# Why is the Stefan-Boltzman formula for this problem like this?

• s3a
In summary: Thank you for clarifying the difference and the reason for not editing initial posts. I'll keep that in mind!Also, I see what you're saying about how the equation I derived doesn't really make sense as an equation. So, I guess I'll just forget about it and stick with the two correct equations we've been talking about.
s3a

## Homework Statement

Consider a naked human body, with a surface area of 1.5 m^2 and an average temperature of 30° C (skin temperature), that stands at the middle of an empty breezy room whose surfaces have a temperature of 10° C.

Assuming the human body as a blackbody (5% error), compute the net rate of heat loss of this person by radiation in watts.

## Homework Equations

Stefan-Boltzman Law:
(i) P = dE/dt = ε σ A T^4
(ii) P = dE/dt = ε σ A ( T^4 - (T_(surrounding environment)^4 )

## The Attempt at a Solution

To actually answer the question of the problem, I'm pretty sure that it just wants me to do as follows.:

P = dE/dt = ε σ A ( T^4 - (T_(surrounding environment)^4 )

P = (1) (5.67E-8) (1.5) (303.15^4 - 283.15^4)

P = 171.6086089452735 W.Having said that, since the surrounding environment is a non-vacuum (meaning that it is a set of stuff that can and do emit radiation), should the equation instead be

P = ε_human σ A_human (T_human)^4 - ε_human σ A_human (T_walls)^4
(the double A_human is not a typo)

P = ε_human σ A_human (T_human)^4 - ε_walls σ A_human (T_walls)^4

P = σ A_human [ε_human (T_human)^4 - ε_walls (T_walls)^4], such that

P = σ A_human [ε_human (T_human)^4 - ε_walls (T_walls)^4] = σ A_human ε [(T_human)^4 - (T_walls)^4] only if ε_human = ε_walls?

Any input would be GREATLY appreciated!

Last edited:
10 oC = 283 K

Tsurrounding environment is what it is regardless of the emissivity which affects the radiated power but not the temperature.

kuruman said:
10 oC = 283 K

Tsurrounding environment is what it is regardless of the emissivity which affects the radiated power but not the temperature.
Oops! I accidentally thought it was -10° C. I corrected it in my main post.

Also, about what you're saying involving the temperature of the surrounding environment, that makes sense, but it doesn't agree with the math I showed in my main post. Am I simply wrong about how that formula gets its ( T^4 - (T_(surrounding environment))^4 ) part, such that it just happens to work out to be the same algebraically by coincidence?

s3a said:
Also, about what you're saying involving the temperature of the surrounding environment, that makes sense, but it doesn't agree with the math I showed in my main post. Am I simply wrong about how that formula gets its ( T^4 - (T_(surrounding environment))^4 ) part, such that it just happens to work out to be the same algebraically by coincidence?
The correct equation you should use is (ii) in post #1. See also here
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html
The emissivity of the surroundings is irrelevant, you are not trying to figure out the power transferred to the surroundings.

kuruman said:
It seems that it's too late for me to edit my opening post, since the website seems to no longer allow it (but I'll keep that in mind for any future posts).

kuruman said:
The correct equation you should use is (ii) in post #1. See also here
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html
The emissivity of the surroundings is irrelevant, you are not trying to figure out the power transferred to the surroundings.
Also, sorry for still dragging this along, but is the difference between (i), (ii) and the one I derived that (ii) is for net power being transferred (not just radiated) to the surrounding environment in general (so not just the walls, but everywhere in the universe other than the human body), the one I derived is for net power being transferred (not just radiated) specifically to the surrounding walls (which is less than or equal to the the total / non-net amount being radiated (where not all of that total power being radiated is transferred) from the human body to the walls), and (i) is the total radiated amount (so not net transferred amount) by the human body?

(I) and (ii) assume different modeling.
In (i) the assumption is that power is lost by radiation and there is no power gain by radiation absorption from something outside the radiating body.
In (iI) the assumption is that power is lost by radiation and there is power gain by radiation absorption from the surroundings assumed to be (although they probably are not) at some average temperature Tsurroundings.

I am not sure what the equation you say derived is supposed to describe. What is the model that you followed to write it down?

For future reference: Editing initial posts is to be avoided. If you wish to take something back, post again and refer to the previous post for any correction or modification you wish to make.

kuruman said:
(I) and (ii) assume different modeling.
In (i) the assumption is that power is lost by radiation and there is no power gain by radiation absorption from something outside the radiating body.
In (iI) the assumption is that power is lost by radiation and there is power gain by radiation absorption from the surroundings assumed to be (although they probably are not) at some average temperature Tsurroundings.

I am not sure what the equation you say derived is supposed to describe. What is the model that you followed to write it down?
That makes sense, but it only clarifies the difference between (i) and (ii). So, basically, what's the difference between (ii) and what I derived?

The equation I derived is

P_(human body) = ε_(human body) σ A_(human body) T^4

P_(walls) = ε_walls σ A_walls T^4

P_(net power) = P_(human body) – P_walls,

P_(net power) = [ε_(human body) σ A_(human body) (T_(human body))^4] – [ε_walls σ A_walls (T_walls)^4],

and, then, assuming that A_(human body) ≤ A_walls (because the walls would need to radiate through at least as much surface area as the human body can take for the walls to have as big an effect as they could have),

P_(net power) = [ε_(human body) σ A_(human body) (T_(human body))] – [ε_walls σ A_(human body) (T_walls)^4],

P_(net power) = [ε_(human body) σ A_(human body) (T_(human body))] – [ε_walls σ A_(human body) (T_walls)^4],

P_(net power) = σ A_(human body) ( [ε_(human body) (T_(human body))] – [ε_walls (T_walls)^4] ).

kuruman said:
For future reference: Editing initial posts is to be avoided. If you wish to take something back, post again and refer to the previous post for any correction or modification you wish to make.
Alright.

s3a said:
That makes sense, but it only clarifies the difference between (i) and (ii). So, basically, what's the difference between (ii) and what I derived?
The difference is that (ii) is correct and what you derived is incorrect. Let me explain why since you say you understand (i).
(a) The power emitted is positive and given by Pemitted = εbody σ T4body.
(b) If the body also absorbs power, then that power has to be proportional to the emissivity of the body, not of the emitting surface. Think about it. A body that is painted black (ε≈1) will absorb more power than a body covered with aluminum foil (ε ≈ 0). Furthermore that power goes as the fourth power of the surroundings that emit the radiation. Thus, the absorbed power is Pabsorbed = εbody σ T4surroundings.
(c) It follows that the net emitted power is
Pnet = Pemitted - Pabsorbed = εbody σ T4body - εbody σ T4surroundings = εbody σ (T4body- T4surroundings).

kuruman said:
The difference is that (ii) is correct and what you derived is incorrect. Let me explain why since you say you understand (i).
(a) The power emitted is positive and given by Pemitted = εbody σ T4body.
(b) If the body also absorbs power, then that power has to be proportional to the emissivity of the body, not of the emitting surface. Think about it. A body that is painted black (ε≈1) will absorb more power than a body covered with aluminum foil (ε ≈ 0). Furthermore that power goes as the fourth power of the surroundings that emit the radiation. Thus, the absorbed power is Pabsorbed = εbody σ T4surroundings.
(c) It follows that the net emitted power is
Pnet = Pemitted - Pabsorbed = εbody σ T4body - εbody σ T4surroundings = εbody σ (T4body- T4surroundings).
You accidentally forgot about ##A_{body}## for both ##P_{emitted}## and ##P_{absorbed}##, right?

That is, you should have typed ##P_{net} = P_{emitted} - P_{absorbed} = ε_{body} σ A_{body} T^4_{body} - ε_{body} σ A_{body} T^4_{surroundings} = ε_{body} σ A_{body} (T^4_{body}- T^4_{surroundings})##, right?

So, can an emissivity be thought of as a "transferability" (meaning both an "emissivity" and a "receivability") (since emission from one perspective can be thought of as reception from another perspective)? (That is, the surface area of the human body can be thought of as radiating power into the human body (that is, "underneath" the human body’s skin) due to the surrounding temperature, right?)

Also, as a thought experiment, let’s say that the surrounding walls had an emissivity of exactly 0. Their temperature alone would cause there to be a (negative) absorption term for the net power transfer of the human body (via the non-zero emissitivity of the human body), right?

s3a said:
You accidentally forgot about AbodyA_{body} for both PemittedP_{emitted} and PabsorbedP_{absorbed}, right?
Yes I did. Thank you for the correction.
s3a said:
So, can an emissivity be thought of as a "transferability" (meaning both an "emissivity" and a "receivability") (since emission from one perspective can be thought of as reception from another perspective)? (That is, the surface area of the human body can be thought of as radiating power into the human body (that is, "underneath" the human body’s skin) due to the surrounding temperature, right?)
Right. A person who suffers from hypothermia is wrapped in a thermal blanket. By being highly reflective (low emissivity) the heat that would have otherwise been lost to the surroundings is reflected back into the body.
https://goneoutdoors.com/thermal-blanket-work-5145153.html
s3a said:
Also, as a thought experiment, let’s say that the surrounding walls had an emissivity of exactly 0. Their temperature alone would cause there to be a (negative) absorption term for the net power transfer of the human body (via the non-zero emissitivity of the human body), right?
In principle, yes. However, if the walls only lost heat without gaining any, eventually they would reach zero absolute temperature which is a violation of the third law of thermodynamics.

kuruman said:
Yes I did. Thank you for the correction.
No problem.

kuruman said:
Right. A person who suffers from hypothermia is wrapped in a thermal blanket. By being highly reflective (low emissivity) the heat that would have otherwise been lost to the surroundings is reflected back into the body.
https://goneoutdoors.com/thermal-blanket-work-5145153.html
kuruman said:
In principle, yes. However, if the walls only lost heat without gaining any, eventually they would reach zero absolute temperature which is a violation of the third law of thermodynamics.
When I said "reception from another perspective," I meant the skin absorbing 95% (so reflecting away 5%) of the thermal energy radiated to the human body by the walls (rather than the human body's own heat being reflected back inward).

And, before you said that "in (ii) the assumption is that power is lost by radiation and there is power gain by radiation absorption from the surroundings assumed to be (although they probably are not) at some average temperature Tsurroundings."

But, I'm still confused as to why the emissivity of the walls doesn't matter.

About investigating further as to why the emissivity of the walls doesn't matter, let's go back to my thought experiment of the walls having an emissivity of exactly 0, and let's say that instead, the walls had an emissivity that was very finitely close to 0 (so not exactly 0, like before) (that way the third law of thermodynamics is not violated, and the thought experiment becomes more realistic), and let's also consider them in another scenario to have an emissivity of exactly 1.

How is the human body keeping up with the rate of energy that the Stefan-Boltzman law's equation says should be collected from the walls (since watts are joules per second, after all) if the walls are giving away heat so slowly, in the case where the emissivity is very finitely close to 0 (unlike in the case where the walls' emissivity is exactly 1).

## 1. Why is the Stefan-Boltzman formula used to calculate thermal radiation?

The Stefan-Boltzman formula, also known as the Stefan-Boltzman law, is used to calculate thermal radiation because it relates the amount of radiation emitted by a blackbody to its temperature. This makes it a useful tool for understanding and predicting the behavior of thermal radiation in various systems.

## 2. How is the Stefan-Boltzman formula derived?

The Stefan-Boltzman formula is derived from the laws of thermodynamics and the theory of blackbody radiation. It was first developed by Austrian physicist Josef Stefan in 1879 and later refined by physicist Ludwig Boltzman.

## 3. What is the significance of the constant in the Stefan-Boltzman formula?

The constant in the Stefan-Boltzman formula, often denoted by the symbol σ, is known as the Stefan-Boltzman constant. It has a value of approximately 5.67 x 10^-8 W/m^2K^4 and represents the proportionality between the temperature of a blackbody and the amount of radiation it emits.

## 4. Can the Stefan-Boltzman formula be used for non-blackbody objects?

While the Stefan-Boltzman formula is derived for blackbodies, it can also be used to approximate the thermal radiation emitted by non-blackbody objects. This is achieved by introducing an emissivity factor, which represents the efficiency of an object in emitting thermal radiation compared to a perfect blackbody.

## 5. How is the Stefan-Boltzman formula applied in real-world scenarios?

The Stefan-Boltzman formula has various practical applications, including in the fields of astronomy, climate science, and engineering. It is used to calculate the radiation emitted by stars and planets, to model the Earth's energy balance and climate, and to design and optimize thermal systems such as solar panels and heat engines.

• Introductory Physics Homework Help
Replies
3
Views
112
• Introductory Physics Homework Help
Replies
41
Views
4K
• Introductory Physics Homework Help
Replies
3
Views
2K
• Introductory Physics Homework Help
Replies
8
Views
2K
• Introductory Physics Homework Help
Replies
3
Views
590
• Introductory Physics Homework Help
Replies
1
Views
881
• Introductory Physics Homework Help
Replies
2
Views
1K
• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
1
Views
2K
• Introductory Physics Homework Help
Replies
1
Views
1K