Radiation by the body. The amount of heat radiated by the...

Radiation by the body. The amount of heat radiated by the body depends on its surface temperature and area. Typically, this temperature is about $30^{\circ} \mathrm{C}$ (although it can vary). The surface area depends on the person's height and weight. Anempirical formula for the surface area of a person's body is $A\left($ in $\mathrm{m}^{2}\right)=(0.202) M^{0.425} h^{0.725}$ where $M$ is the person's mass (in kilograms) and $h$ is his or her height (in meters).(a) What would be the surface area of a 165 lb $(75 \mathrm{kg}), 6 \mathrm{ft}(1.83 \mathrm{m})$ person? (This is a good chance to use the $y^{x}$ key of your calculator.) (b) How much heat would the person radiate away per second at a skin temperature of $30^{\circ} \mathrm{C} ?$ (At the low temperatures of room-temperature objects, nearly all the heat radiated is infrared radiation, for which the emissivity is essentially $1,$ regardless of the amount of skin pigment.) (c) How much net heat would radiation remove from the person's body if the air temperature is $20^{\circ} \mathrm{C} ?$ (d) Take measurements on your own body to test the validity of the area formula. Treat yourself as a sphere and several cylinders.

Radiation by the body. The amount of heat radiated by the body depends on its surface temperature and area. Typically, this temperature is about $30^{\circ} \mathrm{C}$ (although it can vary). The surface area depends on the person's height and weight. Anempirical formula for the surface area of a person's body is $A\left($ in $\mathrm{m}^{2}\right)=(0.202) M^{0.425} h^{0.725}$
where $M$ is the person's mass (in kilograms) and $h$ is his or her height (in meters).(a) What would be the surface area of a 165 lb $(75 \mathrm{kg}), 6 \mathrm{ft}(1.83 \mathrm{m})$ person? (This is a good chance to use the $y^{x}$ key of your calculator.) (b) How much heat would the person radiate away per second at a skin temperature of $30^{\circ} \mathrm{C} ?$ (At the low temperatures of room-temperature objects, nearly all the heat radiated is infrared radiation, for which the emissivity is essentially $1,$ regardless of the amount of skin pigment.) (c) How much net heat would radiation remove from the person's body if the air temperature is $20^{\circ} \mathrm{C} ?$
(d) Take measurements on your own body to test the validity of the area formula. Treat yourself as a sphere and several cylinders.

Key Concepts

Blackbody Radiation

This concept refers to the idealized physical behavior in which an object emits electromagnetic radiation purely as a function of its temperature. Although real objects do not perfectly act as blackbodies, the idea provides a fundamental basis for understanding how all matter emits thermal radiation, particularly in applications such as infrared heat loss from the human body.

Stefan-Boltzmann Law

A key principle in thermodynamics, the Stefan-Boltzmann Law states that the total energy radiated per unit surface area of a blackbody is proportional to the fourth power of its absolute temperature. This law is essential in calculating the radiative heat loss for surfaces by relating temperature and radiated energy.

Empirical Estimation of Body Surface Area

This concept involves using empirically derived formulas to estimate the surface area of a human body based on measurable characteristics like mass and height. Such estimations are important in fields like medicine and physiology, where body surface area is correlated with metabolic rate, heat loss, and drug dosage calculations.

Emissivity

Emissivity is a material-specific property that quantifies how effectively a surface emits thermal radiation compared to a perfect blackbody. Values range between 0 and 1, with an emissivity of 1 indicating total emission efficiency. This concept is critical when applying the Stefan-Boltzmann Law to real-world objects, as it modifies the idealized calculations to account for surface characteristics.

Net Radiative Heat Transfer

This concept describes the overall energy exchange by radiation between an object and its environment, which depends on both the temperature of the object and that of its surroundings. The net transfer is determined by the difference in radiative energy emitted and received, making it a key factor in understanding how bodies lose or gain heat in a thermal environment.

Transcript

00:01 Let's take a look at the first question.

00:02 So the first question was asking us to determine the surface area of a human body, okay, which is like 75 kilogram and the height is 1 .83 meter.

00:18 Okay? so we know this equation final question.

00:24 So all these two is just plugging the value window, which is the mass is 75 kilogram and the height is 1 .83 meter.

00:31 If you're plugging, it should give you the answer is 1 .96 meter square.

00:38 Let's look at the question b.

00:40 Question b was asking us, the heats will radiate on a person per second when the skin temperature is at 30 degrees celsius.

00:53 Well, this is the equation for the heat radiation per second at a certain temperature.

01:00 H is the heat current or not always the power or the heat per second.

01:05 Okay.

01:07 A is the surface area.

01:09 E is the invisivity.

01:14 Okay.

01:15 Invisivity here is usually equal to one.

01:17 So basically you can just, yeah, set equal to one.

01:22 So this is equal to one.

01:24 So we really give us, it just, oh, it would just be like this.

01:35 Just a sigma t2 power of 4.

01:38 Sigma is the step on constant, which is 5 .6.

01:42 6, 7 times 10 to the power negative 8, watts per meter square times kelvin to the power of 4.

01:52 A, we already know it, we are from the previous question, and t is just the temperature.

01:59 Okay, in the question you're saying it's at temperature of 30 degrees celsius.

02:04 So if we convert it to kelvin, it's 30 plus 273, which will give us 30 -3 kelvin.

02:12 Now if we're plugging a value, which should give us, let me go like this.

02:20 So the surface area was 1 .96 meter square times sigma, which is 5 .67 times 10 to 98, watts per meter square times kelvin to the power of 4, and then times 303 kelvin to the power of 4...

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