Thermodynamics of Insulating Water Bottles in Everday Life

Note

This Exemplar will teach the following concepts from the ACS Examinations Institute General Chemistry ACCM:

VI. A. 2. B. In thermodynamic treatments of chemical systems, the definition of the system of interest versus the surroundings is important. 

VI. C. 1. B. Heat flow is quantitatively obtained from \(\Delta T \) via molar heat capacity or specific heat and the mass of the substance involved. 

VI. F. In accord with thermodynamics, energy is conserved in chemical changes, but the change of form in which the energy is present may be harnessed via natural or human-made devices. 

In addition to decreasing the usage of plastic water bottles, reusable water bottles are often far superior in keeping a beverage hot or cold. We will be focusing particularly on insulating water bottles, which can maintain the temperature of a beverage for long periods. We can apply chemistry concepts, including heat transfer and the zeroth law of thermodynamics, to gain an understanding of how these bottles are so effective. 

These concepts fall under the broader topic of thermodynamics, which is the branch of science that studies the relationship between heat, work, temperature, and other forms of energy. Thermodynamics details how heat, a form of energy, is transferred between objects.¹ Oftentimes, heat is said to be moving from a “system” to its “surroundings”. 

Heat can be defined as a transfer of energy between objects due to a difference in temperature.² The relationship between heat, mass, specific heat, and change in temperature can be expressed with the equation below.

 

\[ q=m c \Delta T \]

 

The mass of an object is measured in grams, while the units of specific heat are \({~J} / {g}^{\circ} {C} \). Every substance has a unique specific heat, which can be defined as the energy required to raise the temperature of one gram of a substance by one degree Celsius.³ Thus, the higher the specific heat value, the more energy is required to raise the temperature of a substance. In other words, the higher the specific heat value, the more difficult it is to change a substance’s temperature. Two common materials used in water bottles are plastic and stainless steel, specifically stainless steel (304) and PET plastic. Comparing the specific heat values, stainless steel (304) has a specific heat of 0.5 \({~J} / {g}^{\circ} {C} \).⁴ PET plastic has a specific heat of 1.05 \({~J} / {g}^{\circ} {C} \).⁵ These values give us insight into the properties of each substance. Plastic, for example, is an insulator, which makes sense when looking at its specific heat value; it is more difficult to change the temperature of plastic. The lower specific heat value of stainless steel hints at its conductive properties; it takes less heat to change the temperature of the metal, and thus it is good at transferring heat. 

The first law of thermodynamics states that energy cannot be created nor destroyed, but it can be transferred between forms.⁶ For example, if we have hot coffee inside a water bottle, we could consider the beverage to be the system and the walls of the water bottle to be the surroundings. Following the first law of thermodynamics, we can express the relationship between q_system and q_surroundings as follows. Any heat that is lost or gained by the system must be subsequently gained or lost by the surroundings. 

 

\begin{equation}
-q_{\text {surroundings }}=q_{\text {system }}
\end{equation}

 

This relationship allows us to evaluate the performance of water bottles constructed from different materials. Material is an important consideration when designing a water bottle; we would likely want a material that is an insulator to keep hot drinks warm and cold drinks cool. 

 

Let’s evaluate two common materials used in water bottles, stainless steel (304) and PET Plastic.

Exercise \(\PageIndex{1}\)

Assume we have a 17 oz water bottle made of stainless steel with a mass of 0.7 lbs (specific heat of stainless steel (304) = 0.5 \({~J} / {g}^{\circ} {C} \)). We completely fill the bottle with water at a temperature of 5 \(^{\circ} {C} \) (specific heat of water = 4.2 \({~J} / {g}^{\circ} {C} \)). Assuming the water bottle is left outside on an extremely warm day, reaching a temperature of 35 \(^{\circ} {C} \), calculate the final temperature of the water.

Note

Density of water = 1.00 g/mL

1 lb = 453.592 g

1 oz = 29.5735 mL

Answer

Using the equation that solves for heat (q), we can set \(-q_{\text {surroundings }}=q_{\text {system }}\), then plug the necessary values into the equations after converting them to the correct units. After, we can solve for the missing final temperature value. Recall, we will consider the beverage to be the system and the walls of the water bottle to be the surroundings.

\begin{gathered}
17 \mathrm{~oz} \times \frac{29.574 \mathrm{~mL}}{1 \mathrm{~oz}}=502.758 \mathrm{~mL} \\
\\
\frac{1.00 \mathrm{~g}}{\mathrm{~mL}} \times 502.758 \mathrm{~mL}=502.758 \mathrm{~g} \\
\\
0.7 \mathrm{~lbs} \times \frac{453.592 \mathrm{~g}}{1 \mathrm{~lbs}}=317.515 \mathrm{~g} \\
\\
-(317.515 \mathrm{~g})\left(0.5 \mathrm{~J} / \mathrm{g}^{\circ} \mathrm{C}\right)\left(x-35^{\circ} \mathrm{C}\right)=(502.758 \mathrm{~g})\left(4.2 \mathrm{~J} / \mathrm{g}^{\circ} \mathrm{C}\right)\left(x-5^{\circ} \mathrm{C}\right) \\
\\
-158.7575 x+5,556.5125=2,111.5836 x-10,557.918 \\
\\
16,114 .4305=2,270.3411 x \\
\\
x=7.10^{\circ} \mathrm{C}
\end{gathered}

Exercise \(\PageIndex{2}\)

Instead, assume we have a 17 oz plastic water bottle with a weight of 24 g (specific heat of PET plastic = 1.05 \({~J} / {g}^{\circ} {C} \)) at 35 \(^{\circ} {C} \). Assuming all other conditions are the same as outlined in Exercise #1, calculate the final temperature of the water.

Answer

Repeat the steps outlined in Exercise #1 with the new values.

\begin{aligned}
-(24 \mathrm{~g})\left(1.05 \mathrm{~J} / \mathrm{g~}^{\circ} \mathrm{C}\right)\left(x-35^{\circ} \mathrm{C}\right) & =(502.758 \mathrm{~g})\left(4.2 \mathrm{~J} / \mathrm{g~}{^\circ \mathrm{C}}\right)\left(\mathrm{x}-5^{\circ} \mathrm{C}\right) \\
\\
-25.2 x+882 & =2,111.5836 x-10,557.918 \\
\\
11,439.918 & =2,136.7836 x \\
\\
x & =5.35^{\circ} \mathrm{C}
\end{aligned}

Exercise \(\PageIndex{3}\)

Compare the materials used in Exercise #1 and Exercise #2 in terms of the temperatures calculated.

Answer

Based on the calculations, plastic appears to be a better material to use because the temperature of the water increases less compared to in the stainless-steel water bottle. Without actually performing the calculations, this can be deduced by simply comparing the specific heat values; the value for the plastic is higher than that of the stainless steel. This is seemingly contradictory because most insulating water bottles are made of stainless steel. 

Design:

Let’s consider the zeroth law of thermodynamics. This law states that when two objects are in thermal contact with one another, energy in the form of heat will be transferred between the two objects until thermal equilibrium is reached. This equilibrium is reached when both objects are at the same temperature.⁷ 

Though material may seem like the most important factor in constructing a water bottle, the science of insulating water bottles is more dependent on design. The calculation performed above was for a water bottle with a single wall, not for a true insulating water bottle. Regarding the design of an insulating water bottle, there are two stainless-steel walls separated by a vacuum, or a space absent of any atoms or molecules.⁸ As a result, heat cannot move between the walls. 

Figure \(\PageIndex{1}\): From this cut-open view of an insulating water bottle, you can see the way in which design contributes to how these bottles work.

As dictated by the zeroth law of thermodynamics, two objects must be in thermal contact for heat to be transferred such that thermal equilibrium is reached. Thus, since the two walls of the water bottle are not in contact, heat cannot be transferred between the inner wall and the outer wall. The liquid is only in contact with the inner wall. The outer wall is exposed to air molecules. In the exercises above, we made the assumption that no heat was lost to the air, which was reasonable since we wanted to focus on evaluating the interaction between the liquid and the material of the water bottle based on the specific heat value. A similar assumption is made in calorimetry. Of course, in reality, heat can dissipate into the air or be transferred to the outer wall of the water bottle, depending on the temperature of the outer wall and the beverage. Recall, heat will be transferred such that thermal equilibrium is achieved. In an insulating water bottle, the liquid is only in contact with the inner wall, which will not interact with the air. Yes, as dictated by thermodynamics, heat will likely be transferred between the liquid and the inner wall depending upon each respective temperature; however, the change will be insignificant to the transfer of heat that would occur if the inner wall was in contact with the air. 

These insulating water bottles can keep beverages roughly the same temperature for far longer periods of time compared to the standard water bottle. Eventually, the temperature of the liquid will change. No insulating water bottle is constructed with a perfect vacuum, thus eventually heat will be transferred to or be gained from the air, affecting the temperature of the beverage. Furthermore, the use of stainless steel in the construction of these water bottles is more out of concern for sustainability, safety, and durability. Unlike plastic water bottles, with stainless steel, chemicals such as BPA will not leach into the liquid. Furthermore, stainless-steel water bottles are reusable, preventing the build up of plastics, and extremely durable, allowing your water bottle to take a beating over the long run.⁹ 

References:

1. Drake, G. W. F. Thermodynamics. https://www.britannica.com/science/thermodynamics (accessed 2022-11-10)

2. Zumdahl, S. S.; DeCoste, D. J. Chemical Principles; Cengage Learning, 2017. 

3. Belford, R.; Kattoum, R. Specific Heat Capacity. https://chem.libretexts.org/Courses/...%20its%20phase (accessed 2022-10-24).

4. Matweb. https://www.matweb.com/search/datasheet_print.aspx?matguid=abc4415b0f8b490387e3c922237098da (accessed 2022-10-24).

5. Frankland, J. The Importance of Specific Heat in Screw and Extruder Design. https://www.ptonline.com/articles/the-importance-of-specific-heat-in-screw-and-extruder-design (accessed 2022-10-24).

6. Zumdahl, S. S.; DeCoste, D. J. Chemical Principles; Cengage Learning, 2017. 

7. OpenStax. https://openstax.org/books/physics/pages/12-1-zeroth-law-of-thermodynamics-thermal-equilibrium (accessed 2022-10-24).

8. Flobottle. https://flobottle.com/blogs/news/how-do-insulated-bottles-work#:~:text=An%20insulated%20bottle%20has%20two,transfer%20of%20heat%20via%20convection. (accessed 2022-12-5).

9. Greens Steel. https://greenssteel.com/blogs/news/s...ottle-benefits (accessed 2022-11-10). 

 Figure 1. Wikimedia. https://commons.wikimedia.org/wiki/F...ewar_Flask.svg (accessed 2022-12-6)