You are certainly right. SSDD believes that the presence of a warmer body will keep the colder body from radiating anything toward it even thought the surface charges are vibrating like mad.
No, an IR sensor at ambient temperature would not detect energy flowing in all directions.
What do I mean? It's easy.that is the SB equation for radiating bodies, not in a vacuum, and in the presence of other matter. Set T and Tc to the same number....what does P then equal?
Both steel balls will be radiating in all directions. Do you dispute this?
And this
Stefan-Boltzmann Law
States your equation describes a hot body, T, radiating into cooler surroundings Tc. But back to your comment. I don't disagree
Set T and Tc to the same number....what does P then equal?
The answer must be zero. But both steel balls will still radiate, in my example. Going to my link, it implies the emitter is the same temperature as its surroundings. But the emitter will still emit if it is above absolute zero.
P= the amount of radiation emitting from the radiator which is dependent upon the temperature difference between T and Tc. When T and Tc are the same and P=zero...that means something. The equation is making a statement about reality. Alter the statement and you must alter the equation...alter the equation and it no longer represents the physical law.
Where is the logical fallacy?
Really...Which expression within that equation allows you to calculate net? I see T-Tc...that expression describes a gross change, not a net change.
No, an IR sensor at ambient temperature would not detect energy flowing in all directions.
What do I mean? It's easy.that is the SB equation for radiating bodies, not in a vacuum, and in the presence of other matter. Set T and Tc to the same number....what does P then equal?
Both steel balls will be radiating in all directions. Do you dispute this?
And this
Stefan-Boltzmann Law
States your equation describes a hot body, T, radiating into cooler surroundings Tc. But back to your comment. I don't disagree
Set T and Tc to the same number....what does P then equal?
The answer must be zero. But both steel balls will still radiate, in my example. Going to my link, it implies the emitter is the same temperature as its surroundings. But the emitter will still emit if it is above absolute zero.
P= the amount of radiation emitting from the radiator which is dependent upon the temperature difference between T and Tc. When T and Tc are the same and P=zero...that means something. The equation is making a statement about reality. Alter the statement and you must alter the equation...alter the equation and it no longer represents the physical law.
And I say the equation represents the NET change in energy between the object and environment. The object will still be radiating.
From my link:
If the hot object is radiating energy to its cooler surroundings at temperature Tc, the net radiation loss rate takes the form
Heat Transfer Through Conduction: Equation & Examples - Video & Lesson Transcript | Study.com
Q over t is the rate of heat transfer - the amount of heat transferred per second, measured in Joules per second, or Watts. kis the thermal conductivity of the material - for example, copper has a thermal conductivity of 390, but wool has a thermal conductivity of just 0.04. T1 is the temperature of one object, and T2 is the temperature of the other. Since it's a temperature difference, you can actually use Celsius or Kelvin, whichever is most convenient. And d is the thickness of the material we're interested in. So the rate of heat transfer to an object is equal to the thermal conductivity of the material the object is made from, multiplied by the surface area in contact, multiplied by the difference in temperature between the two objects, divided by the thickness of the material.
Calculation Example
Okay, let's go through an example. Let's say you're going to a water park, and you're going to take a Styrofoam cooler with you. The cooler has a total surface area of 1.2 meters squared, and the walls are 0.03 meters thick. The temperature inside the cooler is 0 Celsius, and at the hottest part of the day it's 38 degrees Celsius. During this time of day, how much heat energy per second is lost by the cooler? And how much heat energy is lost in three hours at the water park assuming the temperature stays at 38 degrees? (Note: The thermal conductivity of Styrofoam is 0.01.)"
So you're saying their equation is incorrect, that the computation is not T2-T1, but T2-T1 + imaginary heat radiated from the cooler to the warmer
Heat Transfer Through Conduction:
Thanks for the link.
We're discussing radiation at the moment.
Not conduction.
Doesn't matter whether you are talking about conduction, radiation, free electrons flowing down a wire, or a bowling ball rolling down a hill...it is all energy exchange and all obeys the second law of thermodynamics. You miss the boat with your terribly flawed belief that radiation has some special exemption from obeying the second law of thermodynamics.
ehs
Really...Which expression within that equation allows you to calculate net? I see T-Tc...that expression describes a gross change, not a net change.
No, an IR sensor at ambient temperature would not detect energy flowing in all directions.
What do I mean? It's easy.that is the SB equation for radiating bodies, not in a vacuum, and in the presence of other matter. Set T and Tc to the same number....what does P then equal?
Both steel balls will be radiating in all directions. Do you dispute this?
And this
Stefan-Boltzmann Law
States your equation describes a hot body, T, radiating into cooler surroundings Tc. But back to your comment. I don't disagree
Set T and Tc to the same number....what does P then equal?
The answer must be zero. But both steel balls will still radiate, in my example. Going to my link, it implies the emitter is the same temperature as its surroundings. But the emitter will still emit if it is above absolute zero.
P= the amount of radiation emitting from the radiator which is dependent upon the temperature difference between T and Tc. When T and Tc are the same and P=zero...that means something. The equation is making a statement about reality. Alter the statement and you must alter the equation...alter the equation and it no longer represents the physical law.
And I say the equation represents the NET change in energy between the object and environment. The object will still be radiating.
From my link:
If the hot object is radiating energy to its cooler surroundings at temperature Tc, the net radiation loss rate takes the form
ehs
Really...Which expression within that equation allows you to calculate net? I see T-Tc...that expression describes a gross change, not a net change.No, an IR sensor at ambient temperature would not detect energy flowing in all directions.
What do I mean? It's easy.that is the SB equation for radiating bodies, not in a vacuum, and in the presence of other matter. Set T and Tc to the same number....what does P then equal?
Both steel balls will be radiating in all directions. Do you dispute this?
And this
Stefan-Boltzmann Law
States your equation describes a hot body, T, radiating into cooler surroundings Tc. But back to your comment. I don't disagree
Set T and Tc to the same number....what does P then equal?
The answer must be zero. But both steel balls will still radiate, in my example. Going to my link, it implies the emitter is the same temperature as its surroundings. But the emitter will still emit if it is above absolute zero.
P= the amount of radiation emitting from the radiator which is dependent upon the temperature difference between T and Tc. When T and Tc are the same and P=zero...that means something. The equation is making a statement about reality. Alter the statement and you must alter the equation...alter the equation and it no longer represents the physical law.
And I say the equation represents the NET change in energy between the object and environment. The object will still be radiating.
From my link:
If the hot object is radiating energy to its cooler surroundings at temperature Tc, the net radiation loss rate takes the form
Wuwei just gave it to you but I see you are in denial
The first two expressions of emission and absorption are subtracted. I told you many times. A vacuum has nothing to do with thermal emission or absorption. A non-GHG between object and background would give the same results.
Nope. Nobody believes that.
