equation in green, then what we need to do is divide both sides of the equation by 𝑚𝑐. This leaves us with 𝑄 divided by 𝑚𝑐 is equal to Δ𝑇. So all that remains is the substitution of values of 𝑄, 𝑚, and 𝑐 to find out Δ𝑇. So we’ve plugged in all of values here, and let’s quickly look at the units. First of all, we can look at the denominator. The kilogram here cancels with the per kilogram here. And then secondly, this kilocalorie cancels with this kilocalorie here. So essentially, what we’re left with is one over degrees Celsius in the denominator and that is the same as if we had degrees Celsius in the numerator.

So our answer is going to be in degrees Celsius as we’d expect because we’re trying to find a change in temperature. Now evaluating the fraction gives us a changing temperature of 0.1387 dot dot dot and so on and so forth degrees Celsius. But if we look carefully, all of the values that we’ve used to solve the problem are given to three significant figures. Therefore, we should give our answer to three significant figures as well. So here is the third significant figure. It’s this eight. It’s the one afterwards that will tell us what will happen to the eight. So this significant figure, the fourth one, is a seven. That’s larger than five, so the third significant figure will round up. The eight will become a nine, and this gives us our final answer.

If a healthy young 70-kilograms man does 20 push-ups in one minute, then its temperature should rise by 0.139 degrees Celsius to three significant figures.