Let's put this in a theoretical box which is perfectly insulated, so no energy by any form can be lost from the box.
3W comes in. What happens to it? Some is converted to kinetic energy to spin the disk, some is converted to light energy to display LEDs, some is converted to heat energy as the chip components "work" (which is effectively kinetic energy at an atomic (?) level converting into heat).
Assuming the box is at 0degC at 00:00am. What temperature is that box at 24 hours later?
Matt
That's a tricky one. 3W is 3 joules/second, so over 24 hours you'll have pumped 259200 joules of energy into the box.
The temperature rise you get depends on both the mass of the box, and what it is made of (i.e. the specific heat capacity of the material). For example, 4200 joules will raise 1 Kg of water by 1 degree C, but to raise 1Kg of Iron by 1 degree C only needs about a tenth of that energy, 449 joules.
I'd hate to try and estimate the theoretical specific heat capacity of a Freesat box
That's a tricky one. 3W is 3 joules/second, so over 24 hours you'll have pumped 259200 joules of energy into the box.
The temperature rise you get depends on both the mass of the box, and what it is made of (i.e. the specific heat capacity of the material). For example, 4200 joules will raise 1 Kg of water by 1 degree C, but to raise 1Kg of Iron by 1 degree C only needs about a tenth of that energy, 449 joules.
I'd hate to try and estimate the theoretical specific heat capacity of a Freesat box
OK then, its a perfectly insulated box which contains exactly 1kg of water less the mass of the freesat box. The components in a freesat box a) will work underwater (!) and b) have exactly the same specific heat capacity as water. So you have a mass weighing 1kg to be heated at the same rate as water.
End result, with my theoretical constraints, are that the water would rise in temperature by (3 joules x 60s x 60m x 24hrs)/4200 = 61.7 degrees.
Which just goes to show that Freesat boxes are no use as kettles
Arrived in Somerset, so pleased we now have a genuine and friendly discussion over a perplexing and apparently contradictory set of ideas. No idea which point of view is right (may be neither :eek:). At least it;s now possible to discuss it without rancour and insults
Long time lurker on these forums but very rare poster. I appreciate that this is well off the original topic but I have to put my two pennyworth in here. When I got my degree in engineering (40 years ago) it was taught that the first law of thermodynamics states that (short of nuclear fusion/fission) energy cannot be created but simply converted from one form to another. Furthermore the lowest form of energy is heat, and all energy eventually ends up as heat (the law of increasing entropy – or second law of thermodynamics) . Electricity can be converted into light, sound, mechanical but eventually all energy degenerates to heat. This almost always be traced back to friction.
In the case of the small 15W incandescent light bulb vs the low energy light bulb (we are talking a true 15W low energy here I hope not a 15W “Equivalent”), the incandescent bulb will have a much smaller surface area hence to dissipate 15W of heat the surface of the bulb will have to get much hotter. If you put them both in your identical box, my money will be on an identical temperature rise over a given time.
The satellite’s in space have a high KE which they are able to keep most of since there is very little air friction to (generate heat and ) slow them down. However in low earth orbit there is a little bit of air and over time they lose the KE to friction and their orbit decays. Then all of the KE is transferred to heat in spectacular fashion.
We shall see what happens. Anyone else fancy a play ?
We need a Phsyicist, where are you Brian Cox
Looked at this forum loads of times and it's been very informative, but I've never been able to contribute anything useful. However I can't resist posting on a thread that starts with a simple question about connecting a satellite box, has a slanging match between two engineers about thermodynamics in the middle and then moves onto a debate about the effects of sealing light bulbs in boxes!
I'm no Brian Cox but I do have a physics degree so for what it's worth, here is my go. The relective surfaces inside a box would not be 100% efficient so some photons are absorbed each time light strikes them, which is why you would not get a blinding flash when you opened the box. The energy of the absorbed photons will be converted to heat. The box itself will pass heat to its surroundings by conduction into the air in contact with its exterior surfaces and by emitting infrared radiation.The box will heat up until it reaches thermal equilibrium with its surroundings. Assuming both types of bulb are receiving 15W of power, then both boxes should reach the same final stable temperature all other things being equal.
Looked at this forum loads of times and it's been very informative, but I've never been able to contribute anything useful. However I can't resist posting on a thread that starts with a simple question about connecting a satellite box, has a slanging match between two engineers about thermodynamics in the middle and then moves onto a debate about the effects of sealing light bulbs in boxes!
I'm no Brian Cox but I do have a physics degree so for what it's worth, here is my go. The relective surfaces inside a box would not be 100% efficient so some photons are absorbed each time light strikes them, which is why you would not get a blinding flash when you opened the box. The energy of the absorbed photons will be converted to heat. The box itself will pass heat to its surroundings by conduction into the air in contact with its exterior surfaces and by emitting infrared radiation.The box will heat up until it reaches thermal equilibrium with its surroundings. Assuming both types of bulb are receiving 15W of power, then both boxes should reach the same final stable temperature all other things being equal.
Interesting but still doesn't explain it fully as far as i can see.
Firstly the incandescent bulb in producing more heat than light will increase the internal temperature of the box to the equilibrium temperarure presumably in a much shorter period than the low energy bulb of the same wattage.
After it's turned off it continues to produce heat until it cools down, the low energy bulb ceases to pruduce photon output instantly (An ac flourescent design goes out every 1/100 second). The heat lost will surely be more simply because of the higher average internal temperature.
In a real room situation a significant proportion of the photons will simply go out the windows. It's usually obvious to an outside observer that the room is ilumminated even when the curtaiins are drawn.
Why do we wrap satellites in pretty gold foil, why do we paint houses white in hot countries.
Consider what a mirror does to photons, a surface silvered one reflects them all. That's why we collect tiny amounts of photons from distant stars and galaxies, bounce them off a surface silvered parabolic mirror and focus then onto a photographic plate, tracking the source for hours. The photons create a chemical change in the phptographic emulsion (heat doesn't work). For decades the worlds greatest reflecting telescopes have used this principle.
Consider what a mirror does to photons, a surface silvered one reflects them all. That's why we collect tiny amounts of photons from distant stars and galaxies, bounce them off a surface silvered parabolic mirror and focus then onto a photographic plate, tracking the source for hours. The photons create a chemical change in the phptographic emulsion (heat doesn't work). For decades the worlds greatest reflecting telescopes have used this principle.
There is no such thing as a perfect reflector the best mirrors only achieve about 97% - surface film optics achieve about 99.9999% but nothing achieves 100%. There is some absorption and conversion to another form of energy (usually heat) in every case.
There is no such thing as a perfect reflector the best mirrors only achieve about 97% - surface film optics achieve about 99.9999% but nothing achieves 100%. There is some absorption and conversion to another form of energy (usually heat) in every case.
So presumably with photons travelling at the speed of light (obviously!) the reflection @ 99.99% effectiveness quite quickly decays into zero emitted light and all the energy of the light converted to heat?
Bottom line is, that energy has got to go somewhere or be stored as a potential ready for release.
There is no such thing as a perfect reflector the best mirrors only achieve about 97% - surface film optics achieve about 99.9999% but nothing achieves 100%. There is some absorption and conversion to another form of energy (usually heat) in every case.
99.9999% is pretty close to 100%. About the loss of 10 photons in 10000.
According to this a ccd can actually count individual photons
My maths may be very wrong here, but with 99.9999% efficient reflectivity, a single burst of photons would be reduced to 0.000000000009554% of its initial "strength" after just 1/10th of a second, assuming a mirrored box measuring 1m x 1m x 1m.
(My formula in excel =power(0.999999,29979245.8) where light has a speed of 299,792,458 m/s))
Be wary of what you learn - my only physics capability is that I got a B at GCSE (I was robbed, should have had an A), but that was 22 years ago!! My sister has a PhD in astrophysics if that helps my credibility at all...
My maths may be very wrong here, but with 99.9999% efficient reflectivity, a single burst of photons would be reduced to 0.000000000009554% of its initial "strength" after just 1/10th of a second, assuming a mirrored box measuring 1m x 1m x 1m.
(My formula in excel =power(0.999999,29979245.8) where light has a speed of 299,792,458 m/s))
Matt
I may be wrong but a photon is a fundamental particle (like the supposedly faster than light neutrinos - whole new thread ). You can't have a bit of a particle, it either exists or it doesn't. In the case of the mirror the particle must either be diverted or destroyed. If you look at the ccd link you will see you can count the individual photons.
I suppose I am saying that (assuming my maths is right!) that if 0.000001% of the photons per reflection are "diverted or destroyed" each time the burst of light hits the reflective surface, then the percentage I've quoted will be the amount left. Pure theory of course, because for an individual photon it will either be reflected or something else will happen to it, so when the burst of light gets down to a very weak level, the 99.99999% efficiency will no longer hold true.
I may be wrong but a photon is a fundamental particle (like the supposedly faster than light neutrinos - whole new thread ). You can't have a bit of a particle, it either exists or it doesn't. In the case of the mirror the particle must either be diverted or destroyed. If you look at the ccd link you will see you can count the individual photons.
Wot about virtual particles? Do they exist, or not ?
I may be wrong but a photon is a fundamental particle (like the supposedly faster than light neutrinos - whole new thread ). You can't have a bit of a particle, it either exists or it doesn't. In the case of the mirror the particle must either be diverted or destroyed. If you look at the ccd link you will see you can count the individual photons.
CCDs can indeed detect individual photons - the best CCDs are over 90% efficient. A photon can be thought of as a small wave packet or a particle but someone has already mentioned particle wave duality and that's another can of worms.
The simplest way to see the absorption of light by silvered surfaces is to stand between two mirrors. As the reflections recede into the distance the images become dimmer.
Comments
The temperature rise you get depends on both the mass of the box, and what it is made of (i.e. the specific heat capacity of the material). For example, 4200 joules will raise 1 Kg of water by 1 degree C, but to raise 1Kg of Iron by 1 degree C only needs about a tenth of that energy, 449 joules.
I'd hate to try and estimate the theoretical specific heat capacity of a Freesat box
Matt
OK then, its a perfectly insulated box which contains exactly 1kg of water less the mass of the freesat box. The components in a freesat box a) will work underwater (!) and b) have exactly the same specific heat capacity as water. So you have a mass weighing 1kg to be heated at the same rate as water.
End result, with my theoretical constraints, are that the water would rise in temperature by (3 joules x 60s x 60m x 24hrs)/4200 = 61.7 degrees.
Which just goes to show that Freesat boxes are no use as kettles
But they do heat things!!
Matt
In the case of the small 15W incandescent light bulb vs the low energy light bulb (we are talking a true 15W low energy here I hope not a 15W “Equivalent”), the incandescent bulb will have a much smaller surface area hence to dissipate 15W of heat the surface of the bulb will have to get much hotter. If you put them both in your identical box, my money will be on an identical temperature rise over a given time.
The satellite’s in space have a high KE which they are able to keep most of since there is very little air friction to (generate heat and ) slow them down. However in low earth orbit there is a little bit of air and over time they lose the KE to friction and their orbit decays. Then all of the KE is transferred to heat in spectacular fashion.
Matt
Looked at this forum loads of times and it's been very informative, but I've never been able to contribute anything useful. However I can't resist posting on a thread that starts with a simple question about connecting a satellite box, has a slanging match between two engineers about thermodynamics in the middle and then moves onto a debate about the effects of sealing light bulbs in boxes!
I'm no Brian Cox but I do have a physics degree so for what it's worth, here is my go. The relective surfaces inside a box would not be 100% efficient so some photons are absorbed each time light strikes them, which is why you would not get a blinding flash when you opened the box. The energy of the absorbed photons will be converted to heat. The box itself will pass heat to its surroundings by conduction into the air in contact with its exterior surfaces and by emitting infrared radiation.The box will heat up until it reaches thermal equilibrium with its surroundings. Assuming both types of bulb are receiving 15W of power, then both boxes should reach the same final stable temperature all other things being equal.
Interesting but still doesn't explain it fully as far as i can see.
Firstly the incandescent bulb in producing more heat than light will increase the internal temperature of the box to the equilibrium temperarure presumably in a much shorter period than the low energy bulb of the same wattage.
After it's turned off it continues to produce heat until it cools down, the low energy bulb ceases to pruduce photon output instantly (An ac flourescent design goes out every 1/100 second). The heat lost will surely be more simply because of the higher average internal temperature.
In a real room situation a significant proportion of the photons will simply go out the windows. It's usually obvious to an outside observer that the room is ilumminated even when the curtaiins are drawn.
Why do we wrap satellites in pretty gold foil, why do we paint houses white in hot countries.
Consider what a mirror does to photons, a surface silvered one reflects them all. That's why we collect tiny amounts of photons from distant stars and galaxies, bounce them off a surface silvered parabolic mirror and focus then onto a photographic plate, tracking the source for hours. The photons create a chemical change in the phptographic emulsion (heat doesn't work). For decades the worlds greatest reflecting telescopes have used this principle.
There is no such thing as a perfect reflector the best mirrors only achieve about 97% - surface film optics achieve about 99.9999% but nothing achieves 100%. There is some absorption and conversion to another form of energy (usually heat) in every case.
So presumably with photons travelling at the speed of light (obviously!) the reflection @ 99.99% effectiveness quite quickly decays into zero emitted light and all the energy of the light converted to heat?
Bottom line is, that energy has got to go somewhere or be stored as a potential ready for release.
Matt
99.9999% is pretty close to 100%. About the loss of 10 photons in 10000.
According to this a ccd can actually count individual photons
See the use in Astronomy here
http://en.wikipedia.org/wiki/Charge-coupled_device
(My formula in excel =power(0.999999,29979245.8) where light has a speed of 299,792,458 m/s))
Matt
Matt
I may be wrong but a photon is a fundamental particle (like the supposedly faster than light neutrinos
I suppose I am saying that (assuming my maths is right!) that if 0.000001% of the photons per reflection are "diverted or destroyed" each time the burst of light hits the reflective surface, then the percentage I've quoted will be the amount left. Pure theory of course, because for an individual photon it will either be reflected or something else will happen to it, so when the burst of light gets down to a very weak level, the 99.99999% efficiency will no longer hold true.
Matt
Wot about virtual particles? Do they exist, or not ?
http://www2.slac.stanford.edu/vvc/theory/virtual.html
Ah, but what about wave-particle duality?
CCDs can indeed detect individual photons - the best CCDs are over 90% efficient. A photon can be thought of as a small wave packet or a particle but someone has already mentioned particle wave duality and that's another can of worms.
The simplest way to see the absorption of light by silvered surfaces is to stand between two mirrors. As the reflections recede into the distance the images become dimmer.