Calorimetry - Emitted Joules 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! 
 A: 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: $$Q\left(t\right)=\frac{{\left(3\sigma A\right)}^{2}}{{C}_{A}}\left({T}_{0}-T\left(t\right)\right)$$.
