# Calorimetry - Emitted Joules [closed]

How can one calculate the total amount of emitted joules from an object with a temperature that isn't constant? A great start is this formula:

P = σ • A • (T^4)


If the formula is translated from sybmols to units it will look like this:

J/s = σ • (m^2) • (K^4)


(J/s) is joules emitted per second from an object. ( σ ) is the Stefan Boltzmann constant, 5.67•(10^−8). (m^2) is the area of the object. (K^4) is the temperature of the object, in Kelvin, to the power of 4. Now, if I transform the formula:

J = σ • (m^2) • (K^4) • s


Now one can get total amount of joules emitted ( J ) during a certain time ( s ), if one know the temperature and area. To the tricky part: what if the temperature isn't constant? The temperature will depend on how many joules that has already been emitted. And the joules that are being emitted will depend on the temperature.

How can I solve this? Do I need to combine this with an other formula?

Best regards! Please comment if something is unclear!

## closed as off-topic by ACuriousMind♦, Kyle Kanos, Danu, HDE 226868, Ryan UngerAug 16 '15 at 3:02

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• Hi JakeTheSnake. Welcome to Phys.SE. If you haven't already done so, please take a minute to read the definition of when to use the homework-and-exercises tag, and the Phys.SE policy for homework-like problems. – Qmechanic Aug 18 '15 at 12:21

I am unsure of the solution below because of my free manipulation of $P$ and $Q$, so if anyone has a comment or better way, let me know.

Some assumptions: I assume that in the question you meant that we have a body with a surface area $A$, which is at an initial temperature $T_{0}$ and is losing energy by means of radiation, for which you used the Stefan-Boltzmann law. It doesn't have another heat source, so the temperature is constantly falling.

So, there are two more parameters you need to include in your question: the initial temperature of the surface $T_{0}$, and the heat capacity of the surface: let's denote it $C_{A}$ and assume it's constant throughout time, on any point of the surface and on any temperature (makes the calculations easier).

The definition of heat capacity implies $dQ={C}_{A}dT$. We also know $dQ=P\left(t\right)dt$. Note that we may use $dQ$ instead of $\partial Q$ since we're assuming the only way the body loses heat is by radiation. By combining these 2 statements and using Stefan-Boltzmann's law in place of $P$ we get the differential equation $$\frac{dT}{dt}=-\frac{\sigma A}{{C}_{A}}{T}^{4}$$ (the minus sign coming from the fact that the body is losing heat), which after solving and putting $T\left (0\right )=T_{0}$ yields the surface temperature:

$$T\left(t\right)=\frac{1}{\sqrt[3]{\frac{1}{{T}_{0}^{3}}+\frac{3\sigma A}{{C}_{A}}t}}$$.

By using Stefan-Boltzmann's law again, along with the newfound temperature expression, in the equation $dQ=P\left(t\right)dt$ you may get what you seek: the formula to compute the amount of energy released by radiation as a function of time. By integrating the above combination we get the final expression: $$Q\left(t\right)=\frac{{\left(3\sigma A\right)}^{2}}{{C}_{A}}\left({T}_{0}-T\left(t\right)\right)$$.

• Brilliant, thank you! I am a little confused with all the symbols, but can I take the final expression of T(t), and put it in this: P = σ • A • (T^4)? And by that make a function that will look like this: P( T( t ) )? – JakeTheSnake Aug 13 '15 at 8:48
• The symbols are from standard calculus, if you haven't learned the math, don't worry, it's easy :) Yes, the general idea is that you get the expression for the temperature and be able to plug it into the law. But, you have to be careful here, because the power depends on time, $Q\ne Pt$ ! You have to integrate $dQ=P\left(t\right)dt$ in order to get the expression. I'll ammend that to my answer. – Soba noodles Aug 13 '15 at 8:53
• U da best man! One more question tho, what does Q(t) stand for? The emitted joules? :) – JakeTheSnake Aug 13 '15 at 11:35
• Yes, you plug in the time t, you get the amount of energy radiated :) – Soba noodles Aug 13 '15 at 20:28
• @JakeTheSnake - Also keep in mind that the formula assumes infinite thermal conductivity: in other words, that the temperature stays uniform throughout the body, which is obviously not true for real materials. A real body will cool more slowly than the formula suggests, since the core of the body will remain hot while the surface cools, with a gradual transfer of heat from core to surface. – WhatRoughBeast Sep 8 '15 at 23:46