Which pan bakes more efficiently, glass or metal?
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- 418907
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Chemistry concepts covered:
VI. Energy and Thermodynamics: Energy is the key currency of chemical reactions in molecular-scale systems as well as macroscopic systems.
C. 1. Heat exchange is measured via temperature change.
C. 1. b. Heat flow is quantitatively obtained from ∆T via molar heat capacity or specific heat and the mass of the substance involved.
C. 1. d. d. When observed in the laboratory, exothermic processes will show warming (heat evolved), while endothermic processes will show cooling (heat absorbed)
Question: Which pan bakes more efficiently, glass or metal?
Introduction:
When baking brownies, you may have noticed that using a different pan requires a different amount of baking time, as illustrated in the baking instructions for Ghirardelli brownies (1). If one looks closely, the bake time for glass pans is longer than metals, regardless of pan size. Bakers will often adjust their types of pans depending on their needs, and more advanced bakers will keep a variety of pans available for their use (2). The question of why baking time differs for glass versus metal pans is rooted in the material’s chemical properties. What are these properties, and do they make one type of pan more effective than the other? To answer these questions, we will investigate the branch of thermodynamics, and more specifically thermal conductivity, specific heat, heat capacity, and heat transfer (3).
Figure \(\PageIndex1\). A metal pan containing brownies. (Wikimedia; Wulfratt, J) (4)
Thermal Conductivity:
Thermal conductivity is how quickly an element will absorb energy and release energy. It is measured in Watts/meter*Kelvin. When a pan is placed in an oven, it starts to absorb energy in the form of heat from the air around it. As the pan heats up, the internal kinetic energy of the pan starts to increase. Pans with a higher thermal conductivity heat up faster. Once it has absorbed this heat, it will transfer it to areas around it that are lower in temperature - other parts of the pan, as well as the food it is cooking. Certain areas of the pan might heat up quicker than other areas, called heat spots. Heat spots are not desired when baking, since it causes certain parts of the dessert to cook more than others. If you want brownies that are baked all the way through, you want a pan that is uniform in heat throughout, and no heat spots. You would also want a pan to have a high thermal conductivity, so that it heats up quickly (5).
| Material of Pan | Thermal Conductivity |
| Glass Pyrex | 1.1 W/mK |
| Copper | 401 W/mK |
| Tin | 67 W/mK |
| Aluminum | 237 W/mK |
| Stainless Steel | 16 W/mK |
Figure \(\PageIndex2\). Source: Engineering Toolbox (6, 7).
Specific Heat and Heat Capacity
When looking at the difference in baking times for glass versus metal pans, specific heat, or the amount of energy needed to raise the temperature of a 1 gram of a substance by 1 degree celsius, must be considered. Specific heat is independent of the quantity of a substance, and therefore is an intensive property. On the contrary, heat capacity is an extensive property, meaning it is dependent on the amount of a substance. Thus, 100 grams of a substance will have double the heat capacity as 50 grams of that same substance. Heat capacity refers to the amount of energy needed to raise the temperature of a substance by one degree celsius. While specific heat will be the same for the same substance or material, heat capacity will be the same for the same substance/material of the same mass.
To find the pan that bakes the most efficiently, we must look at wchich has the lower specific heat, as a lower specific heat means less energy is needed to raise the temperature of the pan by 1 degree celsius. Less energy needed to raise the temperature means the pan will heat up faster. With the specific heat of the material, we can find the heat capacity of the pan by multiplying the mass of the pan by the specific heat of the pan.
With this knowledge of specific heat, we can look at the specific heat of different pan materials, which can be used in our calculations to determine how much energy is needed to heat up the pan.
| Material of Pan | Specific Heat (J/kg C°) |
| Glass Pyrex | 753 |
| Copper | 385 |
| Tin | 228 |
| Aluminum | 897 |
| Stainless Steel | 468 |
Figure \(\PageIndex3\). Source: Engineering Toolbox (6, 8).
From looking at these specific heats, put in the units J/kg C°, we can see that it takes a lot more energy to raise 1 kg of aluminum than 1 kg of tin. However, to determine which pan heats up the fastest, we also must take into consideration the mass of the pan.
How to Determine Heat Transfer:
We just learned that specific heat is the amount of energy needed to raise the temperature of 1 gram of a substance by one degree Celsius. We can use this value, symbolized by C, in the equation for heat transfer to find out how many joules are needed to raise the temperature of the pan. The less joules needed, the faster the pan will heat up, making bake time more efficient. The heat transfer equation is q=mCΔt, where q = heat added, m = mass, C = the specific heat capacity constant which is different for each pan material, and ΔT is the change in temperature of the material. It is important to note that all of these reactions will have a positive q value because baking is endothermic; the pan absorbs the heat from it’s surroundings in order ot heat up and transfer to the dessert being baked.
-q surrounding = q system
Imagine you have two pans. One pan is made of glass pyrex and weighs 1.00 kilogram. The other pan is made of 70% stainless steel and 30% copper, and also weighs 1.00 kilogram. We can find how much heat is needed to raise the temperature of each pan by 1 degree celsius using our heat transfer equation.
Solution
\[ q = mc ΔT \]
Glass pyrex: q = m x c x ΔT
\[ q = (1 kg)(753 J/kg℃)(1℃) \]
\[ q = 753 J \]
Copper/stainless steel: q = mcΔT
\[ q = (1 kilogram)(.70)(385 J/kg℃)(1℃) \]
\[ q = 269.5 J \]
\[ q = (1 kilogram)(.30)(468 J/kg℃)(1℃) \]
\[ q = 140.4 J \]
\[ 269.5 J + 140.4 J = 409.9 J \]
\[ \dfrac{753 J} {409.9 J} = 1.83 J \]
It takes 1.84 times as much energy to heat the glass than it does to heat the copper and stainless steel pan. Additionally, copper has a greater thermal conductivity than stainless steel, and so contributes to an evenly heated pan. The thermal conductivity of glass pyrex is way lower, further showing that a stainless steel pan would be more efficient at baking brownies.
To find the amount of heat necessary to raise the temperature of one of our pans to 350º F, the typical temperature used to bake brownies at, we can use the equation q = mc ΔT, as well. For example, if we set the temperature of the oven to 350º F, the amount of energy required to increase a stainless steel pan, say one that weighs 1.75 pounds or 0.794 kg, can be found by plugging in stainless steel’s specific heat into the equation. To use this equation, we must first convert the change in temperature (room temperature of 70ºF to 350º F) to Celsius, as units need to be consistent throughout in order to cancel and be left with only joules.
Solution
\[ \dfrac{ºF - 32} {1.8} = ºC \]
\[ T1 = \dfrac{70ºF - 32} {1.8} = 21.1 ºC \]
\[ T2 = \dfrac{350ºF - 32} {1.8} = 176.7 ºC \]
\[ q = (0.794 kg)(468 J/kg C°)(176.7ºC-21.1ºC) \]
\[ q = 57819.7152 J = 57800 J \]
Thus, it takes 57800 J to raise the temperature of a 1.75 pound stainless steel pan from room temperature to 350ºF, the temperature at which we bake brownies.
If we were to look at the value of q for a glass pyrex pan, with a specific heat of 753 J/kg C°, of the same weight and the same ΔT, then this value would be 93030 J.
\[ q = (0.794 kg)(753 J/kg C°)(176.7ºC-21.1ºC) \]
\[ q = 93030 J \]
This value is 1.61 times bigger than the energy required to heat up the stainless steel pan by the same amount. Thus, the stainless steel pan heats up a lot faster, and therefore the brownies require less time in the oven because the pan gets hotter faster
Determining Heat Added:
If you wanted to determine how much heat was required to raise the temperature of a material, you could further use the q = mcΔT equation. As long as you know the value of the specific heat capacity of the material, you can substitute the other values in to determine the heat required.
How much heat is needed to raise the temperature of 200 mL of water from 30 degrees celsius to 50 degrees celsius? The specific heat capacity of water is 4.184 J/g
Solution
\[ q= m c ΔT \]
q= (200 grams)(4.184 J/g)(50 - 30 degrees)
q= 16,736 Joules, or 16.736 kJ
Conclusion:
When looking at the different materials of baking tools, it’s important to know what makes a good pan a good pan. From our investigation we have found that thermal conductivity and heat capacity contribute to how fast a pan can heat up and how effectively it can transfer that heat to the dessert being baked. Glass has a low thermal conductivity and a high specific heat. When compared to a metal pan, glass is less efficient at baking brownies due to its longer time to heat up and transfer energy.
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