Cooling Utensils

Background

I have come across an interesting question in a kashrut course that I’m taking. What I find particularly interesting is the fact that since then, I have shared the question with many people and barring a solitary exception, the result has always been the same. The intuitively answer is given, which is incorrect, despite the person knowing all the relevant physics facts that would lead to the correct answer. But for some reason, it’s not put together. I myself first answered the question incorrectly, and when it was pointed out to me had to think about the relevant facts before I was able to really get it. Since I have already shared this with several people and found it interesting, I thought I’d write it up here.

The question we had talked about was this: hot food is taken off the stove and moved into utensils. I put some on a plate and some in a bowl. Which one cools off to an edible temperature faster, the food on the plate or in the bowl?

Building Intuition

To make this problem tractable, let’s add a few details and make some assumptions. Let’s start with the food. Imagine something that’s liquid-y enough to be able to assume the shape of the container, but thick enough it can hold its shape on the plate for long enough to solve our problem. Think something like thick mashed potatos, but with the “spherical cow”-style assumption of it being homogenous so we don’t have to deal with pockets of different densities in the food. Let’s label our scenarios:

1. Scenario I: I put the food inside a hemispherical bowl whose internal diameter is $r$, so my food takes on a half-sphere shape.
2. Scenario II: I turn it upside down onto a plate so that I still have a half-sphere of hot food cooling down, just on a flat surface now.

For temperature purposes, imagine the cooked food being very hot – 70C (158 F) – and edible temperature being about 50C (122 F). Air is at a “room temperature” of 20C (68 F). Image both utensils – the plate and bowl – are made from the same homogenous material with thickness $d$.

The main thing we worry about with cooling is surface area. We know the rate that heat leaves our system is the surface intergral of the heat flux over the objects area, i.e.

$$ \dot{Q} = \oint_A q \,dA.$$

So to make cooling quicker, we either need to manipulate the surface area or change the heat flux, which depends on the temperature gradient and the materials involved.

For our purposes, we have two interfaces to deal with: the circular interface that touches the air in scenario I or the plate in scenario II, and the round surface that either touches the inside of the bowl in scenario I and the air in scenario II. By construction of the example those are nice values: $A_\text{circ} = \pi r^2$ and $A_\text{hemi}= 2 \pi r^2$. With that in hand:

$$\begin{aligned} \dot{Q}_{I} &= q_b\, 2 \pi r^2 + q_a \,\pi r^2 \\
\dot{Q}_{II} &= q_{b}\, \pi r^2 + q_a \, 2 \pi r^2 \end{aligned}$$

where $q_b$ is the heat flux at the utensil interface and $q_a$ the heat flux at the air interface. Since I’m mainly interested in which is faster, I can write this as a fraction:

$$\frac{\dot{Q}_{I}}{\dot{Q}_{II}} = \frac{2 q_b + q_a}{q_b + 2 q_a} $$

Virtually every one I have talked to about this assumes that the plate cools faster, which for our purposes means the above fraction is less than 1, which is fulfilled when $q_a < q_b < 0$ (remember that heat is leaving the system, so the fluxes are negative).

And there is our first snag: most everyone knows that air, like gases in general, are terrible conductors compared to solid materials, particularly ceramics and the like that are often used for utensils. As it turns out $|q_b| \gg |q_a|$ so we could frankly neglect the air interface when thinking about the cooling time and simply declare the bowl in scenario I the winner without further work.

Go Newton

If we really wanted to, we could try to go further and solve something more precise. Newton’s law of cooling, familiar to all calculus students, let’s me calculate the rate of heat loss using only temperature as

$$ \dot{Q} = - h A \left(T(t) - T_\text{env}\right).$$

With $\dot{Q} = m c \frac{dT}{dt}$ where $m$ is the mass and $c$ is the specific heat capacity, this gives me an ODE in temperature only as

$$\frac{dT}{dt} = - {\frac{h A}{m c}} \left(T(t) - T_\text{env}\right)$$

which has the solution

$$T(t) = T_\text{env} + \left(T_0 - T_\text{env}\right)\,\mathrm{exp}(-K t)$$

where $K=hA/mc$. We’re only interested in the comparison of $K_I$ and $K_{II}$, so it mostly focuses on finding $h$. The heat transfer coefficient of air for relevant scenarios can be looked up. The question is how to treat the utensil. It’s interesting to know that we can deal with heat transfer through walls by by an inverse additive relation like this:

$$ \frac{1}{h_\text{b}} = \frac{1}{h_{a}} + \frac{d}{k_b} + \frac{1}{h_f}$$

In this case, $h_a$ is the coefficient for air, $k_b$ is a coefficient for the utensil material, and $h_f$ the relevant coefficient for the food. Recall that $d$ is the thickness of the utensil.

While this is fine for a back-of-the-envelope calculation, technically it doesn’t apply. If the Biot number

$$\mathrm{Bi} = h L / k, $$

is less than about 0.1 it is fine to ignore thermal processes inside of bodies and make the isothermal assumption inherent in the formulation of Newton’s law of cooling.

As per the literature the thermal conductivity of mashed potatoes near our target temperatures is approximately $k =0.58 \mathrm{W} \mathrm{m}^{-1} \mathrm{K}^{-1}$. For our purposes, the characteristic length can be taken to be the hemisphere volume over area:

$$L = V/A = \left(\frac 23 \pi r^3 \right) \left( 3 \pi r^2 \right)^{-1} = \frac 2 9 r. $$

The value $h$ should be averaged over the surfaces and I don’t feel like doing that. Let’s see how far we got with this:

$$\begin{aligned} \mathrm{Bi} &< 0.1 \\ \frac{100}{261} h\,r &< \frac {1}{10} \\ \Rightarrow h &< 261 / (1000 r) \end{aligned}$$

An Ikea bowl is about 5-6 in in length, so let’s use $r=0.07 \mathrm{m}$. With that we get the Biot number we want when $h<3.7$. Since even for air we have $\mathcal{O}\left(h_\text{air}\right)= 10$ this looks like it won’t really hold. Oh well. Being more precise than this is getting to be overkill and I already spent too much time on this problem.

Long story short: if you want your food to cool faster, maximize surface area but also maximize surface area in contact with the utensil, not with air.