# Thermodynamics equation

1. Mar 11, 2007

### douds

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

What is it possible to do with?

Signification-Integration-ploting

Sorry for my english tank's

2. Mar 11, 2007

### lalbatros

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. Mar 12, 2007

### douds

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?

Last edited: Mar 12, 2007
4. Mar 12, 2007

### dextercioby

Don't use the differential notation "d", when you mean the "finite difference" $\Delta$.

Also, you might want to check the arithmetics...

5. Mar 12, 2007

### douds

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. Mar 12, 2007

### dextercioby

Again you forget that 0

(1.00694)^4 \simeq 1.028005. Not to mention that the percentage is incorrect.

7. Mar 12, 2007

### douds

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 ?

8. Mar 13, 2007

### douds

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. Mar 14, 2007

### eaboujaoudeh

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. Mar 14, 2007

### douds

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. Mar 14, 2007

### eaboujaoudeh

i don't understand what u mean, what difference does it make if u put 100constants ? its the same principle !

12. Mar 14, 2007

### douds

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.

13. Mar 14, 2007

### eaboujaoudeh

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. Mar 15, 2007

### douds

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

Last edited: Mar 15, 2007
15. Mar 16, 2007

### eaboujaoudeh

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.