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, $C_v$ of2.75 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.
SheldonCooper
?:P
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