So first off, i'm going to have to break to you that this \Delta E_\text{cond}(t) + \Delta E_\text{conv}(t)$\Delta E_\text{cond}(t) + \Delta E_\text{conv}(t)$
is by far the most complicated bit. There is a simple equation [tex]\frac{q}{\A} = \h\Delta\T[\tex]$\frac{q}{A} = h\Delta T$ where [ itex ] \q [ /itex ]$q$ is the heat flux (W/m^2), [ itex ] \A [ /itex ]$A$ is the surface area of the object, [ itex ] \Delta T [ /itex ]$\Delta T$ is the temperature difference between the bulk of the fluid and the object and \h$h$ is the heat transfer coefficient. Unfortunately [ itex ] \h [ /itex ]$h$ is quite difficult to determine as it depends on the properties of the fluid and its flow regime as well as the geometry of the problem.
In order to determine the heat transfer coefficient we define a number of dimensionless quantities such as [ itex ] \Pr [ /itex ];$Pr$; the Prandtl number. This is the ratio of momentum diffusivity andto thermal diffusivity for the fluid. Effectively this describes the relative size of the thermal and velocity boundary layers for the fluid. For low [ itex ] \Pr [ /itex ]$Pr$ conduction through the fluid dominates (Mercury has Pr of around 0.015). In the opposite case of engine oil momentum diffusivity is much larger so convection dominates (Pr of 100 to 40000). The Prandtl number for air and most other gases is around 0.7-0.8.
Obviously the Reynolds number for the fluid, [ itex ] \Re _\text{L} [ /itex ]$\Re _\text{L}$ is going to be important and is defined as [ tex ] \Re = \frac{\v L}{nu} [ /tex ]$\Re = \frac{v L}{\nu}$
This describes the flow regime around the object where [ itex ] \L [ /itex ]$L$ is a characteristic length and can be taken to be the diameter of the sphere in this case. [ itex ] \nu [ /itex ]$\nu$ is the kinematic viscosity and [ itex ] \v [ /itex ]$v$ is the velocity of the object relative to the bulk of the fluid. For the appropriate substance and temperature you can find the kinematic viscosity of your gas and then calculate the appropriate Reynolds number.
Lastly we define the Nusselt number. [ tex ] \Nu _\text{L} = \frac{\h L}{k} [ /tex ]$Nu _\text{L} = \frac{h L}{k}$
where [ itex ] \L [ /itex ]$L$ is again the characteristic length, [ itex ] \h [ /itex ]$h$ is the heat transfer coefficient and [ itex ] \k [ /itex ]$k$ is the thermal conductivity of the fluid which is evaluated at the film temperature. This is defined as the arithmetic mean of the temperature of the bulk of the fluid [ itex ] \T _text{g} [ /itex ]$T _\text{g}$ also known as [ itex ] \T _text{infty} [ /itex ]$T _\infty$ and the temperature at the solid boundary.
As a small aside I will mention the Biot number. [ tex ] \Bi = \frac{\h L}{k _text{b} [ /tex ]$Bi = \frac{h L}{k _\text{b}}$
where the symbols have their usual meanings except [ itex ] \k _text{b} [ /itex ]$k _\text{b}$ is the thermal conductivity of the body (your sphere in this case). If this is 0.1 or less then the conduction throughout the solid body is sufficient to assume that the body has no internal temperature gradient (which simplifies your problem).
Now on to solving the problem. We want to express the Nusselt number in terms of the Prandtl and Reynolds numbers in order to find an expression for [ itex ] \h [ /itex ]$h$. Unfortunately this is very difficult but luckily large cohorts of experimentalists have slaved away at this problem for us.
For external flow over a sphere, T. Yuge (I don't have an explicit reference for this see textbook link at the end) found the following correlation [ tex ] \Nu = \2 + 0.43 Re ^\frac{1}{4} [ /tex ]$Nu = 2 + 0.43 Re ^\frac{1}{4} $.