I am hosting a dinner tonight for which I'll be serving white wine (Riesling to be more specific). Generally white wine is best served chilled (not COLD!) at around 50 F or 10 C.
Just for kicks, I thought I'd treat this as a problem of transient conduction. I assume that (forced)convection is negligible since I will leave my wine bottle in my kitchen which has still air.
The following assumptions are made:
- The wine bottle is assumed to be cylindrical with an outside to inside radii, $r_o/r_i$ ratio of
1.10 - The only mode of heat transfer for this bottle is conduction (perhaps a poor assumption?). The kitchen air is considered to be still and at
25 C - The un-open bottle of wine is a closed thermodynamic system. The glass material has a conductivity $k$ of
1.0 W/m-Kand the wine itself has a specific heat at constant volume, $C_v$ of2.75 kJ/kg-kas per this - The volume of the bottle of wine is
750 mLor $750 \times 10^{-6} m^3$ - The wine is at a temperature of
5 Cand hence needs to be warmed for a while. The entire wine is assumed to have a lumped capacitance (all the wine is at the same temperature with little variation with radius). - The temperature difference between the wine and the bottle wall is assumed to be $\sim 10 C$ and so is the temperature difference between the bottle wall and the room (just a rough order of magnitude).
The first law of thermodynamics (transient) is applied to this closed system bottle of wine:
$$\frac{\mathrm{d}{E}}{\mathrm{d}t} = \delta\dot{Q} – \delta\dot{W}$$
The $\delta\dot{W}$ term is zero for this closed system as only heat is exchanged with the kitchen atmosphere.
$$\frac{m C_v \Delta T_\text{wine-bottle}}{\Delta t} = \frac{2 \pi k \Delta T_\text{bottle-kitchen}}{ln(r_o/r_i)}$$
This gives me the time the bottle of wine needs to be placed in my kitchen outside the fridge as:
$$\Delta t \approx 0.025 \frac{\Delta T_\text{bottle-air}}{\Delta T_\text{wine-bottle}} C_v \approx 68 \text{ seconds}$$
This seems to be a rather small amount of time!!! Are my assumptions wrong? Should I improve this with convective heat transfer between the bottle and the kitchen air? Will my guests be disappointed? :P
EDIT::Including convective heat transport:
$$\underbrace{\frac{m C_v \Delta T_\text{wine-bottle}}{\Delta t}}_\text{Change in total/internal energy w.r.t time} = \underbrace{\frac{2 \pi k \Delta T_\text{bottle-kitchen}}{ln(r_o/r_i)}}_\text{conduction} + \underbrace{h A \Delta T_\text{bottle-kitchen}}_\text{convection}$$.
Here $h$ is the heat transfer coefficient $\sim 1 W/m-K$, $A$ is the surface area of the cylinder. Based on the volume of the cylinder being $70 mL = \pi r_i^2 h$. The height of the bottle is about $1 foot$ or $0.3048 m$ and the generally close assumption that $r_o \approx 1.1 r_i$, I have (all $\Delta T$'s are close and cancel out):
$$\Delta t = \frac{m C_v ln(r_o/r_i)}{\left[ 2 \pi k + 2\pi r_o(r_o + h) ln(r_o/r_i)\right]} \\
\Delta t \approx 260.76 \text{ seconds} \approx 4 \text{ minutes}
$$
This seems more plausible….. But I start doubting myself again.
Best Answer
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?