Understanding Heat Radiation and Heat Change in Thermodynamics

In summary, the conversation discusses an equation (dQ/Q=((1+dT/T)^4)-1) and its possible applications in relation to radiation and thermodynamics. The participants discuss the accuracy and limitations of the equation and explore its implications for predicting temperature changes in a black body. They also consider the effects of adding additional thermal energy to the system and the differences between calculating surface radiation and heat change in the whole body.
  • #1
douds
8
0
May somebody tell me something about this equation ?

dQ/Q=((1+dT/T)^4)-1

What is it possible to do with?

Signification-Integration-ploting

Sorry for my english tank's
 
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  • #2
The power four suggests me it could be related to a radiation emission problem.
Therefore, this is probably not really thermodynamics since heat exchange by radiation seems to be the topics.
In addition, if my guess is correct, it may be strange that there is no additional factor in this equation. Indeed this equation would imply that the heat lost by radiation is proportional to the heat content in this system. First, the heat content Q is an ambiguous concept. Second, the radiation heat flux is usually proportional to a surface instead of an internal energy.

It would be necessary to have more information to clarify these ambiguities.
 
  • #3
Stefan' law : Q=a.(T^4) (1) not nul for a system

If Q is augmented of dQ, in the system, is it legitime to write that

equation (1) becomes Q+dQ= a.(T+dT)^4 (2), dT being new equilibium's temperature?

If yes, It follows then by dividing (2) by (1) member to member, then :
1+dQ/Q=(1+(DT)/T)^4 constant a containing (sigma and A aera) disappear.

at least the given equation (0) dQ/Q=(1+(DT)/T)^4-1


Let's assume T= 288K for the system
( 15°C) et let's seek energy fraction dQ necessary to drop the température of 2°. dT = 2 K

1+(dQ)/Q=(1+(2/288))^4=(1.00694)^4 => dQ/Q=(1.00694)^4-1=(1.00278)-1=0.00278

in other words, 0,0027788% of Q is enough to drop temperature of 2°. Is that correct?
 
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  • #4
Don't use the differential notation "d", when you mean the "finite difference" [itex] \Delta [/itex].

Also, you might want to check the arithmetics...
 
  • #5
stefans'law

ok for the response

Stefan' law : Q=a.(T^4) (1) not nul for a system

If Q is augmented of deltaQ, in the system, is it legitime to write that

equation (1) becomes Q+deltaQ= a.(T+deltaT)^4 (2), deltaT being new equilibium's temperature?

If yes, It follows then by dividing (2) by (1) member to member, then :
1+deltaQ/Q=(1+(deltaT)/T)^4 constant a containing (sigma and A aera) disappear.

at least the given equation (0) deltaQ/Q=(1+(deltaT)/T)^4-1


Let's assume T= 288K for the system
( 15°C) et let's seek energy fraction dQ necessary to drop the température of 2°. deltaT = 2 K

1+(deltaQ)/Q=(1+(2/288))^4=(1.00694)^4 => deltaQ/Q=(1.00694)^4-1=(1.00278)-1=0.00278

in other words, 0,0027788% of Q is enough to drop temperature of 2°. Is that correct?

And now??
 
  • #6
Again you forget that 0

(1.00694)^4 \simeq 1.028005. Not to mention that the percentage is incorrect.
 
  • #7
ok, thank you for the mistake

Now in other words: 0.028068 of Q is enough to drop temperature of 2°.

Comment the physical interpretation of this equation?

If the temperature of a black body is at the temperature T, the radiated energy is Q. If the temperature increases of 2K then the radiated energy is

Q+0.028068XQ=1.0280684XQ

Is there any physical truth behind this equation ?:confused:
 
  • #8
thermo equation

thermo equation

"“The theory, it is when all is known and that nothing functions. The practice, it is when all functions and that nobody knows why. Here, we joined together theory and practical: nothing functions… and nobody knows why! ”
 
  • #9
if the body is a black body then the equation should stand. just to check the a u mentioned in the beginning is the Boltzmann's constant?
why shouldn't there be a physical truth behind the equation? a small change in temperature will only change the amount of energy emmitted by a fraction!.
from my view pt i think it stands
 
  • #10
thermo equation

yes, "a" is a constant grouping sigma, the emissivity coefficient; A, the radiating area and k, the Boltzman's constant of the system.

Now listen, let's assume we add Q' to our system where Q' is different of delta Q, but Q'= l+M+N+O+delta Q, where deltaQ appears in equation 2

possible ?
 
  • #11
i don't understand what u mean, what difference does it make if u put 100constants ? its the same principle !
 
  • #12
Ok, if the system is at a temperature T, it radiates Q by stefan's law
if we have T+deltaT, it will radiates Q+deltaQ where delta is 0.028XQ.

if now, we had thermal energy Q' to the system, with Q' <Q, (but notQ'<<Q), only one part of Q' will be lost by radiation, an other one goes in stock (M.C.(T1-T2)), an other may be lost in convection. I think I cannot say that all Q' will be lost by radiation, but only a fraction of Q', this is the sens of my question.o:)
 
  • #13
there is no stock here to talk about. the original meaning of the equation is that a body at temperature T will emit a heat Q. to calculate the heat loss you went to another formula here.
 
  • #14
I try to make a formal system to appreciate the impact of additionnal energy Q' on a black body radiating Q. The real bodies have a mass, how would you articulate the two equations for the same body?

deltaQ/Q=(1+(deltaT/T)^4)-1
Q'=m.c.deltaT'

because I initially believed that it was possible to predict the temperature of a body according to its energy,(in other words that the equation was reversable), but somebody told me that Stefan's law may not predict any drop of temperature deltaT according to this scheme, for example,may be wrong

(deltaQ/Q)+1=(1+(deltaT/T)^4) ->(deltaQ/Q+1)^(1/4)=1+(deltaT/T)

if deltaQ=0.05Q and T=300K deltaT=303K
 
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  • #15
the first equation calculates how much a SURFACE radiates heat at different temperatures. while the second one calculates the heat change in the whole body. so its a different formula.
 

1. What is the basic equation for thermodynamics?

The basic equation for thermodynamics is known as the First Law of Thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system.

2. How is the Second Law of Thermodynamics expressed in equation form?

The Second Law of Thermodynamics is often expressed using the equation: ΔS = Q/T, where ΔS is the change in entropy, Q is the heat added to the system, and T is the temperature.

3. What is the significance of the thermodynamic equilibrium equation?

The thermodynamic equilibrium equation, also known as the Zeroth Law of Thermodynamics, states that if two systems are in thermal equilibrium with a third system, they are also in thermal equilibrium with each other. This allows for the measurement of temperature and the development of the concept of thermal equilibrium.

4. How does the Carnot cycle relate to thermodynamics?

The Carnot cycle is a theoretical thermodynamic cycle that serves as an idealized model for heat engines. It follows the principles of thermodynamics, specifically the Second Law, and helps to determine the maximum efficiency of a heat engine.

5. What is the role of Gibbs free energy in thermodynamics equations?

Gibbs free energy is an important thermodynamic potential that takes into account both the enthalpy and entropy of a system. It is used to determine the spontaneity of a reaction or process, with a negative value indicating a spontaneous reaction and a positive value indicating a non-spontaneous reaction.

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