# Is there an equation relating distance travelled with a point of heat?

Watch
Announcements
Thread starter 6 years ago
#1
Hi,

A physics-ey question I've been thinking about for a long time.
I was thinking about an iron (as in ironing iron) after I was told to turn it off, and after using it to warm my hands a little I was wondering if there was an equation that would relate the temperature of the iron with the distance the heat would have to travel.

Naturally, we have to assume normal conditions (NTP) where the particles in the air do not move and are static (so as not to have wind blowing - this would reduce temperature for a given distance).
I'm familiar with the inverse square law where:

Intensity =

1________
distance^2

(sorry, a bit crudely done)
I was also thinking that as heat is pretty much IR radiation (in this case as it is we are simply thinking about the transfer of the energy from the iron to its surroundings) and so the only radiation being emitted is IR (other frequencies can be counted as negligible. We can use:

Intensity =

Power
Area

where Power = Luminosity (again, assumption can be made as we're assuming only radiation is IR) and so L = k(T^4)4pi(r^2)
(where k = is the Stefan Boltzmann constant 5.67x10^-8) (sorry, can't do a sigma sign).
and so we can sub that in and so:

k(T^4)4pi(r^2)
A
proportional to:

1________
distance^2

But here's where it's a bit weird. The Power equation is at max power and at max Power P = k T^4 (again k is constant) so:

k T^4 proportional to:

1________
distance^2

But that can't be correct can it? As it can be solved simultaneously and find a constant which I don't think exists?
Backtracking to:

Intensity =
P
A
P_______=
4pi(r^2)

(P_)r^2
(4pi )
where P over 4pi sounds like a more reasonable constant. Can someone clear this up for me please?
0
reply
6 years ago
#2
(Original post by Icicle Man)
Hi,

A physics-ey question I've been thinking about for a long time.
I was thinking about an iron (as in ironing iron) after I was told to turn it off, and after using it to warm my hands a little I was wondering if there was an equation that would relate the temperature of the iron with the distance the heat would have to travel.

Naturally, we have to assume normal conditions (NTP) where the particles in the air do not move and are static (so as not to have wind blowing - this would reduce temperature for a given distance).
I'm familiar with the inverse square law where:
1
Intensity = --------------
distance^2

(sorry, a bit crudely done)
I was also thinking that as heat is pretty much IR radiation (in this case as it is we are simply thinking about the transfer of the energy from the iron to its surroundings) and so the only radiation being emitted is IR (other frequencies can be counted as negligible. We can use:
Power
Intensity = ----------
Area

where Power = Luminosity (again, assumption can be made as we're assuming only radiation is IR) and so L = k(T^4)4pi(r^2)
(where k = is the Stefan Boltzmann constant 5.67x10^-8) (sorry, can't do a sigma sign).
and so we can sub that in and so:

k(T^4)4pi(r^2) 1
----------------- proportional to --------------
A distance^2

But here's where I'm a little confused, the Power equation is at max power and at max Power P = k T^4 (again k is constant) and so the A and the 4pi(r^2) cancel and so we're left with:
1
k(T^4) proportional to -----------
distance^2

But this doesn't look right as the equation above can be solved relatively easily simultaneously and so the constant of proportionality will be a constant that doesn't change for any material.
Going back to:

I = ---
It's really hard to read through this. Do you know that TSR has LaTeX built into it? if you enclose math statements with {latex}{/latex} tags (use [square brackets] though), then you can write fractions easily! The code for a fraction is 1/2 is written as \dfrac{1}{2}.

Not quite sure what your question is at the moment. As far as I'm aware heat transfer has a different equation for each of the three possible modes: Fourier's law for conduction, Newton's law for convection and Stefan-Boltzmann law for radiation.
0
reply
Thread starter 6 years ago
#3
(Original post by lerjj)
It's really hard to read through this. Do you know that TSR has LaTeX built into it? if you enclose math statements with {latex}{/latex} tags (use [square brackets] though), then you can write fractions easily! The code for a fraction is 1/2 is written as \dfrac{1}{2}.

Not quite sure what your question is at the moment. As far as I'm aware heat transfer has a different equation for each of the three possible modes: Fourier's law for conduction, Newton's law for convection and Stefan-Boltzmann law for radiation.
Sorry, I know, it looks terrible but I've done my best to clear it up now.
I've never used LaTex before so would probably have to spend some time learning it before I use it, but I will keep that in mind, thanks.
0
reply
Thread starter 6 years ago
#4
(Original post by lerjj)
It's really hard to read through this. Do you know that TSR has LaTeX built into it? if you enclose math statements with {latex}{/latex} tags (use [square brackets] though), then you can write fractions easily! The code for a fraction is 1/2 is written as \dfrac{1}{2}.

Not quite sure what your question is at the moment. As far as I'm aware heat transfer has a different equation for each of the three possible modes: Fourier's law for conduction, Newton's law for convection and Stefan-Boltzmann law for radiation.
Please see: http://www.thestudentroom.co.uk/show....php?t=3051031
0
reply
6 years ago
#5
The total power radiated doesn't obey an inverse square law - just the power per unit area.

e.g. if you had an iron radiating 100W in a pokey bedsit the rate at which heat was hitting the inner surfaces of the room... the 4 walls, ceiling and floor combined would be 100W

if you stuck the same iron in the royal Albert hall the rate at which heat was hitting inner surface of the room would also be 100W, but spread over a much larger area.

apologies if this doesn't answer the question, I'm not really sure what you're asking tbh.
0
reply
Thread starter 6 years ago
#6
(Original post by Joinedup)
The total power radiated doesn't obey an inverse square law - just the power per unit area.

e.g. if you had an iron radiating 100W in a pokey bedsit the rate at which heat was hitting the inner surfaces of the room... the 4 walls, ceiling and floor combined would be 100W

if you stuck the same iron in the royal Albert hall the rate at which heat was hitting inner surface of the room would also be 100W, but spread over a much larger area.

apologies if this doesn't answer the question, I'm not really sure what you're asking tbh.
That's exactly what I mean, collectively the energy dissipated would be the same, but is there a way to measure the energy, e.g. say a 1 metre radius away from an iron, and say "hey, let's use this equation and find the temperature at 1m from the iron". I mean, because the energy given out is 100w so we should technically be able to find the energy closer to the iron, right?

P.S.HAPPY NEW YEAR ALL
0
reply
6 years ago
#7
(Original post by Icicle Man)
That's exactly what I mean, collectively the energy dissipated would be the same, but is there a way to measure the energy, e.g. say a 1 metre radius away from an iron, and say "hey, let's use this equation and find the temperature at 1m from the iron". I mean, because the energy given out is 100w so we should technically be able to find the energy closer to the iron, right?

P.S.HAPPY NEW YEAR ALL
do you mean the equilibrium temperature of an object at a given distance from the iron - i.e. the temperature at which it's radiating away all the heat it's recieving from the iron?

fwiw you do this type of calc in planetary science to estimate the surface temperature of planets - course there you don't have air in the room to transfer heat away from the body receiving the radiation by convection and you can reasonably assume outer space is only radiating negligable amounts of heat into your system (in buildings you'll have walls etc which are actually going to be quite hot in absolute temperature - even in January)

a lot of the assumption that you can make in astronomy are actually simplifying the calculation for you - planets & stars tend to be nearly spherical which is a handy shape. I'd suggest thinking about planets if you want to keep it fairly simple but you could do it for irons in rooms if you took all the additional variables and heat transfer mechanisms into consideration.
0
reply
Thread starter 6 years ago
#8
(Original post by Joinedup)
do you mean the equilibrium temperature of an object at a given distance from the iron - i.e. the temperature at which it's radiating away all the heat it's recieving from the iron?

fwiw you do this type of calc in planetary science to estimate the surface temperature of planets - course there you don't have air in the room to transfer heat away from the body receiving the radiation by convection and you can reasonably assume outer space is only radiating negligable amounts of heat into your system (in buildings you'll have walls etc which are actually going to be quite hot in absolute temperature - even in January)

a lot of the assumption that you can make in astronomy are actually simplifying the calculation for you - planets & stars tend to be nearly spherical which is a handy shape. I'd suggest thinking about planets if you want to keep it fairly simple but you could do it for irons in rooms if you took all the additional variables and heat transfer mechanisms into consideration.
Not entirely sure what you mean by equilibrium temperature, I just meant the temperature it's at at 'x' distance from the iron which I think is the same thing as 'radiating away all the heat it's receiving from the iron'.

I can see how this could be applied to a planets and that's why I posed my question as a point of heat as that would be far simpler to model than a surface like an iron.

Actually, modelling as a planet makes everything a lot easier to think about as external factors like walls don't need to be considered. So, is there any equations that allow stuff like this to be calculated as I've never come across anything like this.

As a follow up in thinking; say our atmosphere, would the temperature at 'x' distance from our earth (which has an atmosphere) be significantly greater than that of a planet with no atmosphere and is just surrounded by a vacuum.
My guess is these planetary calculations model a planet to have no atmosphere as otherwise they would have to take into account how greenhouses gases can sustain the heat and so make things warmer at a closer distance and keep the atmosphere warmer than anything outside the atmosphere.
0
reply
6 years ago
#9
(Original post by Icicle Man)
Not entirely sure what you mean by equilibrium temperature, I just meant the temperature it's at at 'x' distance from the iron which I think is the same thing as 'radiating away all the heat it's receiving from the iron'.

I can see how this could be applied to a planets and that's why I posed my question as a point of heat as that would be far simpler to model than a surface like an iron.

Actually, modelling as a planet makes everything a lot easier to think about as external factors like walls don't need to be considered. So, is there any equations that allow stuff like this to be calculated as I've never come across anything like this.

As a follow up in thinking; say our atmosphere, would the temperature at 'x' distance from our earth (which has an atmosphere) be significantly greater than that of a planet with no atmosphere and is just surrounded by a vacuum.
My guess is these planetary calculations model a planet to have no atmosphere as otherwise they would have to take into account how greenhouses gases can sustain the heat and so make things warmer at a closer distance and keep the atmosphere warmer than anything outside the atmosphere.
Well space with radiation going through it doesn't have a temperature.

you could assume the planet absorbs radiation through the disk it presents to the star and radiates from its entire surface

Flux at distance d from source

Flux F (W/m^2)
Luminosity L (W)
Distance d (m)

F=L/(4πd^2)  (from surf. area of a sphere)
----
Power absorbed over the planetary disk facing the sun

Planet radius R (m)
Pin=FπR^2  (from surf. area of a circle)
-----
Power radiated from entire planet surface at Temperature T
Pout=4πR^2σT^4  (from surf. area of a sphere & S-B’s)
-----
There will be a Teqm at which heat in and heat out are balanced

At equilibrium Pin=Pout

FπR^2=4πR^2σTeqm^4 (πR^2 on either side can be cancelled)
F=4σTeqm^4 (rearrange for Teqm)
Teqm=(F/4σ)^0.25 
-----

So the only variable affecting Teqm is the flux and the variables affecting the flux are luminosity and distance - the radius of the planet has cancelled during calculation telling us that the temp doesn't depend on the radius of the planet... all spherical objects at the same distance from the same star will have the same temperature... (which is why you might sometimes hear people talking *as if* there is a temperature in the space around the star - the equilibrium temperature of something at that distance from the star)

I've assumed all incident radiation on the planet is absorbed - in more realistic models an albedo would be included
0
reply
Thread starter 6 years ago
#10
That is some serious level of physics going on in that answer. Thank you for taking the time to write it.
A few parts I have a few questions on:
In your description "from surf. area of a sphere & S-B's", what are S-B's?

Also, are we allowed to model a planet radius using pi(r^2) as that would mean we assume the planet (in this case star) is a completely flat disc. The fact that it's 3d would surely mean the energy reached by the closest part vs 2 most distant is different; are we assuming the energies are equal purely because of the distance the energy has already travelled (i.e. thousands of miles)? Is there a way to model this in a way we can take into account the semi-spherical shape of the planet?

Also, how comes we can assume there's a point where there is an equilibrium? Are we saying "At x distance from P and y distance from Q" the energies are the same? Have I understood that right?
We've also assumed that the radii of the planets are same seen as we cancelled out pi(r^2), but as heat spreads out in more than a single direction, surely the incident area where the energy strikes has a larger radii than the initial planet? Or are we altering the calculation to only include the same radii by making Pin = Pout?

Thanks.
0
reply
6 years ago
#11
(Original post by Icicle Man)
That is some serious level of physics going on in that answer. Thank you for taking the time to write it.
A few parts I have a few questions on:
In your description "from surf. area of a sphere & S-B's", what are S-B's?
I'm using S-B to stand for Stefan-Boltzmann law E=σT^4 which applies to planets radiating heat as well as stars - radiation is the only way for a body to lose heat in outer space.

Also, are we allowed to model a planet radius using pi(r^2) as that would mean we assume the planet (in this case star) is a completely flat disc. The fact that it's 3d would surely mean the energy reached by the closest part vs 2 most distant is different; are we assuming the energies are equal purely because of the distance the energy has already travelled (i.e. thousands of miles)? Is there a way to model this in a way we can take into account the semi-spherical shape of the planet?
It's valid to model the area of a planet exposed to solar radiation as a disk for looking at it's total heat input since it's radius is very much smaller than it's distance from the sun.

e.g. mercury
average radius 2.440x10^6m
average distance to sun 5.791x10^10m

each square meter of surface receives an amount of sunlight proportional to the sine of the angle at which the incoming radiation hits the surface - this is why the poles on earth (and mars) are colder than the equator... not because they are further away from the sun.
modelling this would be more tricky.

Also, how comes we can assume there's a point where there is an equilibrium? Are we saying "At x distance from P and y distance from Q" the energies are the same? Have I understood that right?
well if it doesn't reach an equilibrium we're in trouble, the alternatives would be that it carried on getting hotter and hotter forever which is an absurdity because we're assuming the rate of heat arriving at the planet is constant but the rate at which the planet loses heat by radiation is proportional to T^4

We've also assumed that the radii of the planets are same seen as we cancelled out pi(r^2), but as heat spreads out in more than a single direction, surely the incident area where the energy strikes has a larger radii than the initial planet? Or are we altering the calculation to only include the same radii by making Pin = Pout?

Thanks.

not quite sure what you mean - in this model photons from the sun go in straight lines and they either hit the planet (and are absorbed) or they don't and continue off into outer space, it just depends on whether they pass further than r from the centre of the planet or not.
0
reply
X

### Quick Reply

Write a reply...
Reply
new posts Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

#### What factors affect your mental health the most right now?

Anxiousness about lockdown easing (64)
4.58%
Uncertainty around my education (216)
15.45%
Uncertainty around my future career prospects (153)
10.94%
Lack of purpose or motivation (199)
14.23%
Lack of support system (eg. teachers, counsellors, delays in care) (60)
4.29%
Impact of lockdown on physical health (81)
5.79%
Loneliness (125)
8.94%
Financial worries (48)
3.43%
Concern about myself or my loves ones getting/having been ill (64)
4.58%
Exposure to negative news/social media (67)
4.79%
Lack of real life entertainment (72)
5.15%
Lack of confidence in making big life decisions (120)
8.58%
Worry about missed opportunities during the pandemic (129)
9.23%

View All
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.