A 68 kg cyclist is pedaling down the road at 15 km / h, using a total metabolic power of 480

A 68 kg cyclist is pedaling down the road at 15 km / h,...

A $68 \mathrm{~kg}$ cyclist is pedaling down the road at $15 \mathrm{~km} / \mathrm{h}$, using a total metabolic power of $480 \mathrm{~W}$. A certain fraction of this energy is used to move the bicycle forward, but the balance ends up as thermal energy in his body, which he must get rid of to keep cool. On a very warm day, conduction, convection, and radiation transfer little energy, and so he does this by perspiring, with the evaporation of water taking away the excess thermal energy. Being able to exhaust this thermal energy is very important. If he isn't able to get rid of any of the excess heat, by how much will the temperature of his body increase in 10 minutes of riding, to the nearest $0.1^{\circ} \mathrm{C}$ ? A. $0.3^{\circ} \mathrm{C}$ B. $0.6^{\circ} \mathrm{C}$ C. $0.9^{\circ} \mathrm{C}$ D. $1.2^{\circ} \mathrm{C}$

A $68 \mathrm{~kg}$ cyclist is pedaling down the road at $15 \mathrm{~km} / \mathrm{h}$, using a total metabolic power of $480 \mathrm{~W}$. A certain fraction of this energy is used to move the bicycle forward, but the balance ends up as thermal energy in his body, which he must get rid of to keep cool. On a very warm day, conduction, convection, and radiation transfer little energy, and so he does this by perspiring, with the evaporation of water taking away the excess thermal energy.
Being able to exhaust this thermal energy is very important. If he isn't able to get rid of any of the excess heat, by how much will the temperature of his body increase in 10 minutes of riding, to the nearest $0.1^{\circ} \mathrm{C}$ ?
A. $0.3^{\circ} \mathrm{C}$
B. $0.6^{\circ} \mathrm{C}$
C. $0.9^{\circ} \mathrm{C}$
D. $1.2^{\circ} \mathrm{C}$

Key Concepts

Metabolic Power

Metabolic power refers to the rate at which energy is produced by the body through metabolic processes. In physics problems like this one, the given metabolic power is used to determine how much energy is generated over a period of time, which can then be converted into thermal energy affecting the body’s temperature.

Energy Accumulation Over Time

This concept involves calculating the total energy produced over a certain period by multiplying the power (energy per unit time) by the duration of the activity. This accumulated energy, if not dissipated, would be fully absorbed by the body, potentially raising its temperature.

Specific Heat Capacity

Specific heat capacity is a property of a material that quantifies the amount of energy required to raise the temperature of a unit mass of the substance by one degree Celsius. In the context of the human body, it is used to calculate how much the body temperature will increase when a given amount of energy is absorbed.

Transcript

00:01 Okay, so here we're going to look at somebody riding a bicycle and look at the temperature rise of the body when they are doing so.

00:08 So in part a, we have a 53 kilogram woman that is going to be cycling for one hour at 15 kilometers per hour.

00:28 We know that the metabolic rate is 1 ,650 kilojoules per hour at 15 kilometers per hour.

00:43 Okay, so the total amount of heat that is being burned by this person riding the bicycle queue here, or heat, is going to be equal to the metabolic rate times time.

00:59 So here we'll have 1650 kilojoules per hour times how long they were doing the activity, which is one hour.

01:15 So that will give us 1 ,650 kilojoules.

01:18 So this is how much heat in energy that was burned and given off.

01:28 And your body is what is absorbing all of that heat.

01:32 So while you're exercising, you're burning energy and your body is absorbing that release heat.

01:39 So here, for this equation, q equals mc delta t.

01:48 This is an equation used to calculate generally the temperature change of the material when heat is input or output.

01:58 So in this equation, q equals heat, m equals mass, and c is the heat capacity, and then delta t is your change in temperature.

02:12 So for somebody riding this bicycle, that weighs 53 kilograms, and the overall q is 1 ,650 kilojoules.

02:22 We'll set this up like this.

02:24 Q equals 1 ,650 kilojoules, and that's going to be equal to m, which is the mass, so we'll put 53 kilograms, and we'll multiply by the heat capacity, which is 3 .47 kilojoules per calvin kilogram.

02:50 And then we'll just put delta t, since that's all we're looking for is what the change or the rise in the body temperature is.

02:57 So we're just looking for how the temperature changes.

02:59 So when we calculate for delta t here, we'll get that delta t is 8 .97 kelvin, so about 9 kelvin, which is about 9 degrees celsius as well.

03:19 So that's a lot of temperature rise if there was no heat exchange to the surroundings.

03:23 So in part b, we want to explain how is this possible? if you have to maintain your body temperature within plus or minus about 1 degrees celsius, how could you possibly go on a bike ride? and do something where the theoretical temperature change is 9 degrees c.

03:42 Well, this all comes from the body's control mechanisms, and in particular here we're talking about sweating.

03:52 And what sweating does is sweating is just the evaporation of water from your skin.

03:58 So you have h2o liquid, and what you do is an endothermic process, where you add heat, and that comes from your body, and that heat energy turns the liquid water into water vapor.

04:14 So while you're increasing, in theory, you would be increasing the temperature of your body.

04:19 This process is pulling heat away and helping you regulate your temperature.

04:25 And so this is an endothermic process that is helping you do that.

04:33 So for part c here, we're going to talk about obesity and how the difference between the heat capacity for water and fat can have an effect on the change in temperature in your body and how your body needs to regulate differently during an activity...

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