A Walk in the Sun. Consider a poor lost soul waking at 5 km / h on a hot day in the desert, wear

Name: Problem 127, Chapter 17: University Physics with Modern Physics
Uploaded: 2020-12-14T18:55:28-08:00
Duration: 26 min 14 s
Channel: Matthew Miranda

Question

A Walk in the Sun. Consider a poor lost soul waking at 5 $\mathrm{km} / \mathrm{h}$ on a hot day in the desert, wearing only a bathing suit. This person's skin temperature tends to rise due to four mechanisms: (i) energy is generated by metabolic reactions in the body at a rate of 280 $\mathrm{W}$ , and almost all of this energy is converted to heat that flows to the skin; (ii) heat is delivered to the skin by convection from the outside air at a rate equal to $k^{\prime} A_{\mathrm{skin}}$ $\left(T_{\text { air }}-T_{\text { skin }}\right),$ where $k^{\prime}$ is 54 $\mathrm{J} / \mathrm{h} \cdot \mathrm{C}^{\circ} \cdot \mathrm{m}^{2}$ the exposed skin area $A_{\mathrm{skin}}$ is $1.5 \mathrm{m}^{2},$ the air temperature $T_{\text { air }}$ is $47^{\circ} \mathrm{C}$ the exposed skin area $A_{\mathrm{skin}}$ is $36^{\circ} \mathrm{C}$ (iii) the skin absorbs radiant energy from the sun at a rate of 1400 $\mathrm{W} / \mathrm{m}^{2}$ , (iv) the skin absorbs radiant energy from the environment, which has temperature $47^{\circ} \mathrm{C} .$ (a) Calculate the net rate (in wats) at which the person's skin is heated by all four of these mechanisms. Assume that the emissivity of the skin is $e=1$ and that the skin temperature is initially $36^{\circ} \mathrm{C} .$ Which mechanism is the most important? (b) At what rate (in $L / h )$ must perspiration evaporate from this person's skin to maintain a constant skin temperature? (The heat of vaporization of water at $36^{\circ} \mathrm{C}$ is $2.42 \times 10^{6} \mathrm{J} / \mathrm{kg} . )(\mathrm{c})$ Suppose instead the person is protected by light-colored clothing $(e \approx 0)$ so that the exposed skin area is only $0.45 \mathrm{m}^{2} .$ What rate of perspiration is required now? Discuss the usefulness of the traditional clothing worn by desert peoples.

   A Walk in the Sun. Consider a poor lost soul waking at 5 $\mathrm{km} / \mathrm{h}$ on a hot day in the desert, wearing only a bathing suit. This person's skin temperature tends to rise due to four mechanisms: (i) energy is generated by metabolic reactions in the body at a rate of 280 $\mathrm{W}$ , and almost all of this energy is converted to heat that flows to the skin; (ii) heat is delivered to the skin by convection from the outside air at a rate equal to $k^{\prime} A_{\mathrm{skin}}$ $\left(T_{\text { air }}-T_{\text { skin }}\right),$ where $k^{\prime}$ is 54 $\mathrm{J} / \mathrm{h} \cdot \mathrm{C}^{\circ} \cdot \mathrm{m}^{2}$ the exposed skin area $A_{\mathrm{skin}}$ is $1.5 \mathrm{m}^{2},$ the air temperature $T_{\text { air }}$ is $47^{\circ} \mathrm{C}$ the exposed skin area $A_{\mathrm{skin}}$ is $36^{\circ} \mathrm{C}$ (iii) the skin absorbs radiant energy from the sun at a rate of 1400 $\mathrm{W} / \mathrm{m}^{2}$ , (iv) the skin absorbs radiant energy from the environment, which has temperature $47^{\circ} \mathrm{C} .$ (a) Calculate the net rate (in wats) at which the person's skin is heated by all four of these mechanisms. Assume that the emissivity of the skin is $e=1$ and that the skin temperature is initially $36^{\circ} \mathrm{C} .$ Which mechanism is the most important? (b) At what rate (in $L / h )$ must perspiration evaporate from this person's skin to maintain a constant skin temperature? (The heat of vaporization of water at $36^{\circ} \mathrm{C}$ is $2.42 \times 10^{6} \mathrm{J} / \mathrm{kg} . )(\mathrm{c})$ Suppose instead the person is protected by light-colored clothing $(e \approx 0)$ so that the exposed skin area is only $0.45 \mathrm{m}^{2} .$ What rate of perspiration is required now? Discuss the usefulness of the traditional clothing worn by desert peoples.

University Physics with Modern Physics

Hugh D. Young, Roger… 12th Edition

Chapter 17, Problem 127

Instant Answer

Solved by Expert Matthew Miranda

12/14/2020

Step 1

Since the metabolic rate is given as 280 W, this is the rate at which heat is generated by the body and transferred to the skin. Show more…

Show all steps

Thanks for your feedback!

A Walk in the Sun. Consider a poor lost soul waking at 5 $\mathrm{km} / \mathrm{h}$ on a hot day in the desert, wearing only a bathing suit. This person's skin temperature tends to rise due to four mechanisms: (i) energy is generated by metabolic reactions in the body at a rate of 280 $\mathrm{W}$ , and almost all of this energy is converted to heat that flows to the skin; (ii) heat is delivered to the skin by convection from the outside air at a rate equal to $k^{\prime} A_{\mathrm{skin}}$ $\left(T_{\text { air }}-T_{\text { skin }}\right),$ where $k^{\prime}$ is 54 $\mathrm{J} / \mathrm{h} \cdot \mathrm{C}^{\circ} \cdot \mathrm{m}^{2}$ the exposed skin area $A_{\mathrm{skin}}$ is $1.5 \mathrm{m}^{2},$ the air temperature $T_{\text { air }}$ is $47^{\circ} \mathrm{C}$ the exposed skin area $A_{\mathrm{skin}}$ is $36^{\circ} \mathrm{C}$ (iii) the skin absorbs radiant energy from the sun at a rate of 1400 $\mathrm{W} / \mathrm{m}^{2}$ , (iv) the skin absorbs radiant energy from the environment, which has temperature $47^{\circ} \mathrm{C} .$ (a) Calculate the net rate (in wats) at which the person's skin is heated by all four of these mechanisms. Assume that the emissivity of the skin is $e=1$ and that the skin temperature is initially $36^{\circ} \mathrm{C} .$ Which mechanism is the most important? (b) At what rate (in $L / h )$ must perspiration evaporate from this person's skin to maintain a constant skin temperature? (The heat of vaporization of water at $36^{\circ} \mathrm{C}$ is $2.42 \times 10^{6} \mathrm{J} / \mathrm{kg} . )(\mathrm{c})$ Suppose instead the person is protected by light-colored clothing $(e \approx 0)$ so that the exposed skin area is only $0.45 \mathrm{m}^{2} .$ What rate of perspiration is required now? Discuss the usefulness of the traditional clothing worn by desert peoples.

Play audio

Feedback

Matthew Miranda and 84 other subject Physics 101 Mechanics educators are ready to help you.

Ask a new question

Labs

Want to see this concept in action?

NEW

Explore this concept interactively to see how it behaves as you change inputs.

View Labs

Key Concepts

Metabolic Heat Generation

Metabolic heat generation refers to the heat produced by the body's metabolic processes, which convert chemical energy into thermal energy. This internal source of heat is significant because it continuously contributes to the overall heat load on the body, and its rate can vary with activity level and physiological state.

Thermal Energy Balance

This concept involves accounting for all sources of heat gain and loss in a system—in this case, the human body. The net heat on the skin results from the sum of energy inputs from sources like metabolic heat, convective transfer from ambient air, and radiative absorption from the sun and the environment, balanced against cooling processes. It is crucial for understanding how body temperature changes under different environmental conditions and activity levels.

Convection Heat Transfer

Convection is the process by which heat is transferred between a surface and a fluid (such as air) moving over it. In this context, the heat exchange between the skin and the surrounding air is driven by the temperature difference and a convective heat transfer coefficient, influencing how quickly the skin temperature can rise or fall.

Radiative Heat Transfer

Radiative heat transfer involves the emission or absorption of electromagnetic radiation, primarily in the infrared. The skin absorbs radiant energy both directly from the sun and from environmental sources. This form of heat transfer is influenced by factors such as the intensity of the radiation, the temperature of the surroundings, and the emissivity of the skin or clothing.

Evaporative Cooling

Evaporative cooling is the process by which the body loses heat through the evaporation of sweat from the skin. This mechanism uses the latent heat of vaporization, which is the energy required to convert liquid water into vapor, thus helping to regulate the body's temperature even when external conditions promote heating.

Influence of Clothing on Heat Transfer

Clothing can significantly alter the heat balance of the body by changing the effective area exposed to the environment and by modifying the absorption and emission of radiant energy. Light-colored or reflective clothing, for example, can reduce radiant heat absorption and provide insulation, thereby affecting how much heat is gained or lost through the skin.

Recommended Videos

Transcript

00:01 Question number 127 is asking us about a person walking through the desert in the heat of the day wearing a bathing suit.

00:09 We're asked in the first part of our question to find the rate at which the person's skin is heated up by the following four mechanisms.

00:18 The first of these mechanisms is the heat of metabolic reactions within a person's body.

00:27 This is where the energy is converted to heat underneath the person's skin and is acting to heat it up.

00:37 So that's one of the sources of heat.

00:41 The second source of heat is the heat from air convection within the desert that's acting to heat up the person's skin.

00:52 One of the ways in which heat is transferred from one object to another is through convection.

01:01 And so the person's skin will feel the heat by this method.

01:07 Another method by which the person's skin will be heated up is by means of solar radiation.

01:16 The sun in the sky definitely act to heat up the person's skin very well.

01:25 And the final source of heat will be the radiation from the environment, the environmental radiation.

01:43 So we have our four sources of heat impacting this person.

01:54 Let's see which one contributes most significantly.

01:59 We're told that the rate at which the person's skin is heated by metabolic reactions is valued at 280 watts, so that rate is given to us.

02:16 The heat from air convection is given to us via an expression of the following nature.

02:26 The heat h is equal to k prime times the area of exposed skin, i'm just going to call a sub -skin, times the temperature of the air minus the temperature of the skin.

02:49 We're given values for all of these terms.

02:58 K -prime is valued at 54 joules per hour per degree celsius per meter squared.

03:15 The area of skin exposed to the environment by this person wearing the bathing suit is 1 .4.

03:24 Meters squared, the temperature of the air is valued at 47 degrees celsius, and the temperature of the person's exposed skin is 36 degrees celsius.

03:47 Performing our calculations, we can cancel out the meters squared and the degrees celsius from our problem to get our answer in joules per hour.

04:02 And the heat from air convection turns out to be 891 joules per hour.

04:12 Now our problem wants all these values in watts.

04:16 Watts is a measure of joules per second.

04:22 So we're going to need to convert that 891 joules per hour into joules per second.

04:31 The way we do that is multiply 891 joules per hour by one hour over 3 ,600 seconds, which is how many seconds there are in one hour.

04:48 Now we can cancel out the hours, and we find that the heat from air convection is valued at 0 .2.

05:02 2475 watts.

05:06 So that's the answer to that source of heat.

05:16 Now for solar radiation, all we need to do to find the rate at which the sun heats up the person's skin is take the energy from the sun per meters squared incident on the person's skin.

05:38 This is also known as the solar constant.

05:40 1 ,400 watts per meter squared times the area of the person's exposed skin.

05:52 1 .5 meters squared.

05:58 We multiply that out.

06:00 We get a value of 2 ,100 watts for the heat energy coming from the sun.

06:14 Now, the radiant energy from the environment, the heat from that, can be determined using this expression.

06:24 H is equal to a times e, sigma, times t to the 4, minus t sub s to the 4.

06:43 A is the area of the exposed skin, which we know already.

06:50 E is the emissivity of the skin, which is valued to 1, and sigma is the stefan boltzmann constant, which has a value of 5 .67 times 10 to the minus 8 watts per meter squared times kelm.

07:15 To the fourth.

07:19 Notice that in the denominator of this expression, we have kelvin's, or the units of temperature.

07:27 So what we're going to need to do, we'll still be using 47 degrees celsius for t and 36 degrees celsius for t sub s, but we're going to need to convert them to kelvin's for using that expression there.

07:44 To do that, all we need to do is is add 273 to both of them.

07:52 So t will become 320 kelvin's, and t sub s will be 36 plus 273 gives us 309 kelvin's.

08:17 So now we can begin to plug everything in to find the heat from environmental radiation on this person's skin.

08:26 It'll be equal to 1 .5 meters squared times one, the emissivity of the person's skin, times the stefan boltzmann constant of 5 .67 times 10 to the minus 8 watts, watts, per meter squared times kelvin's to the fourth times 320 kelvin's to the fourth power, minus 309 kelvin's to the fourth power.

09:17 That works out to give the heat of radiation from the environment a value of 116 .45 watts.

09:33 So now we have the rate of heat from these four different mechanisms.

09:42 And considering the sun heats up the person's skin at a rate of 2100 watts, we can consider this source of heat the most important.

09:56 Now, the second part of this question wants us to determine how much the person must perspire in order to maintain a constant body temperature.

10:11 The way we can figure this out is to know that the heat, absorbed by the skin, q, is equal to the mass of person.

10:23 Inspiration developed by the person times the latent heat l.

10:30 Q, rate of heat absorption, is also equal to the total power of each of these four sources of heat from the previous part of the problem times one hour.

10:54 So we have two expressions for q.

10:58 Let's go ahead and equate them to one another, p total times one hour is equal to ml.

11:16 The thing is, we want to find how much this person must perspire in order to maintain this constant skin temperature...

Ace is your personal tutor. It breaks down any question with clear steps so you can learn.

Start Using Ace