• #26
PeterDonis
Mentor
Insights Author
2020 Award
35,940
14,000
where did you get that formula from?

That formula is a general relationship between the amount of heat emitted by an object and the change in its temperature. However, if you don't know either of those things in advance, you can't use that formula to solve anything. You need to do the integral I gave in post #15.
 
  • #27
501
66
If you mean, for finding ΔT\Delta T without having to solve the integral I gave in post #15, no, I don't have one. There isn't one. If you don't already know ΔT\Delta T in advance, the only way to solve for it is to know T as a function of t and then solve the integral I gave. You also have to know the length of time you are interested in (since that's what gives you the limits of integration).

I tried this - assuming time limits are known - on integrating dT and dt using the the two known relations P = ... and Q = ... , you get t as a complex function of T , terms of which include ln() and arctan .

Although you have said there is no solution , can this relation not be solved ?

*Please excuse me if this question sounds repetitive .
 
  • #28
PeterDonis
Mentor
Insights Author
2020 Award
35,940
14,000
on integrating dT and dt using the the two known relations P = ... and Q = ... , you get t as a complex function of T

You're going to have to show your work, this is too vague for me to understand what you're doing.
 
  • #29
501
66
You're going to have to show your work, this is too vague for me to understand what you're doing.

dQ/dt = m*c*dT/dt ( dividing both sides by dt ) .
and
dQ/dt = σ*A*(T4 - T04) (where T0 is temperature of surroundings ) .

∴m*c*dT/dt = σ*A*(T4 - T04) - then integrate it .
 
  • #30
PeterDonis
Mentor
Insights Author
2020 Award
35,940
14,000
dQ/dt = m*c*dT/dt ( dividing both sides by dt )

Differentiation is not an algebraic operation; you can't just "divide by dt", you have to have functions of ##t## on both sides that you are differentiating. In the original equation, you don't; you have ##\Delta T## on the RHS, not ##T##. ##\Delta T## is not temperature as a function of time; it's the change in temperature between two states of the object that you happen to be interested in. That's not a function of time, and you can't differentiate it with respect to time.
 
  • #31
501
66
Differentiation is not an algebraic operation; you can't just "divide by dt", you have to have functions of t on both sides that you are differentiating.

I have not done this - You can look at this in two ways - Q = m*c*ΔT and then I differentiate with respect to t ,
or , for small change in temperature , dQ = m*c*dT , and now I just simply divide both sides by dt .

In the original equation, you don't; you have ΔT\Delta T on the RHS, not TT. ΔT\Delta T is not temperature as a function of time; it's the change in temperature between two states of the object that you happen to be interested in. That's not a function of time, and you can't differentiate it with respect to time.

Yes , I made a mistake here . You could use this only for small temperature difference in T and T0 . Sorry .
 
  • #32
501
66
It could be, where did you get that formula from? However, I think that the area of the object will matter, so I think that should be included.

A molecule of an object , loses heat on fall in temperature . So , greater number of molecules , greater heat lost .

Also , greater the loss in temperature , greater the loss of heat .

∴ heat lost depends on physical dimensions , density ,and temperature .

Physical dimensions and density together are replaced by m and ,
ΔQ ∝ m ;
ΔQ ∝ ΔT .

∴ΔQ = c*m*ΔT with c the constant of proportionality and equal to specific heat capacity .
 
  • #33
PeterDonis
Mentor
Insights Author
2020 Award
35,940
14,000
Q = m*c*ΔT and then I differentiate with respect to t

But, as you go on to note, you can't do this unless ##\Delta T## is infinitesimal, i.e., unless you really have ##dQ## and ##dT##. You can't differentiate ##\Delta T## by ##t## because it isn't a function of ##t## (##T## is, but ##\Delta T## is not).

for small change in temperature , dQ = m*c*dT , and now I just simply divide both sides by dt

This is the equivalent of differentiating by ##t##, because ##dQ## and ##dT## are both infinitesimals (strictly speaking, you would take the limit as ##dt \rightarrow 0##).

Also, this only works if both ##m## and ##c## are constant. Constant ##m## is no problem, but constant ##c## is; most substances have a heat capacity that changes with temperature. So the full formula would have to be

$$
\frac{dQ}{dt} = m c \frac{dT}{dt} + m T \frac{dc}{dT} \frac{dT}{dt} = m \frac{dT}{dt} \left( c + T \frac{dc}{dT} \right)
$$

Finally, as I said before, this formula doesn't help unless you already know ##T## as a function of ##t## (and ##c## as a function of ##T##, if it's not constant). In the scenario under discussion in this thread, we don't.
 

Related Threads on Thermal Radiation

Replies
1
Views
2K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
5
Views
912
  • Last Post
Replies
2
Views
646
  • Last Post
Replies
10
Views
2K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
3
Views
1K
Top