What's the heat capacity of pizza? I'm working on designing a pizza oven. What's a good estimate of the expected heat capacity of a pizza given its radius?
A range might be the best way to answer this based on the type of pizza. The capacitance would imaginably be significantly larger for an extra thick supreme vs a thin cheese.
Maybe one approach could be to look at the capacity of ingredients: cheese, sauce, yeasty bread, ... I don't know how to calculate the interaction with or effects of thickness and layers. If anyone has a guess or some resources to find it that would be fantastic.
 A: I would imagine the high heat capacity of water, its high latent heat of vaporization and the large amount of it present would dwarf everything else present in the pizza. Water has 10x the heat capacity of steel for goodness sake.
One recipe is 320mL of water for 500g of flour for dough. Cheese is probably the next heaviest thing on the pizza and is 30-40% water. Or pizza sauce which is 70% water.
So treat the pizza as a cup of water with 50-60% the mass of the pizza. It's up to you to order a bunch of pizzas to measure, weigh, and eat to determine the average density of a pizza so you can translate to weight to diameter. Or maybe just read frozen pizza boxes at the grocery store. Or forget diameter and just go by weight.
A: See, e.g., https://www.engineeringtoolbox.com/specific-heat-capacity-food-d_295.html
Flour - 0.38 kcal/kg/degC,
Sausage, Bologna - 0.71 kcal/kg/degC,
Tomatoes, red -0.95 kcal/kg/degC, etc.
A: Chemical Engineer here (not a physicist).  I also do a lot of baking.  If you are designing a pizza oven and are preoccupied with the heat capacity of the pizza you are looking in the wrong place.
Baking a pizza is going to be predominately a heat transfer problem and not a thermodynamics problem.  It might be different with a Chicago-style pizza (their pizzas are quite thick), but with a traditional (non-Chicago) American pizza, the design of the oven will be guided by getting heat into the pizza from both the bottom and the top at the right temperature, fast enough to cook the pizza efficiently without burning either the crust or the topping (though you want to get it almost burned).
The main reason for this is that a pizza is very thin and gets heat in two directions, from the top and the bottom.  It's not like a boiling potatoes question (where the source temperature is constant (100 degC) and the surface of the potato is small compared to its bulk).  That said, boiling potatoes is still a heat transfer problem as well - you need to make sure the center of the potato is fully cooked.
The heat transfer from the bottom of the pizza will be dominated by conduction if you are placing the pizza on a hot surface (a pizza stone or the bottom of a ceramic line oven).  If you are putting the pizza on a grill with a heat source underneath and an air gap, then you will have a combination of radiant and convective transfer - which predominates depends on the design of the oven (which is up to you).  Heat transfer to the top will very likely be some combination of radiant and convective heat transfer (a good pizza oven has a grill on the top).
Assuming this is a ceramic bottom surface and a radiant top (where radiant transfer predominates), the question is "how hot should the bottom surface be and how hot and how far away should the top's radiant heat source be in order to completely cook the the pizza in an appropriate amount of time (fully cooked to not quite burned)?"  You want to get the bottom and the top to take the same amount of time to do their cooking "work".
Radiant heat transfer is described by the Stefan-Boltzmann equation (which means that heat transfer is proportional to the difference of the fourth powers of the absolute temperatures of the source and sink (the heat source and the pizza top)).  Conductive heat transfer is simply proportional to the difference in temperatures (the ceramic and the pizza bottom).
You may also want to familiarize yourself with the Maillard Reaction.  It will determine the temperature you want to get the bottom and top surfaces to.
I think this might have been an undergraduate exam question way back in the day (I do remember the infinitely long circular cross-section roast beef in an infinitely long square cross section oven question on the final).
A: It is important to know, that the heat capacity of the pizza will change during the baking process, because water is leaving the system due to evaporation. Therefore we may restrict the calculation to the initial state when the heat capacity is maximal. I would go and deal with this in the following, experimental way:
What you want to know is the heat capacity per mass.
Mass is defined as volume times density.
$$ m = V \cdot \rho $$
Volume is defined for a pizza as area times thickness. We assume, that pieces that are higher like paprika bits and areas that are only covered with sauce and cheese will average each other out and result in a averaged thickness $t_{avg}$.
$$ m = \pi   r^{2}   t_{avg} \cdot \rho $$
If you want to have precise values for the volume instead of using an approximation value for the thickness, one could freeze the pizza, put it into water and immediately measure the volume increase in a closed container of the water after complete submergence. Alternatively the application of a thin polymer coating at the whole surface area of the pizza could be tried to avoid melting during the measurement.
In our case, the density of an averaged size pizza is unknown and it is needed to be determined by external measurement of dimension and weight. Assuming, we would go with a margarita pizza of 26 cm in diameter and 2 cm in thickness and a weight of 350 g, a density would be calculated as the following:
$$ \frac{0,350 kg}{0,13m^2 \cdot \pi \cdot 0,02m} = 329,79 \approx 330 \frac{kg}{m^3} $$
With that value we can determine the mass of any pizza with similar density and surface structure in dependence of the radius:
$$ m= 330 \frac{kg}{m^3} \cdot  \frac{d}{2}^2 \cdot   \pi \cdot   0,02m $$
For the next step you would need to determine the heat capacity in a calorimeter. Before this step, the complete pizza would be homogenized with a device like a blender until it is absolute homogeneous.
A predefined mass, for example 30 g of the slurry would be placed into the measurement cell without any remaining air (with a different heat capacity). In a simple calorimeter a well defined amount of heating energy is applied and the temperature change is being measured. The heat capacity of a system is defined as:
$$ C = \frac{dQ}{dT} $$
For a homogeneous pizza slurry:
$$ C = m \cdot c $$
$c$ is the specific heat capacity which is a material constant.
We assume, our calorimeter is applying 480 J of heating energy for a temperature increase of 5 K during the experiment. We use the simplification, that the specific heat capacity of pizza slurry may not be a function of the temperature and is constant. With this information we can calculate the heat capacity of the pizza per mass. Still, we have to consider the instrumental constant of heat capacity for the calorimeter.
$$ C_{Sys}=C_{pizza}+C_{cal} $$
$$ C_{Sys}=c_{pizza}  \cdot  m_{pizza} + c_{cal} \cdot  m_{cal} $$
$$ \frac{c_{sys}-c_{cal} \cdot   m_{cal}} {m_{pizza}}=c_{pizza} $$
We assume for the calorimeter, that the heat capacity is only determined by the measurement cell which has a weight of 10 g and a specific heat capacity of $0,42 \frac{J}{gK}$ (common value for steel). Under real life conditions one would need to determine the heat capacity of the calorimeter by own experiments, for example by the application of a defined amount of heating energy to the water filled calorimeter and the subsequent measurement of the temperature of the water, another approach would compare the expected temperature of a resulting water body after mixing two exactly measured volumes of water with defined temperatures (the discrepancy would be caused by the calorimeter). For the sake of the argument we may stick here to the simplification.
Therefore examplaric we receive:
$$ \frac{\frac{480J}{5K} - 2,1 \frac{J}{5K g} 10 g}{30g}=3,06 \frac{J}{gK} $$
Therefore we can set up the final relation between the diameter of an averaged margarita pizza and its heat capacity. One may replace the numeric values with the results of individual experiments.
$$ C_{pizza}=c_{pizza} \cdot \frac{d}{2}^2 \cdot   \pi \cdot   t_{avg} \cdot  \rho $$
$$ C_{pizza}=\frac{c_{sys}-c_{cal} \cdot m_{cal}}{m_{pizza}} \cdot \frac{d}{2}^2  \cdot \pi \cdot   t_{avg} \cdot  \rho $$
A: My two cents, feel free to downvote- I think it would be nearly impossible to theoretically calculate that. So experimental calculation would be the best shot.
I have zero clue how to experimentally find the heat capacitance but I will try.
First, make a metallic short cylinder in which you are going to be placing your pizza. Secondly put the empty metallic cylinder in a pot of water and heat it for about 20 minutes maybe (you can change the time). By recording the temperature changes in the metal and the water and knowing their respective heat capacitance you must be able to find the energy spend in 20 minutes (alternatively, you could Google how must heat is produced by the stove per minute if such a data exists)
Thirdly repeat the same experiment but with pizza in the container. Now hypothetically the same amount of heat must be produced. Calculate the total heat absorbed by the metal and water. Subtract this from the earlier heat result we found. The difference between the two observation mist be because some amount of heat is absorbed by the Pizza. Now you know how much heat is absorbed by the Pizza. Calculate the temperature difference and you may be able to find the heat capacity. I don't if this will work practically however...
A: I agree with others that pizza is not likely the primary loss of heat in a pizza oven.  However, to estimate a value, I'd bake a pizza and weigh it in and out of the oven to calculate the evaporative loss. Calling those numbers in kg $m_{in}$ and $m_{out}$, you can calculate the heat absorbed by a pizza as:
$Q = C_p \Delta T m_{in}+\lambda(m_{in}-m_{out})$
Where:
$C_p=0.8 \frac{kJ}{kg K}$ -- A ballpark specific heat for pizza ingredients (see various proposals above).
$\lambda=2260 \frac{kJ}{kg}$ -- Latent heat of evaporated water.
$\Delta T=70 K$ -- Rise in temperature from room (20C) to bread being cooked (90C).
Note that 2260 is much greater than 70*0.8, so once a few percent mass is lost to evaporation, that will dominate the answer.
That's thermodynamics, the subject of this forum.  Calculating the value of the evaporation falls under what engineering calls convective heat transfer, a complicated subject with large uncertainties.  In this case, I think any reasonable estimate would be far more work and far less tasty than the measuring the real thing.
