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 <PERSON> and <PERSON>:
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 <PERSON>'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. 707
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. 15