The Thermodynamics of Pizza - ChemPRIME

# The Thermodynamics of Pizza

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Harold Morowitz writes in "The Thermodynamics of Pizza" [1]:

"We were chatting at lunch one day, when one of my companions complained of burns on the roof of his mouth from the previous evening's pizza party. Talk drifted in two directions:...focusing on why pizza stays so hot for so long."

Pizza stays hot (and many things feel hot or cold) because of factors other than just their temperature. One of the most important factors is their heat capacity.

When we supply heat energy from the brick pizza oven at about 570 K (~300 oC) to the pie, some of the heat is converted into bond energy as water evaporates and chemical reactions occur [2]. But most of the heat energy causes a rise in temperature in the cheese and sauce proportional to the quantity of heat energy supplied. If q is the quantity of heat supplied and the temperature rises from T1 to T2 then

q = C × (T2T1)      (1)

OR

q = C × (ΔT)      (1b)

where the constant of proportionality C is called the heat capacity of the sample. The sign of q in this case is + because the sample has absorbed heat (the change was endothermic), and (ΔT) is defined in the conventional way.

If we add heat to any homogenous sample of matter of variable mass (say the cheese on the pizza), the quantity of heat needed to raise its temperature is proportional to the mass as well as to the rise in temperature. That is,

q = C × m × (T2T1)      (2)

OR

q = C × m × (Δ T)      (2b)

The new proportionality constant C is the heat capacity per unit mass. It is called the specific heat capacity (or sometimes the specific heat), where the word specific means “per unit mass.”

So why does pizza burn the roof of your mouth? The cheese layer on top is well insulated on the layer of pizza crust, so it stays fairly hot. But the crust is at approximately the same temperatue, and it isn't responsible for nearly as severe a "burn". That's because the cheese has a high heat capacity, probably about 3.7 kJ/kg*K [3][4], and the crust has a heat capacity that depends on its moisture content and other factors [5], but it's probably close to 1.7 (similar to wood).

Example 1: How much heat is transferred to the roof of your mouth by a. a teaspoon (15 g) of mozarella when it cools from 140.0 oC to body temperatue (37.0 oC), given that the specific heat capacity of mozarella is 3.7J K–1 g–1? b. the same mass of bread at the same temperature, with heat capacity 1.7 J K–1 g–1?

Solution: a.

q = 3.7 J/goC × 15 g × (140.0 - 37.0)
q = 5.7 kJ.

b.

q = 1.7 J/goC × 15 g × (140.0 - 37.0)
q = 2.6 kJ.

But the bread is also a good insulator, so the heat is transferred to your mouth much more slowly. The inside of a piece of bread remains cool in a toaster, even when the hot filaments burn the surface of the bread!

Note that air at 25oC (77 oF) feels warm, while water at 25oC feels cold. This example shows, in part, why. Note that the mass and specific heat are both important factors. [6] Example 2: a. How much heat is removed from your finger when you immerse it in a test tube containing 12 mL of water at 25oC, and the temperature of the water increases to 26oC? b. How much heat is removed from your finger when you immerse it in a test tube containing 12 mL of air (D = 0.0012 g/mL) at 25oC, and the temperature of the air increases to 26oC?

Solution:

a.

q = 12 mL x 0.997 g/mL x 4.18 J/goC × 1.0 oC
q = 50 J.

b.

q = 12 mL x 0.0012 g/mL x 1.01 J/goC × 15 g × 1.0 oC
q = 0.015 J.

Specific heat capacities provide a convenient way of determining the heat added to, or removed from, material by measuring its mass and temperature change. As mentioned [|previously], James Joule established the connection between heat energy and the intensive property temperature, by measuring the temperature change in water caused by the energy released by a falling mass. In an ideal experiment, a 1.00 kg mass falling 10.0 m would release 98.0 J of energy. If the mass drove a propeller immersed in 0.100 liter (100 g) of water in an insulated container, its temperature would rise by 0.234oC. This allows us to calculate the specific heat capacity of water:

98 J = C × 100 g × 0.234 oC
C = 4.184 J/goC

At 15°C, the precise value for the specific heat of water is 4.184 J K–1 g–1, and at other temperatures it varies from 4.178 to 4.218 J K–1 g–1. Note that the specific heat has units of g (not the base unit kg), and that since the Centigrade and kelvin scales have identical graduations, either oC or K may be used.

Specific heat capacities (25 °C unless otherwise noted)
Substance phase Cp(see below)
J/(g·K)
air, (Sea level, dry, 0 °C) gas 1.0035
argon gas 0.5203
carbon dioxide gas 0.839
helium gas 5.19
hydrogen gas 14.30
methane gas 2.191
neon gas 1.0301
oxygen gas 0.918
water at 100 °C (steam) gas 2.080
water at 100 °C liquid 4.184
ethanol liquid 2.44
water at -10 °C (ice)) solid 2.05
copper solid 0.385
gold solid 0.129
iron solid 0.450
lead solid 0.127

## Electrical Energy Conversion

The most convenient way to supply a known quantity of heat energy to a sample is to use an electrical coil. The heat supplied is the product of the applied potential V, the current I flowing through the coil, and the time t during which the current flows:

q = V × I × t      (2)

If the SI units volt for applied potential, ampere for current, and second time are used, the energy is obtained in joules. This is because the volt is defined as one joule per ampere per second:

1 volt × 1 ampere × 1 second = 1$\begin{matrix}\frac{\text{J}}{\text{A s}}\end{matrix}$ × 1 A × 1 s = 1 J

EXAMPLE 2: An electrical heating coil, 230 cm3 of water, and a thermometer are all placed in a polystyrene coffee cup. A potential difference of 6.23 V is applied to the coil, producing a current of 0.482 A which is allowed to pass for 483 s. If the temperature rises by 1.53 K, find the heat capacity of the contents of the coffee cup. Assume that the polystyrene cup is such a good insulator that no heat energy is lost from it.

Solution The heat energy supplied by the heating coil is given by

q = V × I × t = 6.23 V × 0.482 A × 483 s = 1450 V A s = 1450 J

However,

q = C × (T2T1)

Since the temperatue rises, T2 > T1 and the temperature change ΔT is positive:

1450 J = C × 1.53 K

so that

$\begin{matrix}C=\frac{\text{1450 J}}{\text{1}\text{.53 K}}=\text{948 J K}^{-\text{1}}\end{matrix}$

Note: The heat capacity found applies to the complete contents of the cup-water, coil, and thermometer taken together, not just the water.

As discussed in other sections, an older, non-SI energy unit, the calorie, was defined as the heat energy required to raise the temperature of 1 g H2O from 14.5 to 15.5°C. Thus at 15°C the specific heat capacity of water is 1.00 cal K–1 g–1. This value is accurate to three significant figures between about 4 and 90°C.

If the sample of matter we are heating is a pure substance, then the quantity of heat needed to raise its temperature is proportional to the amount of substance. The heat capacity per unit amount of substance is called the molar heat capacity, symbol Cm. Thus the quantity of heat needed to raise the temperature of an amount of substance n from T1 to T2 is given by

q = C × n × (T2T1)      (4)

The molar heat capacity is usually given a subscript to indicate whether the substance has been heated at constant pressure (Cp)or in a closed container at constant volume (CV).

EXAMPLE 3: A sample of neon gas (0.854 mol) is heated in a closed container by means of an electrical heating coil. A potential of 5.26 V was applied to the coil causing a current of 0.336 A to pass for 30.0 s. The temperature of the gas was found to rise by 4.98 K. Find the molar heat capacity of the neon gas, assuming no heat losses.

Solution The heat supplied by the heating coil is given by

q = V × I × t
= 5.26 V × 0.336 A × 30.0 s
= 53.0 V A s
= 53.0 J

Rearranging Eq. (4), we then have

$\begin{matrix}C_{m}=\frac{q}{n\text{(T}_{\text{2}}-\text{T}_{\text{1}}\text{)}}=\frac{\text{53}\text{.0 J}}{\text{0}\text{.854 mol }\times \text{ 4}\text{.98 K}}=\text{12}\text{.47 J K}^{-\text{1}}\text{ mol}^{-\text{1}}\end{matrix}$

However, since the process occurs at constant volume, we should write

CV = 12.47 J K–1 mol–1

## References

1. Morowitz, H.J. The Thermodynamics of Pizza, Rutgers University Press, New Brunswick, 1991, p. 3.
2. Those energies are explained in Enthalpy of Fusion and Enthalpy of Vaporization and Bond Enthalpies and Exothermic or Endothermic Reactions
3. http://www.springerlink.com/content/6479372833621247/
4. http://www.patentstorm.us/patents/6156356/description.html
5. http://www3.interscience.wiley.com/journal/119439846/abstract
6. Kimbrough, D.R. J. Chem. Educ. 75, 48 (1998)