I'm no expert, but here goes...
I assume the bottle diameter $d$ is 80 mm, and its glass thickness $l$ is 2 mm. (Height $h$ cancels out of the result). The surface area $A$ of the glass is then approximately $A = \pi d h $.
Start by estimating the thermal conductivities for the 3 heat transfer processes:
Conduction:
The thermal conductivity of glass is $k=1 W/m/K$.
$$q = (k A / l) \Delta T_{glass} = 500 A \Delta T = G_{cond} A \Delta T \quad, \quad G_{cond} = 500 W/m^2/K$$
Convection:
The convective heat transfer coefficient of air is $h=5-25 W/m^2/K$, according to one reference (a lot of wiggle room here):
http://www.engineeringtoolbox.com/convective-heat-transfer-d_430.html
$$ q = h A \Delta T \quad, \quad G_{conv} = h W/m^2/K$$
Radiation (larger than I expected, hat tip to Chris White):
The Stefan-Boltzmann constant is $\sigma = 5.67x10^{-8} W/m^2/K^4$.
$$ q = \epsilon \sigma A \left(T_1^4-T_2^4 \right) = \epsilon \sigma A (T_1^3 + T_1^2 T_2 + T_1 T_2^2 + T_2^3)\Delta T \approx 4 \sigma T_{avg}^3 A \Delta T \approx 5 \epsilon A \Delta T = G_{rad} A \Delta T$$
$$ G_{rad} = 5 \epsilon W/m^2/K$$
Here $\epsilon$ is the emissivity, which is $1$ for a black body and zero for a perfect reflector. Per Schaum's "Heat Transfer" $\epsilon = 0.94$ for smooth glass, which is $1$ at this level of accuracy.
Radiation occurs in parallel with convection, so their conductivities add, while conduction through the glass is in series with the other $2$, so its (relatively very large) conductivity adds in parallel. The total conductivity $G$ is then determined by:
$$q = G A \Delta T$$
$$\frac{1}{G} = \frac{1}{G_{cond}} + \frac{1}{G_{conv}+G_{rad}} \approx \frac{1}{G_{conv}+G_{rad}}$$
$$G \approx 10 - 30 W/m^2/K$$
[Conduction through the glass is much easier than convection + radiation, so the latter two form the heat transfer "bottleneck" (hee hee); conduction is negligibly large.]
Now, for the wine, $q = m C_v dT/dt$ , where $m = \rho V$ , $\rho = 978 kg/m^3 \text{ and } C_v = 4.3 kJ/kg/K$ , according to a report I found on-line:
http://www.gwrdc.com.au/wp-content/uploads/2012/11/WineryB-CaseStudyReport2.pdf
Equating the $2$ expressions for $q$, we get a nice first order diff eq:
$$ \frac{dT}{dt} = (T-T_{amb})/\tau $$
where the time constant $\tau$ is:
$$ \tau = \frac{m C_v}{G A} = \frac{\rho C_v }{G} \frac{V}{A} = \frac{\rho C_v }{G} \frac{d}{4} = \frac{84,100}{G} = 2800-8410 \text{ seconds, or } 47 \text{ to } 140 \text{ minutes}.$$
We're only warming the bottle by $5$ out of the initial temperature difference of $20$ degrees, so we don't need to compute logarithms and instead use a linear expression (equivalent to assuming constant heat flow $q$). The time $t_{warm-up}$ required to achieve optimum serving temperature is just:
$$ t_{warm-up} = \tau \frac{5}{20} = 12 \text{ to } 35 \text{ minutes}.$$
Update:
Here's some data for water in a wine bottle (hey, I'm not wasting good wine). I used one of those vacuum storage stoppers; the hole was just right to poke through a kitchen thermometer. Two diffferent ones, actually. The third curve is an exponential with a 30 minute time constant, which looks to be in the ballpark. It looks like I'm underestimating something, maybe convection?
Other small effects that you may be ignoring are heat conduction to the air and temperature drop along the pipe.
If the fluid is a gas, its temperature may drop by some small amount due to pressure drop along the pipe. The fluid temperature will also drop due to the heat transfer that you calculate. An old reference for these effects is Chapter 8 of Compressible Fluid Flow by Ascher Shapiro (1953).
Some practical data on heat transfer may be found in Marks' Standard Handbook for Mechanical Engineers. In the 2007 edition, Eqn 4.4.15 is applicable to practical calculations. This equation uses a combined convective and radiative heat transfer coefficient approach. A table of combined coefficients is also given.
Best Answer
So first off, i'm going to have to break to you that this $\Delta E_\text{cond}(t) + \Delta E_\text{conv}(t)$
is by far the most complicated bit. There is a simple equation $\frac{q}{A} = h\Delta T$ where $q$ is the heat flux (W/m^2), $A$ is the surface area of the object, $\Delta T$ is the temperature difference between the bulk of the fluid and the object and $h$ is the heat transfer coefficient. Unfortunately $h$ 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 $Pr$; the Prandtl number. This is the ratio of momentum diffusivity to thermal diffusivity for the fluid. Effectively this describes the relative size of the thermal and velocity boundary layers for the fluid. For low $Pr$ 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, $\Re _\text{L}$ is going to be important and is defined as $\Re = \frac{v L}{\nu}$
This describes the flow regime around the object where $L$ is a characteristic length and can be taken to be the diameter of the sphere in this case. $\nu$ is the kinematic viscosity and $v$ 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. $Nu _\text{L} = \frac{h L}{k}$
where $L$ is again the characteristic length, $h$ is the heat transfer coefficient and $k$ 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 $T _\text{g}$ also known as $T _\infty$ and the temperature at the solid boundary.
As a small aside I will mention the Biot number. $Bi = \frac{h L}{k _\text{b}}$
where the symbols have their usual meanings except $k _\text{b}$ 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 $h$. 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 $Nu = 2 + 0.43 Re ^\frac{1}{4} $.
This is for Pr approximately equal to 1 and Reynolds number between 1 and 10000.
(I could not find another correlation when Pr is closer to the range expected for air and most other gases. You will have to either accept this approximation or find your own information sorry.)
Decide on appropriate characteristics of your fluid and solid. Calculate the Reynolds and hence Nusselt numbers and then the heat transfer coefficient. Use this to solve for the heat flux for an appropriate temperature difference and then put this in your integral with your radiation term to solve for the total change in energy. I think that about does it. Here is a textbook that goes into a lot of detail:
http://www.unimasr.net/ums/upload/files/2012/Sep/UniMasr.com_919e27ecea47b46d74dd7e268097b653.pdf
I hope this is helpful.