Suppose a body cools from 70°C to 60°C and then again cools from 60°C to 50°C. If surrounding temperature is 30°C then the rate of heat loss from 70° to 60°C would be different to that from 60° to 50°C.

Since the temperature difference between the surrounding and the body is less we can write rate of cooling :-
$$-\frac{\mathrm dT}{\mathrm dt} =k(T-T_0),$$ where $T_0$ is the surrounding temperature which remains constant and $T$ is the temperature of the body at any instant $t$.

Since $T$ does not remain constant ($-\mathrm dT/\mathrm dt$) varies, my question is that if the temperature difference between the body and surrounding and not negligible, then we apply Stefan's law to calculate the initial rate of heat loss. In the Stefan's law, we have to take the difference between body's absolute temperature and surrounding temperature, so if the body is continuously losing heat do we have to always take the starting Temperature of the body at which it was kept or we have change that temperature continuously?

For example, if the initial temperature of the body is $T_1$ and after some time it cools and temperature becomes $T_2$, so the initial heat loss will be $Aeρ(T_1-T')$; then will the rate of heat loss changes when the temperature of the body becomes $T_2$ and rate changes to :- $[Aeρ(T_2-T')]$


closed as off-topic by By Symmetry, stafusa, ZeroTheHero, Jon Custer, Sebastian Riese May 28 '18 at 14:09

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  • 1
    $\begingroup$ Show us your work so far. Can you state what the equation for conduction is? Start with that. $\endgroup$ – Eric Lippert May 24 '18 at 22:13
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    $\begingroup$ What is the rate of heat loss when the body has cooled to 30 C? $\endgroup$ – Chet Miller May 24 '18 at 22:35
  • $\begingroup$ @EricLippert question updated $\endgroup$ – user190625 May 25 '18 at 9:51
  • $\begingroup$ Note that we have an equation editor built into the site which helps for readability of your post. I've edited some of it for you, but you may want to check it as I suspect the $e$ are exponential functions and the $\rho T$ terms the arguments. $\endgroup$ – Kyle Kanos May 25 '18 at 10:03

The rate varies, because cooling is an exponential function. Newton's law of cooling.


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