Physics of the Giant Muffin
I spent part of last night thinking about the scaling of muffins. If you have a recipe for one muffin of characteristic size r, you can scale up the recipe to make a muffins of size r by scaling the volume of ingredients accordingly: instead of 1 egg, you put in a eggs, and instead of 2 cups of water, you put in 2a cups, etc. (Here a is just some numerical scale factor, say 10 or 15.)
Suppose, however, that you wanted to instead make a single muffin of characteristic length a*r. You would scale the volume of ingredients as above, but how would you scale the cooking time? The cooking time would certainly not be the same. Would you just cook for a*t minutes, where t is the amount of time for a single muffin of characteristic length r? No–because the outside of the giant muffin will get burnt before the insides are baked. Anyone who has made jumbo sized food will know that the answer is to lower the temperature and increase the cooking time. This way the heat can penetrate to the center of the muffin before the outsides get burnt.
The question of the hour, however, is can we quantitatively figure out how much longer and how much cooler one should be cooking?
Last night I had thought this would be a simple problem, but I ended up spending more time than I had expected thinking about this. In particular, one has to think a little bit about the mechanism of cooking. Heat is energy, in particular, the energy associated with an ambient temperature is given by scaling by the Boltzmann constant.
Stupid Model: Latent Heat of Cooking
When one is cooking a muffin, the muffin mix is absorbing part of the energy (heat) and using it for a phase transition, e.g. a latent heat of cooking. This is just like melting ice, as you put in a quantity of heat, ice will warm up until it starts melting, at which point the extra heat you put in will be used to convert ice into water without changing the temperature of the joint system. Does cooking occur on a similar principle?
Probably not. For one, the above model assumed that the object is being heated uniformly over its volume; this is not the case for giant foods. (By the way, this is the reason why bundt cakes have a funny shape, so their insides can bake more uniformly.) Further, the ice-water transition we considered is linear–you hold pressure reasonably constant and go back and forth with temperature. When you cook, you start out with a system in state A (mixture of starting ingredients), you heat it up to a high temperature so that the mixture can explore its phase space, and then it reaches a state B (one that maximizes entropy). When you cool it back down to room temperature, the system remains in state B.
This leads me to believe that state B (the cooked state) is a state of lower total energy than state A. A few possible counter examples:
- Maybe state B has a higher energy, but it has a funny energy-phase graph such that B is at a local minimum of energy? Not likely: if you’re only cooking a small bit of muffin, you can heat it up to arbitrarily high temperatures (before burning it). That is, you can always cook at a higher energy than the size of the local potential well, and the system will try to access the lower energy state A.
- Maybe B has higher energy than A, but the system is relaxed into it as a ground state at room temperature because energy has been absorbed into molecular bonds. If the bonds are stable, then the resulting state is of lower total energy.
Anyway, the flow of heat is given by the diffusion equation. However, I’m not sure if there should be an added term which accounts of the energy absorbed/released by each cooked layer of muffin. I’ll leave this to some speculation.
Imagine a muffin is a sphere.
The sphere sits in a heat reservoir that we assume to be a perfect blackbody. Consider a differential spherical shell. There exists a temperature gradient across it. Heat flows through this layer at a rate roughly proportional to the second derivative of the gradient (the diffusion equation). However, some heat may be added/removed due to the process of cooking. We assume that the thermal conductivity of cooked muffin is the same as uncooked muffin (reasonable). We can write this down as a tidy, modified diffusion equaton.
The necessary boundary conditions are nonsingularity at the origin and a Dirichlet boundary at the surface of the muffin representing the blackbody temperature. We can model the cooking process of a given infinitesimal volume of the muffin as a integral of temperature T(t,x) over time, i.e. a fraction of the total energy at every moment of time. Our final boundary condition is the maximum permissible difference in total energy absorbed (“cooked-ness”) between the center of the muffin and the exterior (i.e. the difference in energy between minimally cooked and just-before-burnt).
Anyway, given these parameters one can solve the modified diffusion equation (an analytic solution shouldn’t be too much trouble, but a numerical solution should be really easy) in three dimensions.
We can also discuss qualitative aspects of the solution. There are roughly two phases in cooking, the first phase is thermal non-equilibrium when the temperature at the center is not equal to the blackbody temperature. It is in this phase that a difference in cooking between the middle and crust of the muffin occur. The second phase is in thermal equilibrium, and hence all parts of the muffin are cooking uniformly. So the differential equation should be solved in the first regime using the boundary condition for maximum permissible difference, and then it should be easy to determine the amount of time cooking in the second phase.
The final solution can be minimized for cooking time.
There is another alternative–it may be optimal to have the blackbody (oven) temperature increase with time, this would decrease the temperature gradient at any given moment in time while also allowing for quicker baking time. The idea would be that you reach equilibrium cooking conditions as soon as possible (or after a specified difference in crust-inner is attained) and then you can increase the heat quasistatically to increase cooking rate.
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