# The Cooking Efficiency of Pots and Pans


Kitchen equipment stores sell two kinds of pots, one kind (like Revereware) constructed with a layer of copper on the bottom of stainless steel pot, and another which is solid aluminum, or in the case of "Dutch Ovens", solid iron. Even Pyrex saucepans are available.[1] People often have strong preferences for one or the other. Is there a scientifically supported difference? Similarly, some recipes give different cooking times for brownies, depending on whether a glass or metal pan is used to bake them. Is one better than the other? Is there a difference in their cooking efficiencies[2]?

## Heat Capacity of Cooking Vessels

When we supply heat energy from the stove burner to a pan, a rise in temperature occurs which is proportional to the quantity of heat energy supplied. (We'll assume at first that the pan is empty, and that it isn't heated enough to melt it!). If q is the quantity of heat supplied and the temperature rises from T1 to T2 then

$\text{q} = \text{C} × \text{(T}_2 – \text{T}_1)$

or

$\text{q} = \text{C} × (\Delta \text{T})$

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, such as a pure substance or a solution, the quantity of heat needed to raise its temperature is proportional to the mass as well as to the rise in temperature. That is,

$\text{q} = \text{C} × \text{m} × (\text{T}_2–\text{T}_1)$

or

$\text{q} = \text{C} × \text{m} × (\Delta \text{T})$

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.”

Example $$\PageIndex{0}$$

Suppose we have two pots, each weighing 1.00 kg, but one is aluminum and the other is 75% stainless steel with a 25% copper clad bottom. We can calculate the amount of heat required to raise the temperature of each pot by 1oC:

Al:

$\text{q} = \text{C} × \text{m} × (\Delta \text{T}) = \text{0.903} \dfrac{J}{g^\circ C} x \text{1000 g} x \text{1}^\circ C = \text{903 J}$

Cu/Stainless (250 g Cu and 750 g stainless steel):

$\text{q} = \text{C} × \text{m} ×(\Delta \text{T})= \text{0.456} \dfrac{J}{g ^\circ C} × \text{750 g} × \text{1}^\circ C = \text{342 J}$

$\text{q} = \text{C} × \text{m} × (\Delta \text{T}) = \text{0.385} \dfrac{J}{g ^\circ C} × \text{250 g} × \text{1}^\circ C = \text{96.3 J}$

Total = 439 J

So it takes 903 J / 439 J or 2.06 times as much energy and time to heat the Aluminum pot than it does the copper clad pot[3].

But the main reason for using the copper clad pot is that copper conducts heat much better than other common metals, so even if the bottom of the pot is heated unevenly (by flames or a burner), the heat is dispersed evenly and efficiently to the contents of the pan. The relative thermal conductivities of Al:Cu:SiO2:Stainless = 200:333:25:1. We also see that Pyrex saucepans are a poor choice, because they conduct heat so poorly, and have a relatively high heat capacity (0.737). This is an advantage for oven cooking (see below).

## Determining Amount of Heat

It is impossible to measure heat directly, because heat is defined as energy that is transferred between bodies of different temperatures. How do we measure the heat output of an oven, or assign energy values see below to foods? "Caloric" food energy values are measures of the heat they produce when burned in our body or in a "calorimeter" designed for that purpose.

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:

$\text{98 J} = \text{C} × \text{100 g} × \text{0.234} ^\circ C$

$\text{C} = \text{4.184} \dfrac{J}{g ^\circ C }$

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 4.219 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.

Example $$\PageIndex{1}$$: Heat

How much heat is required to raise the temperature of 500 mL of water (D = 1.0) from 25.0 oC to 75.0 oC, given that the specific heat capacity of water is 4.184 J K–1 g–1?

Solution:

$\text{q} = \text{4.18} \dfrac{J}{g ^\circ C} × \text{500 g} × \text{(75.0 - 25.0)}$

$\text{q} = \text{104,500 J} or \text{104 kJ}$

## Baking Casseroles

To compare pans for oven baking, assume that an aluminum pan weighs 500 g and a glass dish weighs 2000 g. The aluminum pan would absorb 542 J (calculated as above), but the glass (SiO2) dish would absorb

$\text{q} = \text{0.737} \dfrac{J}{g ^\circ C} \times \text{2000 g} \times \text{1}^\circ C= \text{1474 J}$

The glass pan requires 3.27 more time or energy to heat up than the aluminum pan[4]. The most important reason for using glass casseroles is that they prevent the liquid from boiling off of the contents too fast because of their low thermal conductivity and high specific heat. Heat is supplied very gradually to the contents, which are maintained at 100 °C as the water boils by the latent heat of vaporization of water, and a very even temperature is maintained because of the high heat capacity, in spite of oven temperature variations.[5]

Table $$\PageIndex{1}$$: 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 T=[6] liquid 0.01°C 4.210
15°C 4.184
25°C 4.181
35°C 4.178
45°C 4.181
55°C 4.183
65°C 4.188
75°C 4.194
85°C 4.283
100°C 4.219
water (ice) at T= [7] solid 0°C 2.050
-10°C 2.0
-20°C 1.943
-40°C 1.818
ethanol liquid 2.44
copper solid 0.385
gold solid 0.129
iron solid 0.450
stainless steel solid 0.456
aluminum solid 0.903
Glass (SiO2 solid 0.737

## Electrical Energy Conversion

There is a way to calculate the quantity of heat energy transferred by the electrical coil burner of a stove (or calibrated "calorimeter"). 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:

$q = V * I * t$

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:

$\text{1 volt} × \text{1 ampere} × \text{1 second} = \text{1} \dfrac{J}{A s} × \text{1 A} × \text{1 s} = \text{1 J}$

Example $$\PageIndex{2}$$ : Heat Capacity

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

$\text{q} = \text{V} × \text{I} × \text{t} = \text{6.23 V} × \text{0.482 A } × \text{483 s} = \text{1450 V A s} = \text{1450 J}$

However,

$\text{q} = \text{C} × \text{(T}_2 – \text{T}_1)$

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

$\text{1450 J} = \text{C} × \text{1.53 K}$

so that

$\text{C} = \dfrac{1450 J}{1.53 K} = \text{948} \dfrac{J}{K}$

Note

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

## The Food Calorie

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.

The "dietary Calorie" (with a capital "C") is actually 1000 calories. So with a 3000 Calorie daily diet, we are burning 3 x 106 calories, or 1.26 x 107 J, or 1.25 x 104 kJ of energy.

## Molar Heat Capacities

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

$\text{q} = \text{C} × \text{n} × (\text{T}_2 – \text{T}_1)\label{6}$

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 $$\PageIndex{3}$$ Molar Heat Capacity

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

$$\text{q} = \text{V} × \text{I} × \text{t}$$
= $$\text{5.26 V} × \text{0.336 A} × \text{30.0 s}$$
= $$\text{53.0 V A s}$$
= $$\text{53.0 J}$$

Rearranging Eq. $$\ref{6}$$, we then have

$$\text(C)_m = \dfrac{q}{n(T_2-T_1)} = \dfrac{53.0 J}{0.854 mol * 4.98 K} = \text{12.47} \dfrac{J}{mol*k}$$

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

$$\text{C}_V = \text{12.47} \dfrac{J}{mol*K}$$

From ChemPRIME: 15.1: Heat Capacities

## References

1. Barham, P. "The Science of Cooking". Springer, Berlin, 2001, p.57-8
2. Selco, J.I. Cooking Efficiencies of Pots and Pans", J. Chem. Educ., 71, 1994, p. 1046
3. Selco, J.I. Cooking Efficiencies of Pots and Pans", J. Chem. Educ., 71, 1994, p. 1046
4. Selco, J.I. Cooking Efficiencies of Pots and Pans", J. Chem. Educ., 71, 1994, p. 1046
5. Barham, P. "The Science of Cooking". Springer, Berlin, 2001, p.57-8
6. http://www.engineeringtoolbox.com/water-thermal-properties-d_162.html
7. http://www.engineeringtoolbox.com/ice-thermal-properties-d_576.html

## Contributors and Attributions

This page titled The Cooking Efficiency of Pots and Pans is shared under a not declared license and was authored, remixed, and/or curated by Ed Vitz, John W. Moore, Justin Shorb, Xavier Prat-Resina, Tim Wendorff, & Adam Hahn.