Jogging in the heat of the day. You have probably seen...

Jogging in the heat of the day. You have probably seen people jogging in extremely hot weather and wondered "Why? As we shall see, there are good reasons not to do this! When jogging strenuously, an average runner of mass 68 $\mathrm{kg}$ and surface area 1.85 $\mathrm{m}^{2}$ produces energy at a rate of up to $1300 \mathrm{W}, 80 \%$ of which is converted to heat. The jogger radiates heat, but actually absorbs more from the hot air than he radiates away. At such high levels of activity, the skin's temperature can be elevated to around $33^{\circ} \mathrm{C}$ instead of the usual $30^{\circ} \mathrm{C} .$ (We shall neglect conduction, which would bring even more heat into his body.) The only way for the body to get rid of this extra heat is by evaporating water (sweating). (a) How much heat per second is produced just by the act of jogging? (b) How much net heat per second does the runner gain just from radiation if the air temperature is $40.0^{\circ} \mathrm{C}$ (104 F)? (Remember that he radiates out, but the environment radiates back in.) (c) What is the total amount of excess heat this runner's body must get rid of per second? (d) How much water must the jogger's body evaporate every minute due to his activity? The heat of vaporization of water at body temperature is $2.42 \times 10^{6} \mathrm{J} / \mathrm{kg}$ . (e) How many 750 $\mathrm{mL}$ bottles of water must he drink after (or preferably before!) jogging for a half hour? Recall that a liter of water has a mass of 1.0 $\mathrm{kg}$ .

Key Concepts

Latent Heat of Vaporization

The latent heat of vaporization is the amount of energy required to convert a unit mass of water from liquid to vapor without a change in temperature. This concept is critical in physiological processes like sweating because it quantifies how much heat must be removed from the body to evaporate a given mass of sweat. In scenarios where excessive heat is generated, knowing this value helps to calculate the necessary amount of evaporative water loss to maintain thermal balance.

Evaporative Cooling

Evaporative cooling is the mechanism by which the body loses heat through the phase change of water, typically via sweating. As water evaporates from the skin, it requires a significant amount of energy to change from liquid to vapor, which is taken from the body as heat. This process becomes the primary means of thermal regulation when other heat loss methods, such as radiation, are less effective due to high ambient temperatures.

Metabolic Energy Production

This concept refers to the process by which the human body converts chemical energy from food into mechanical work and heat during physical activity. In strenuous activities such as jogging, most of the energy produced by the body is released as heat rather than useful mechanical work. Understanding this conversion is crucial because the amount of heat generated sets the stage for the body’s thermal regulation needs during exercise.

Radiative Heat Transfer

Radiative heat transfer is the process by which heat is emitted by a body in the form of electromagnetic radiation. In the context of the problem, the runner not only radiates heat from the skin but also absorbs radiation from the environment. The net heat transfer by radiation depends on the temperature difference between the body and its surroundings and is governed by the Stefan–Boltzmann law, which explains the rates of emission and absorption in thermal radiation.

Transcript

00:01 Hey everyone, this is question number 80 from chapter 14.

00:04 In this problem, we're talking about a jogger has massed 68 kilograms.

00:08 We're given the surface area as well.

00:11 We're told that the jogger produces energy at a rate of 1 ,300 watts, and that 80 % of that is converted to heat.

00:18 We're then told, given the skin temperature is 30 degrees celsius, 33 degrees celsius instead of 30, and the air temperature is 40 degrees celsius.

00:28 We're also given the heat of vaporization of water in the body.

00:31 So the premise of this problem is a jogger running outside in the heat, just to give you a little more background.

00:37 So part a asked, this seems like a big problem, but the five parts aren't too bad if you take it one step at a time.

00:44 So part a asks to find the heat produced per second.

00:49 So we're given a energy production rate of 1 ,300 watts in, that's one of our givens.

00:57 And we know that a watt is a jewel per second.

01:02 So all we have to do to find out how much heat is produced per second is figure out how much of this energy produce is heat, and we're told it's 80%.

01:15 So for part a, all you have to do is 0 .8, which is 80 % equals times 1 ,300 watts.

01:31 Which is the total energy produced, but 80 % of it goes to heat.

01:35 And if you solve that, you get 1 .04 times 10 to the third watts, which is joules per second.

01:50 So that's how much heat is produced per second.

01:58 I'll call that p.

02:07 To part b, which is net heat per second gained from radiation.

02:13 So gain from radiation, not is because the air temperature is higher than the runner's body temperature.

02:19 So he's going to gain heat from the air.

02:21 So we just use our radiation equation.

02:24 So b, heat net equals our radiation a surface area, e, which is our emissivity, sigma, which is a constant.

02:41 And then, t to the fourth minus t of the surroundings to the fourth.

02:48 And we can plug in our numbers.

02:51 We're not given an emissivity in the problem, so we're going to assume that it is 1.

02:58 So h -net is equal to surface area, 1 .85 meters squared.

03:09 M -sissivity, like i said, we're just going to assume is 1.

03:12 Sigma is a constant 5 .67 times 10 to the minus.

03:19 8th watts per meter square times k to the fourth.

03:28 And then multiplied by, we have to change our temperatures into calvin...

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