How to cool a cup of coffee faster?

Question

Here is a question I have always wondered about.

• I have a family member that dislikes drinking hot beverages (e.g. coffee, tea). https://www.flickr.com/photos/adamcohn/16290609394
• This family member after preparing the drink will stretch their hand as high up as they can .... and pour it from a large height into another contained, and a lot of steam will come out of the drink.
• Then, they will pour it back into the original cup and repeat this process again and again.
• After doing this several times, the temperature of the drink will have significantly cooled.

My Question: I am wondering if there are some mathematical equations that can describe the relationship between heat transfer and the height at which the coffee is poured from. For example, if I poured a hot cup of coffee from a height of 10 meters - given the temperature of the surroundings, what would the "average" temperature of the coffee be by the time it finished travelling 10 meters?

In more mathematical terms:

Given:

• A 250 mL cup of coffee initially at 80 degrees Celsius.
• An ambient room temperature of 23 degrees Celsius.
• The coffee is poured from a certain height of $Z$ meters.

I found the following 3 equations which might be helpful

1. Newton's Law of Cooling:

$$\frac{dT}{dt} = -k(T - T_{\text{env}}),$$

where:

• $T$ is the temperature of the coffee,
• $T_{\text{env}}$ is the ambient temperature,
• $k$ is a heat constant

2. Bernoulli Principle

$$P + \frac{1}{2}\rho v^2 + \rho gh = \text{constant},$$

where:

• $P$ is the pressure of the fluid (coffee),
• $\rho$ is the density of the fluid,
• $v$ is the velocity of the fluid,
• $g$ is the acceleration due to gravity,
• $h$ is the height from which the coffee is poured.

3. Fluids in Free Fall: Question about fluid in free fall flow

But I am not sure this coffee situation can be simplified (e.g. model it based on Brownian Motions) and these mathematical principles be applied to this situation to understand the relationship between the height at which coffee of a certain temperature is poured and the rate of heat transfer.

Does anyone know if this is possible?

Answer 1

There is is no simple mathematical relationship between heat transfer and pouring height. The more relevant parameter is actually the time-integrated beverage-air surface area exposure, but that is hard to model for the dynamic surface of a poured liquid.

Your question is a bit more specific, but much of the relevant physics is already discussed in:

• "Per unit area, is there more heat transfer through the open top of a mug or the side walls?"

• "Why does water get significantly colder while falling through the air?"

• "Why does splitting hot tea from one glass in two glasses makes it cool faster?"

Simple home experiments demonstrate that a hot beverage in an uncovered cup cools primarily via convection and evaporation from its surface, in a roughly 2:1 ratio. Pouring the beverage temporarily (and often dramatically) increases the exposed surface area of the beverage.

As the coffee falls, its rate of cooling will increase even faster than the increase in its surface area. This is because the coffee's movement through the air increases the cooling rate per unit surface area in the same way that blowing across a hot beverage does. Air at a coffee-air interface is warmed and humidified by the coffee which reduces the conduction and evaporation rates, unless it is replaced by cooler and dryer air. This can be done by either moving the air over the coffee or the coffee through the air.

There is clearly no simple relationship between height and cooling. For example, if you pour the coffee quickly in a thick mass it will cool less than if dribbled out slowly over time in a thin stream from the same height.

To see how challenging it is to model all this mathematically from first principles, one can look at the literature on the simpler case of how a falling spherical drop of water cools. According to a classic 1951 paper by Kinzer and Gunn on "The evaporation, temperature and thermal relaxation-time of freely falling waterdrops", the thermal relaxation time for the drop to reach $1-1/e=63$% of its equilibrium temperature is roughly

$$\tau = \frac{a^2 d c}{3 \left[ \sigma + L D (d\rho/dT)_s\right]\left[1+Fa/s\right]}$$

where

• $a$: radius of the drop
• $d$: density of water
• $c$: specific heat water
• $\sigma$: thermal conductivity of air
• $L$: latent heat of evaporation of the water
• $D$: diffusivity constant of water in air
• $(d\rho/dT)_s$: mean slope of the saturated vapour-density-temperature curve averaged over the temperature of the drop and the surrounding air
• $F$: a dimensionless quantity about 1
• $s$: effective thickness of the spherical shell of air around the drop that is heated and humidified by the drop. Outside this shell lies ambient air.

This last parameter $s$ is where air flow comes in, since at standard temperature and pressure, $a/s \sim 0.2\sqrt{Re}$, where $Re=2a\rho_{\mathrm{air}} u/\eta_{\mathrm{air}}$ is the Reynolds number calculated from the air density $\rho_{\mathrm{air}}$, viscosity $\eta_{\mathrm{air}}$, and relative velocity (falling speed) $u$.

Plausible values of droplet size and velocity can easily increase the cooling rate (i.e. reducing $\tau$) by an order-of-magnitude or more, and I'd expect similar increases for your family member's poured beverage. This sounds like a fun home experiment to me.

Answer 2

The heat stored in the coffee is dependent on the heat mass of the liquid, is it weak coffee or strong. Strong coffee may have more particulates in the water which may increase heat mass.

As the height of the cup increases the stream flowing out from the cup becomes thinner. As the stream becomes thinner, the heat mass of the stream becomes less and therefore heat transfer is faster out of a thinner stream than a thicker stream.

The temperature of the air surrounding the stream is also a major factor. If the temperature of the air surrounding the stream is slightly less than the stream, heat flow out of the stream will be small. If the temperature of the air surrounding the stream is near freezing, then heat flow out of the stream will be very high.

Modeling this cooling scenario above mathematically, is near impossible.