Thermal Radiation: Calc Total Emitted Joules in Certain Temp Range

[Qwertywerty]

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 .

 

[Qwertywerty]

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 .

 

[PeterDonis]

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.