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Navboi
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#1
Hi,
I have some questions surrounding the following question:
![Image]()
The answer is D, but I want to know why the greatest rate of emission of thermal radiation is for S. I would've thought it was Q as the temperature of the room was lower than the temperature of the ball, as there is a greater temperature gradient, there would be a higher rate of emission no? Does the rate of emission not depend on its surroundings and just purely on the temperature of the object?
I have some questions surrounding the following question:
The answer is D, but I want to know why the greatest rate of emission of thermal radiation is for S. I would've thought it was Q as the temperature of the room was lower than the temperature of the ball, as there is a greater temperature gradient, there would be a higher rate of emission no? Does the rate of emission not depend on its surroundings and just purely on the temperature of the object?
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LuigiMario
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#2
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#2
For “just purely on the temperature of the object” I would just add one word to that
“just purely on the” ABSOLUTE “temperature of the object”
-273.15 is rather an important number
“just purely on the” ABSOLUTE “temperature of the object”
-273.15 is rather an important number
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Navboi
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#3
(Original post by LuigiMario)
For “just purely on the temperature of the object” I would just add one word to that
“just purely on the” ABSOLUTE “temperature of the object”
-273.15 is rather an important number
For “just purely on the temperature of the object” I would just add one word to that
“just purely on the” ABSOLUTE “temperature of the object”
-273.15 is rather an important number
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LuigiMario
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#4
no, you ask a reasonable question, but no. (it's one of those tricky questions) where in part one, (physical physics)
the rate of loss of heat from an object depends on the temperature difference from surrounding environment and the surface area
whilst in part two.... (Q.physics) emission of planck (thermal, IR, optical, x-ray...) radiation is purely dependent on absolute temperature and the emissivity of the surface (at the wavelength of interest) and some near constants
https://scienceofdoom.com/2017/02/01...mann-equation/
qualifications: I spent a very nice week at NPL, Teddington where scientists tried to make me learn this, and now I use this stuff every month or two
Q.Phys.... I daredn't use the word Quantum in a sentence, as I know it would be wrong, I would have used it wrongly - so the above answer is 'probably' 'nearly' accurate.....
the rate of loss of heat from an object depends on the temperature difference from surrounding environment and the surface area
whilst in part two.... (Q.physics) emission of planck (thermal, IR, optical, x-ray...) radiation is purely dependent on absolute temperature and the emissivity of the surface (at the wavelength of interest) and some near constants
https://scienceofdoom.com/2017/02/01...mann-equation/
qualifications: I spent a very nice week at NPL, Teddington where scientists tried to make me learn this, and now I use this stuff every month or two
Q.Phys.... I daredn't use the word Quantum in a sentence, as I know it would be wrong, I would have used it wrongly - so the above answer is 'probably' 'nearly' accurate.....
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LuigiMario
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#5
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#5
no, you ask a reasonable question, but no. (it's one of those tricky questions) where in part one, (physical physics)
the rate of loss of heat from an object depends on the temperature difference from surrounding environment and the surface area
whilst in part two.... (Q.physics) emission of planck (thermal, IR, optical, x-ray...) radiation is purely dependent on absolute temperature and the emissivity of the surface (at the wavelength of interest) and some near constants
https://scienceofdoom.com/2017/02/01...mann-equation/
qualifications: I spent a very nice week at NPL, Teddington where scientists tried to make me learn this, and now I use this stuff every month or two
Q.Phys.... I personally daredn't use the word Quantum in a sentence, as I know it would be wrong, I would have used it wrongly - so the above answer is 'probably' 'nearly' accurate.....
the rate of loss of heat from an object depends on the temperature difference from surrounding environment and the surface area
whilst in part two.... (Q.physics) emission of planck (thermal, IR, optical, x-ray...) radiation is purely dependent on absolute temperature and the emissivity of the surface (at the wavelength of interest) and some near constants
https://scienceofdoom.com/2017/02/01...mann-equation/
qualifications: I spent a very nice week at NPL, Teddington where scientists tried to make me learn this, and now I use this stuff every month or two
Q.Phys.... I personally daredn't use the word Quantum in a sentence, as I know it would be wrong, I would have used it wrongly - so the above answer is 'probably' 'nearly' accurate.....
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Navboi
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#6
(Original post by LuigiMario)
no, you ask a reasonable question, but no. (it's one of those tricky questions) where in part one, (physical physics)
the rate of loss of heat from an object depends on the temperature difference from surrounding environment and the surface area
whilst in part two.... (Q.physics) emission of planck (thermal, IR, optical, x-ray...) radiation is purely dependent on absolute temperature and the emissivity of the surface (at the wavelength of interest) and some near constants
https://scienceofdoom.com/2017/02/01...mann-equation/
qualifications: I spent a very nice week at NPL, Teddington where scientists tried to make me learn this, and now I use this stuff every month or two
Q.Phys.... I personally daredn't use the word Quantum in a sentence, as I know it would be wrong, I would have used it wrongly - so the above answer is 'probably' 'nearly' accurate.....
no, you ask a reasonable question, but no. (it's one of those tricky questions) where in part one, (physical physics)
the rate of loss of heat from an object depends on the temperature difference from surrounding environment and the surface area
whilst in part two.... (Q.physics) emission of planck (thermal, IR, optical, x-ray...) radiation is purely dependent on absolute temperature and the emissivity of the surface (at the wavelength of interest) and some near constants
https://scienceofdoom.com/2017/02/01...mann-equation/
qualifications: I spent a very nice week at NPL, Teddington where scientists tried to make me learn this, and now I use this stuff every month or two
Q.Phys.... I personally daredn't use the word Quantum in a sentence, as I know it would be wrong, I would have used it wrongly - so the above answer is 'probably' 'nearly' accurate.....
Since this is A-Level physics, I think its fair to assume that this question doesn't require any of the emissivity stuff you've linked since I'm pretty sure its not in the spec. So am I right in saying that (at my level) just the temperature of the object affects the rate of emission of thermal radiation of the object? as out of S and Q, S has the higher temperature, so S will have the greatest rate of emission?
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LuigiMario
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#7
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#7
not quite, eek
certainly the temperature difference from the surroundings will imply a thermal transfer, specifically avoiding the "emissivity" word,
that's case P & Q, and occours mostly by conduction (to the liquid surrounding, in this case 'the liquid' is air) , and also by convection in the 'liquid' air (and OK still a bit of radiation. as everything emits that's warmer than absolute zero)
Case for R & S, gets you to understand that there's virtually no conduction, convection, but there is (still) radiation. (it's called black-body radiation)
and an infinitely dark object will both absorb and radiate, emit, better than a lighter object. S better approximates "infinitely dark" R better approximates "not infinitely dark"(1)
(1)The most darkest object I ever saw was a stack of razor blades, going into a hydrogen tempering furnace, edge on, those closely stacked very sharp edges did a great dark-object approximation, whilst actually being shiny & razor blade coloured)
if you have one of these covid IR forehead remote readers, you could try reading the temp of similar objects, covered in white paint, black paint, alu foil etc
At work we use a special very dark coloured paint, as we are trying to control the optical reflections in experiments. I noticed that (using a fluke infra red camera) that this paint was grey coloured at 8 μm to 14 μm infra-red bands, whilst in the visible band 0.4 μm to 0.7 μm wavelength, it is much darker.
I think in physics, it doesn't hurt to see that there is slightly more depth to come, if you study it later that GCSE or A-levels - or later work in an area that uses physics. This simple question eventually leads on to the development of the Stealth fighter aircraft,e.g. Lockheed F-117 Nighthawk....
certainly the temperature difference from the surroundings will imply a thermal transfer, specifically avoiding the "emissivity" word,
that's case P & Q, and occours mostly by conduction (to the liquid surrounding, in this case 'the liquid' is air) , and also by convection in the 'liquid' air (and OK still a bit of radiation. as everything emits that's warmer than absolute zero)
Case for R & S, gets you to understand that there's virtually no conduction, convection, but there is (still) radiation. (it's called black-body radiation)
and an infinitely dark object will both absorb and radiate, emit, better than a lighter object. S better approximates "infinitely dark" R better approximates "not infinitely dark"(1)
(1)The most darkest object I ever saw was a stack of razor blades, going into a hydrogen tempering furnace, edge on, those closely stacked very sharp edges did a great dark-object approximation, whilst actually being shiny & razor blade coloured)
if you have one of these covid IR forehead remote readers, you could try reading the temp of similar objects, covered in white paint, black paint, alu foil etc
At work we use a special very dark coloured paint, as we are trying to control the optical reflections in experiments. I noticed that (using a fluke infra red camera) that this paint was grey coloured at 8 μm to 14 μm infra-red bands, whilst in the visible band 0.4 μm to 0.7 μm wavelength, it is much darker.
I think in physics, it doesn't hurt to see that there is slightly more depth to come, if you study it later that GCSE or A-levels - or later work in an area that uses physics. This simple question eventually leads on to the development of the Stealth fighter aircraft,e.g. Lockheed F-117 Nighthawk....
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