# Should I hold a baby formula bottle to cool it down faster?

This is an interesting and somewhat surprising physics problem - holding a hot object in your hand will cool it down faster, even if the air around is colder. I guess that 90% of people would be surprised by this.

The formula is prepared at 60 C and must cooled to 37 C before serving. The ambient temperature is 20 C. My overlord doesn't like to wait - should I hold the bottle in my hand to cool it down faster? Should I let go at some point or keep holding until the end?

I assume the liquid-based plumbing in my hand will eliminate all extra heat and keep the hand at roughly 37 C.

Bonus question: At what ambient temperature should I not hold it at all?

Bonus question 2: I assume the plastic bottle is an average thermal conductor. At what level of thermal resistivity is it better not to hold the bottle?

• Isn't this more of an engineering question? Commented Jun 24 at 7:52
• "Bonus question: At what ambient temperature should I not hold it at all?" Don't ask "bonus questions." Anyways, this question cannot be answered without more details about your hand and the bottle and a bunch of other stuff.
– hft
Commented Jun 28 at 18:50
• – hft
Commented Jun 28 at 18:51
• Focus on a single clear question. That is how this website works best. This is not a place where you just get to do whatever you want.
– hft
Commented Jun 28 at 18:52

Sure! I think I'm qualified to answer, not because I used to teach heat transfer in a Chemical Engineering course, but because I raised four daughters!! Air is an excellent thermal insulator. If you put it in direct contact with your hand, it cools down much faster. You will have the same effect emerging in water, however. If you put the bottle in a turbulent flow of water, it will cool down something like ten times faster. It will be too fast, though, because the liquid inside won't be cooling as fast. So you may also shake the bottle to speed the cooling inside. To calculate the time for cooling from $$60^\circ C$$ to $$40^\circ C$$ of a baby bottle filled with milk in the air, both without convection and with high turbulent convection, we can use the lumped capacitance method. Here's a detailed approach:

### 1. Without Convection

In the absence of convection, heat transfer occurs solely through conduction and radiation. However, for simplicity, we'll consider negligible heat transfer in this scenario due to very high thermal resistance.

### 2. With High Turbulent Convection

For high turbulent convection, we use the given heat transfer coefficient.

### Assumptions and Given Data

• Initial Temperature ( $$T_{\text{initial}}$$): $$60^\circ C$$
• Final Temperature ( $$T_{\text{final}}$$): $$40^\circ C$$
• Ambient Temperature ( $$T_{\text{ambient}}$$): $$25^\circ C$$
• Density of Milk ( $$\rho$$): $$1000 \, \text{kg/m}^3$$
• Specific Heat Capacity ( $$c$$): $$4186 \, \text{J/(kg·K)}$$
• Volume of Bottle: $$0.25 \, \text{L} = 0.00025 \, \text{m}^3$$
• Surface Area of Bottle: $$0.03 \, \text{m}^2$$
• Thermal Conductivity of Bottle: $$0.6 \, \text{W/(m·K)}$$
• Heat Transfer Coefficient for High Turbulent Convection ( $$h$$): $$100 \, \text{W/(m}^2\text{·K)}$$

### Calculations:

1. Mass of Milk: $$\text{Mass} = \rho \times \text{Volume} = 1000 \, \text{kg/m}^3 \times 0.00025 \, \text{m}^3 = 0.25 \, \text{kg}$$

2. Time Constant for Convection: $$\tau = \frac{m \cdot c}{h \cdot A} = \frac{0.25 \, \text{kg} \times 4186 \, \text{J/(kg·K)}}{100 \, \text{W/(m}^2\text{·K)} \times 0.03 \, \text{m}^2} \approx 349.5 \, \text{s}$$

3. Cooling Equation: $$T(t) = T_{\text{ambient}} + (T_{\text{initial}} - T_{\text{ambient}}) \exp\left(-\frac{t}{\tau}\right)$$ $$T(t) = 25 + (60 - 25) \exp\left(-\frac{t}{349.5}\right)$$

### Explanation of Results:

The values calculated will give us the time required to cool from 60°C to 40°C under high turbulent convection.

### Calculation Results for Cooling Time

For cooling from $$60^\circ C$$ to $$40^\circ C$$ under high turbulent convection:

• Time Constant for Convection ( $$\tau$$): Approximately $$349.5 \, \text{s}$$
• Cooling Time: Approximately $$299 \, \text{s}$$ (or about 5 minutes)

### Cooling Time Comparison for Different Scenarios

Here are the results for the cooling times of a baby bottle filled with milk from $$60^\circ C$$ to $$40^\circ C$$ under different conditions:

• High Turbulent Convection (Air at 25°C):
• Cooling Time: Approximately 299 seconds (or about 5 minutes)
• Turbulent Water at 37°C:
• Cooling Time: Approximately 144 seconds (or about 2.4 minutes)
• Turbulent Water at 20°C:
• Cooling Time: Approximately 50 seconds (or about 0.8 minutes) The great advantage of holding the bottle in your hand compared to placing it in running tap water, despite the body temperature being higher, is the possibility of causing agitation in the internal fluid, due to sudden movements. When considering the convection parameters for cooling a baby bottle, you need to account for both the internal and external convection processes. Here’s how to approach the problem:

### External Convection

External convection involves the heat transfer between the outer surface of the baby bottle and the surrounding air. The rate of heat transfer can be described by Newton's Law of Cooling: $$q = h_{\text{ext}} A_{\text{ext}} (T_{\text{surface}} - T_{\text{air}})$$ where:

• $$h_{\text{ext}}$$ is the external convective heat transfer coefficient.
• $$A_{\text{ext}}$$ is the external surface area of the bottle.
• $$T_{\text{surface}}$$ is the temperature of the bottle's outer surface.
• $$T_{\text{air}}$$ is the ambient air temperature.

### Internal Convection

Internal convection involves the heat transfer between the inner surface of the baby bottle and the fluid (milk or formula) inside. This can be described similarly: $$q = h_{\text{int}} A_{\text{int}} (T_{\text{fluid}} - T_{\text{surface}})$$ where:

• $$h_{\text{int}}$$ is the internal convective heat transfer coefficient.
• $$A_{\text{int}}$$ is the internal surface area of the bottle.
• $$T_{\text{fluid}}$$ is the temperature of the fluid inside the bottle.
• $$T_{\text{surface}}$$ is the temperature of the bottle's inner surface.

### Changing Convection Parameters

1. Including Internal Convection:

• Internal Heat Transfer Coefficient ( $$h_{\text{int}}$$): This can be increased by promoting internal mixing, which can be achieved by shaking or stirring the fluid inside the bottle. Higher mixing rates generally increase the internal heat transfer coefficient.
• Fluid Properties: Changing the fluid’s viscosity and thermal conductivity (e.g., using a different liquid) can affect internal convection.
• Surface Area ( $$A_{\text{int}}$$): This is typically fixed by the geometry of the bottle, but using a bottle with internal fins or a different shape can change the effective internal surface area.
2. Including External Convection:

• External Heat Transfer Coefficient ( $$h_{\text{ext}}$$): This can be increased by improving airflow around the bottle. Using a fan or placing the bottle in a cooler with forced air circulation can help. Natural convection can also be enhanced by ensuring the bottle is not in a stagnant air environment.
• External Conditions: Increasing the temperature difference between the bottle’s surface and the ambient air (e.g., by cooling the air) can increase the heat transfer rate.
3. Ignoring Internal Convection:

• Assumption of Well-Mixed Fluid: If the fluid inside the bottle is assumed to be well-mixed, you might treat it as having a uniform temperature. In this case, the internal convection can be simplified to an effective thermal resistance.
• Negligible Internal Resistance: If the internal thermal resistance is very low compared to the external resistance, you can ignore the internal convection effects. This would be the case if the internal mixing is very effective.

To include or exclude internal convection in your calculations, you can use a thermal network approach with resistances in series:

#### Including Internal Convection:

$$\frac{1}{R_{\text{total}}} = \frac{1}{R_{\text{int}}} + \frac{1}{R_{\text{wall}}} + \frac{1}{R_{\text{ext}}}$$ where:

• $$R_{\text{int}} = \frac{1}{h_{\text{int}} A_{\text{int}}}$$
• $$R_{\text{wall}} = \frac{L}{k A_{\text{wall}}}$$ (thermal resistance of the bottle material)
• $$R_{\text{ext}} = \frac{1}{h_{\text{ext}} A_{\text{ext}}}$$

#### Ignoring Internal Convection:

$$\frac{1}{R_{\text{total}}} = \frac{1}{R_{\text{wall}}} + \frac{1}{R_{\text{ext}}}$$

I am equating the hand firmly holding the bottle to being immersed in perfectly agitated water (approximating human blood as water). Obviously the analysis would have to be much more complex, for example approaching the human body with a heat exchanger with capillary tubes flowing at the speed of blood circulation. The complexity, however, may not compensate for the gain in accuracy obtained. Much more important would be to know how to ensure turbulence inside the bottle to make it a perfect mix. I will tell you my technique. I begin rotating the bottle to impress a rotation of the fluid inside. This flow, however, is laminar, so as soon as the flow reaches a steady state, I revert the direction of rotation.Reversing the direction of rotation will indeed disrupt the flow, causing some mixing. I think it would be clever to introduce obstacles for the internal flow: Adding obstacles or irregularities inside the bottle can promote flow instabilities and turbulence. For example, small fins or ridges could create localized turbulent regions.Actually, some bottles have this feature.

You want to maximize the thermal conductivity from the bottle the outside. Water is much better than air in this regard, so water bath would be good. The bottle will cool from the outside in and occasionally shaking it a bit will help with creating a uniform temperature inside.

If you want to be scientific about it, you can do a few test bottles: Fill them at 60C water, put them in a water bath and check the inside temperature with a food thermometer. This will give a you a ball park time estimate how long to cool it in water.

One issue here is that the temperature of tap water does vary with season, weather, and even time of day. One way around this would be ice water, which is always at 0C. If the water is very cold, you probably want to add a another minute or so in air at room temperature to let the bottle wall come up to a more comfortable temperature and the internal temp stabilizes.

• I was hoping we could answer this with an equation instead of doing an experiment. Commented Jun 24 at 7:26
• The equation is the heat equation: $\frac{\partial ^2}{\partial t^2} T=k(\frac{\partial ^2}{\partial x^2} +\frac{\partial ^2}{\partial y^2} +\frac{\partial ^2}{\partial z^2}) T$ where $T(x,y,z)$. To solve it for your case you need to know a few things, like the area of contact, the heat conductivity of the bottle, the heat conductivity of your skin (have you been sweating or not?). You could simplify things by looking along 1 dimension: one region of water, a small region of plastic and then either air or skin being held at constant temperature. Commented Jun 24 at 9:43
• Still, being held at constant temperature is not a good approximation. The air/skin quickly warms up and then slows down heat transer. If you keep blowing air towards the bottle you are actually getting closer to the ideal case of constant air temperature. My point of this long comment: sometimes an experiment is easier and more reliable than doing the calculation. Commented Jun 24 at 9:49

## Premise and suggestions

Before answering your question, I'd like to do again the premise that this question and the comments in your question make from little to no sense to me.

Anyway, beside the suggestions:

• if you need to cool it down fast, put it in a bottle made with a non-insulating material and put the bottle in cold water; then pour the milk in the feeding bottle;
• don't use insulating bottles if you want to cool it down fast

## Let's do some physics

Let's assume you only have bottles that have good conduction insulation, but no insulation to radiation, as the situation depicted in your comment.

Let's do some computation assuming:

• a cylindrical bottle, with basis radius $$R = 0.05 \, m$$ and height $$h = 0.2 \ , m$$; lateral surface is $$S = 0.0785 \, m^2$$ and volume is $$V = 0.0157 \, m^3$$ (approximately $$1.5 \, l$$)
• milk with specific heat $$c = 3970 \frac{J}{kg K}$$ and density $$\rho = 1000 \frac{kg}{m^3}$$
• maximum emissivity of the medium $$\varepsilon = 1$$, so that the heat flux reads $$\Phi = S \sigma \left(T_{milk}^4 - T_{env}^4 \right)$$

In absence of work done on the system, the first principle of thermodynamics thus reads $$m c \frac{d T_{milk}}{d t} = -\Phi \ ,$$

Let's put ourselves in the most favorable condition, i.e. exchanging with environment at $$T^{env} = 0 \, K$$ and solve analytically the ordinary differential equation

$$\frac{d T}{d t} = - \frac{S \sigma}{m c} T^4 \ ,$$

with initial condition $$T(0) = T_0$$. The solution via separation of variable reads

$$\int_{T_0}^T T^{-4} dT = - \frac{S \sigma}{m c} \int_{0}^t dt \quad \rightarrow \quad -\frac{1}{3} ( T^{-3} - T_0^{-3} ) = - \frac{S \sigma}{m c} \, t$$ $$\quad \rightarrow \quad t = \frac{T^{-3} - T_0^{-3}}{3 \frac{S \sigma}{m c}} \ \quad \rightarrow \quad t = 3027.6 \, s = 50 \, min \ ,$$

that I think it's a bit too much for the habits of your overlord.

## Physics suggests to rely on common sense

Said that, I'd suggest one more time to focus on the suggestions at the beginning of this answer.

Liquid cooling is generally more efficient than air cooling for CPUs and car engines, even though a fan is used in the air cooling versions. In the case of the baby bottle we are not comparing the water cooled version to a fan powered air cooling system, but to a passively air cooled system which depends only on the motion of the air due to convection. The blood vessels in the hand circulate a water based fluid through the skin in contact with the bottle, and release that heat over a larger area (basically the surface area of the human body including the surface area of the lungs.) This means the hand being basically a water cooled system is probably more efficient than passive air cooling even when the temperature of the bottle is close to 37 C. It is probably best to keep the milk in the bottle swirling to promote good heat transfer. Since your hand is unlikely to cover the entire surface of the milk bottle, you are going to get some air cooling anyway when using the hand cooling method.

Obviously at constant thermal conductivity you would want to keep the bottle in contact with the cooler bath, the air, rather than the warmer bath, your hand.

So air has the advantage that it is colder. What is the advantage of the hand then? I'm not sure. I think the question is getting at something like: If I hold the bottle then there will be a larger thermal conductivity so it will cool faster despite the smaller temperature difference. Maybe even squeezing the bottle can increase the thermal conductivity more.

I don't know anything about how the thermal conductivity between one solid and one gas (the bottle's outer surface and the ambient air) compares to the thermal conductivity between two solids (the bottle's outer surface and your hand) and also if the thermal conductivity between two surfaces varies e.g. as a function of the normal force between them.

The basic formula is $$dT_{Bottle}/dt = k(T_{Bath} - T_{Bottle})$$ Assuming that hand/bottle $$k$$ is different than air/bottle $$k$$ then the question presents two option. You can have lower $$k$$ and lower $$T_{Bath}$$ or you can have higher $$k$$ and higher $$T_{bath}$$.

This is really an engineering problem that you could try to optimize. But I guess you just want to, at all times, choose whatever option maximizes this derivative. At short times, $$T_{Bottle}$$ is so great that the difference between $$T_{Air}$$ and $$T_{Hand}$$ is neglibile then you'd want to hold it with your hand to take advantage of the large thermal conductivity. But as the bottle gets close to equilibrium with your hand you'd need to swap to air to sacrifice thermal conductivity for the larger $$\Delta T$$.

This answer assumes that thermal conductivity is different for bottle/air and bottle/hand. I have no idea if this is the case.

This answer also assumes both the air and hand act as a thermal reservoir. That is their temperatures don't change appreciatively as a result of heat transfer from the bottle. I think this is a good assumption for the air, but it might be questionable for the hand.

• >What is the advantage of the hand then? I'm not sure. --- Air is know to be a great insulant, meaning it has very low thermal conductivity. Hand, on the other hand, is 60% water, which is a great thermal conductor. Commented Jun 26 at 11:50
• @daniel.sedlacek that’s a fair point. But I think the interface conductivity is what matters here, not the bulk conductivity? Commented Jun 26 at 13:01

No. Holding bottle is not the best way to cool it down faster. I would suggest you to keep the cover of bottle open and also keep stirring bottle slowly to let the heat escape. Stirring will cool it down much quickly.

I don't think holding a bottle vs. leaving it on a table makes much difference (the latter option has the advantage of being able to take care of the baby while the bottle cools.)

One practical option is leaving the bottle outside (provided that it is not summer), where the wind and colder temperature accelerate cooling (i.e., using convection and higher temperature difference.)

Another practice is placing a bottle in a pot filled with cold/hot water (using better thermal conductivity of water, as compared to air) - it is usually used to heat the bottles or unfreeze milk, but it can be also used for cooling.