When should I take wine out of the fridge - transient heat transfer problem

Question

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-K and the wine itself has a specific heat at constant volume, $C_v$ of 2.75 kJ/kg-k as per this
• The volume of the bottle of wine is 750 mL or $750 \times 10^{-6} m^3$
• The wine is at a temperature of 5 C and 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.

Answer 1

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: $$ t_{warm-up} = \tau \frac{5}{20} = 12 \text{ to } 35 \text{ minutes}.$$

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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?

Answer 2

I'm an engineer, not a physicist. My solution:

I suggest you find a bottle of wine with thesame alcohol content. You chill it in the fridge to same temperature as teh one you want to drink. You take it out way before the dinner, put it in the same position you would you real bottle, and start to monitor the temperature. The moment the temperature is right, you note the time this took (and start drinking). Then, you should know the time it takes your wine to reach optimal drinking tempreature after the fridge. Also, you get drink more wine.