Blocks releasing heat energy If you had two blocks, two different sizes yet the same temperature.
Which one would release the most energy in the shortest amount of time and why?
 A: Your scenario is actually a classic transient conduction problem tackled in undergraduate engineering heat transfer, so we can handle this scenario easily.
I took the figure below and adapted a derivation from a popular heat transfer textbook by Incropera and DeWitt:

In this figure a warm object is placed in a tank filled with a known liquid (the analysis that follows will work for gases as well). The energy balance for this object requires that the rate that the object is losing sensible thermal energy is equal to the rate at which heat is transferred to the surrounding fluid via convection:
$$-\dot{E}_{out} = \dot{E}_{st}$$
Expanding each of these terms in the equation can be expanded. The first term is written in terms of Newton's law of cooling; that is the rate at which energy is leaving is proportional to the temperature difference between the temperature of the object $T$ and the ambient temperature $T_\infty$:
$$\dot{E}_{out} = hA_s(T-T_\infty)$$
where $h$ is the convection coefficient, and $A_s$ is the exposed surface area. To expand the second term we need to know what the sensible energy the object contains is. When the heat capacity is constant we have:
$$E_{st} = mc_pT$$
We can differentiate this expression with respect to time and plug into the above formula:
$$-hA_s(T-T_\infty) = mc_p\frac{dT}{dt}$$
Now this is a differential equation which we can solve to come up with a function which tells us the temperature of our object at any point in time:
$$\frac{T-T_\infty}{T_i-T_\infty} = \exp\left[-\left(\frac{hA_s}{mc_p}\right)t\right] = \exp\left[-Jt\right]$$
Now we have a function that tells us the temperature decays exponentially with time $t$. How fast the temperature decays (how fast our object cools) depends on the coefficient $J=hA_s/mc_p$. The larger $J$ is, the faster the object cools!
So to answer your question which object cools faster we have to decide which object has a bigger $J$ coefficient. For example lets consider two spheres taken out of the oven at the same time. They both have the same initial temperature, but their diameters are different. The big sphere has radius $R$, while the small sphere has radius $r$.
Also, we can write the mass as the product of density and volume ($m=\rho V$). If we calculate the $J$ values for both spheres we have:
$$J_r = \frac{hA_r}{\rho V_r c_p} = \frac{h\left( 4 \pi r^2 \right)}{\rho \left( \frac{4}{3}\pi r^3\right) c_p} = \frac{h}{\rho c_p}\frac{3}{r}$$
Similarly, we can calculate $J_R = \frac{h}{\rho c_p}\frac{3}{R}$. So which sphere cools faster? We can take the ratio of these two coefficients to find out:
$$\frac{J_r}{J_R} = \frac{\frac{h}{\rho c_p}\frac{3}{r}}{\frac{h}{\rho c_p}\frac{3}{R}} = \frac{R}{r}$$
Since $R > r$ the ratio is greater than 1, and the smaller sphere has the bigger $J$ coefficient. Thus, the smaller sphere will cool faster. For most (maybe all) convex objects $A_s/V$ will be on the order of $1/L$ where $L$ is a characteristic length. This analysis then generally concludes that smaller objects will cool faster.
Edit: just realized your question asks which one will loose energy faster. We can state this more formally by asking which object will have the higher rate of heat transfer. To do this we simply compute $\dot{E}_{st}$ using the known temperature profile.
$$\dot{E}_{st} = mc_p \frac{dT}{dt} = mc_p \left( - J \right)(T_i-T_\infty)\exp\left[ -Jt \right] = -hA_s(T_i-T_\infty)\exp\left[ -Jt \right]$$
So which one of our objects transfers heat faster? Again we can take the ratio where we will not assume (for the moment) our objects are spheres, only that the $V_{R} > V_{r}$:
$$\frac{\dot{E}_{st,r}}{\dot{E}_{st,R}} = \frac{-hA_{s,r}(T_i-T_\infty)\exp\left[ -J_rt \right]}{-hA_{s,R}(T_i-T_\infty)\exp\left[ -J_Rt \right]} = \frac{A_{s,r}}{A_{s,R}}\exp\left[ \left(J_R - J_r\right)t \right] = $$
$$ = \frac{A_{s,r}}{A_{s,R}}\exp\left[ \frac{h}{\rho c_p}\left(\frac{A_{s,R}}{V_R} - \frac{A_{s,r}}{V_r}\right)t \right]$$
This is not as clear cut. In general, we can't make any assumptions about which object transfers more heat without knowing more about the geometry. For a sphere, however, we be more confident:
$$\text{Heat Transfer Ratio} =\frac{r^2}{R^2}\exp\left[ \frac{h}{\rho c_p}\left(\frac{3}{R} - \frac{3}{r}\right)t \right]$$
First $r^2/R^2 < 1$ always. In addition $1/R - 1/r < 0$, so the $\exp <= 1$ always. Therefore the larger sphere will always transfer more heat (from start to finish) than the smaller one.
Answer: If the two objects are the same material and shape (different sizes), then at any point in time the larger object will generally have released more heat than the smaller object, simply because it has a greater surface area.
If the shapes are very different (e.g. one is a thin plate and the other one is a sphere) there may be a short period of time where the smaller object is releasing more heat than the bigger object.
Of course, after a very long period of time both objects will reach the environment temperature, and the larger object will have released the most heat because it had more to begin with!
Also, you should know that engineers often like to treat anything small and convex like spheres (even cubes!). Any thing really long is often treated as an infinitely long cylinder, and anything large and flat is treated as a wall which is infinitely wide and tall (but known thickness). Why? because it makes the equations simpler and we can get an approximate answer.

Please note this is a very approximate analysis. The reader should bear in mind the following notes.
Note that the above analysis assumes that the size of the object is generally small, such that the temperature $T$ does not vary across the object. That is, we need a much more elaborate treatment of cases where the temperature can vary in space as well as time. The Biot number will help you determine whether the above approximation is valid.
I have made the (pretty crude) approximation that $h_r = h_R$. This is not generally the case for natural convection, but this answer has enough complexity as it is. In general $h$ should be computed from a suitable Nusselt number correlation. Typically, you will find $h=h(T)$ and you will have to account for this when solving the above differential equation.
A: Your question seems to be about heat transfer through convection.
The formula that describes  this phenomenon is:
$dQ/dt=h*A*(To-Tenv)$ 
where $dQ/dt$ is the heat transferred per unit time, 
$A$ is the area of the object,
 $h$ is the heat transfer coefficient,
 $Ta$ is the object's surface temperature and $Tenv$ is the fluid temperature (temperature of air enviroment in your example.)
Since  one brick is larger than the other one, its obvious that the large one which has a larger surrounding area  loses energy in a bigger rate.
See also :https://en.wikipedia.org/wiki/Convective_heat_transfer
