Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: cassie on 16/02/2007 11:25:57
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will the surface of the mirror get warm, or will all the light (heat) be reflected?
would try the experiment outside today, but it's blowing a hooly out there!
cheers
cassie
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If it were a perfect mirror at all wavelengths (colours including the ones you can't see) you are right it wouldn't heat up as all the energy would be reflective. However a normal mirror is not perfect and probably absorbs 5-10% of the visible light on it and a bit more in the infra-red. So it would heat up less than other objects, but still a bit.
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so would the glass on top of the reflective metal be the retainer?
and if so, is a metallic fabric going to be cooler to the touch because it has no glass infront?
thanks for your previous answer by the way, trying to settle a long standing arguement!
cheers
Caz
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They used to make car headlamps about 20cm diameter with an integral parabolic reflector and tungsten filament.
I have often wondered if one measured the resistance of the filament and pointed it to either the 300°K Earth or the 2.74°K clear sky one could notice a difference in the resistance
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I have measured the resistance of many light bulbs at different temperatures - it's something GCSE Science students all have to do. The resistance more or less doubles as the temperature changes from 300K to about 6000K. This is a rate of about 0.03% per degree C. You would need a pretty big temperature difference to produce a significant/measurable change in resistance. A 'bridge' circuit might help but, still.
How cold could we expect the filament of the light bulb to get?
It would have to rely on cooling by radiation.
If you could rely on significant good cooling with this arrangement there would surely be a refrigerator in every home using the principle. Or even Stirling Engines to tap this free source of energy.
I did hear, once, that the Arabs used to produce ice for their drinks centuries ago using the radiation cooling from large, shallow, ponds at night in the desert. However, the desert sky is very clear - very little pollution or water vapor and a lot of the heat loss could have been through evaporation.
You would have conduction down the connecting wires and some convection through the argon gas in the glass envelope.. Also, the filament is not totally screened from radiation from the Earth. I bet you'd only get a very few degrees of lowering of temperature. at most.
A possible experiment might be to use a large , deep, parabolic reflector and a sensitive thermistor.
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I have measured the resistance of many light bulbs at different temperatures - it's something GCSE Science students all have to do. The resistance more or less doubles as the temperature changes from 300K to about 6000K. This is a rate of about 0.03% per degree C. You would need a pretty big temperature difference to produce a significant/measurable change in resistance. A 'bridge' circuit might help but, still.
How cold could we expect the filament of the light bulb to get?
It would have to rely on cooling by radiation.
If you could rely on significant good cooling with this arrangement there would surely be a refrigerator in every home using the principle. Or even Stirling Engines to tap this free source of energy.
I did hear, once, that the Arabs used to produce ice for their drinks centuries ago using the radiation cooling from large, shallow, ponds at night in the desert. However, the desert sky is very clear - very little pollution or water vapor and a lot of the heat loss could have been through evaporation.
You would have conduction down the connecting wires and some convection through the argon gas in the glass envelope.. Also, the filament is not totally screened from radiation from the Earth. I bet you'd only get a very few degrees of lowering of temperature. at most.
A possible experiment might be to use a large , deep, parabolic reflector and a sensitive thermistor.
I can see another problem with this idea - the only effective way of directly measuring the resistance of a piece of wire is to send a current along the wire, but that would inevitably add to the temperature of the wire (not a lot, but considerably more than the 2.74K you are thinking of measuring - although I don't think you are likely to find a night sky on Earth as cold as 2.74K).
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I think Sophiecentaur has underrated the Resistance change of the Tungsten filament between 300°K and its maximum operating temperature of about 3400°K I believe it is about 20 times.
I realise that a headlamp makes a pretty crude radiometer but I just wondered if anyone had tried it.
PS
It is possible to measure the temperature of a resistor without passing a current thru it by measuring the noise generated in it.
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If you were really worried by heating effects, couldn't you build a system that applied a current, and measured the resulting voltage and alter the current so as the product of voltage and current was a constant. You could then measure the resistance by dividing the measured voltage by the applied current.
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They used to make car headlamps about 20cm diameter with an integral parabolic reflector and tungsten filament.
I have often wondered if one measured the resistance of the filament and pointed it to either the 300°K Earth or the 2.74°K clear sky one could notice a difference in the resistance
To perform such a measure (providing that the sky would really be at 2.74 °K, as George pointed out) you should at least chill all the headlamp but the filament, to (significantly) less than 2.74 °K.
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I think Sophiecentaur has underrated the Resistance change of the Tungsten filament between 300°K and its maximum operating temperature of about 3400°K I believe it is about 20 times.
I realise that a headlamp makes a pretty crude radiometer but I just wondered if anyone had tried it.
I've done this many times with kids. It always seems to give the same sort of ratio, whatever type of low voltage bulb you use and whatever student does it.
A 15W 12V bulb measures about 10 Ohms hot and about 5 Ohms with about 1V supply 'cold' using raw, rectified DC from a School power supply and various meters. The minimum temperature may be a fair bit above 300K - but you can't see it glowing! Also, my temperature estimate when hot was probably twice what it really runs at.
Having looked it up, however, I see where you're coming from. Also, the fuse-blowing properties of lamps when you first switch on seems to cast more doubt on my measurements.
I wonder where the discrepancy keeps coming in. School measurements can be a bit naff but not usually wrong by a factor of ten!
For your radiometer experiment, you would have to be very smart to reduce the heating effect of any resistance measurement. Very low energy impulses with correlation techniques, possibly.
Very low temperature noise measurements are a nightmare and rely on finding how much power you need to add to double the measured noise.
On the original topic, the final temperature of a mirror must depend upon where a balance is reached between absorbed and radiated energy. For example,with a matt black side, facing away from the Sun, its temperature would be quite a bit lower than with a shiny face in the shade. A poor absorber is also a poor radiator at any particular wavelength. The actual colour would have a significant effect. To keep cool, the mirror would need to be a poor absorber at short (visible) wavelengths from the (very hot) sun and a good radiator at longer (Infra red) wavelengths from the (just warm) mirror / object. To get it as warm as possible, you'd have to do the reverse.
The Earth is having this problem at the moment, as a matter of fact! (So they tell us.)
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If you were really worried by heating effects, couldn't you build a system that applied a current, and measured the resulting voltage and alter the current so as the product of voltage and current was a constant. You could then measure the resistance by dividing the measured voltage by the applied current.
Extremely difficult because you would have to allow for feedback effects (i.e. changes in temperature will cause changes in resistance; changes in resistance will cause changes in current; changes in current will cause changes in heat output; changes in heat output will cause changes in resistance; ...).
I suppose you could seek to measure resistance at constant current, variable voltage.
Another problem is that the optics of the system is also designed to work in the visible spectrum (approx 2000K to maybe 20,000K), whereas temperatures in the regions of tens of kelvin will be in the microwave region, well outside of the ability of the optics you are using to focus them.
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The optics would certianly be relevant. A good reflector with very low 'sidelobes' with a choke round the perimeter to stop microwaves creeping round from the back would be a must. The original measurements of background radiation used a 'hog horn' type of receiving antenna, I believe. But, if you're only looking for a significant dip in the received radiation rather than attempting to get near 3K things would be less critical I think. You would still need a beamwidth of only a few degrees - a fair few wavelengths of aperture.