Question

Determine the ratio of the heat flow into a six-pack of aluminum soda cans to the heat flow into a 2.00 - \(\mathrm{L}\) plastic bottle of soda when both are taken out of the same refrigerator, that is, have the same initial temperature difference with the air in the room. Assume that each soda can has a diameter of \(6.00 \mathrm{~cm}\), a height of \(12.0 \mathrm{~cm}\), and a thickness of \(0.100 \mathrm{~cm}\). Use \(205 \mathrm{~W} /(\mathrm{m} \mathrm{K})\) as the thermal conductivity of aluminum. Assume that the 2.00 - \(\mathrm{L}\) bottle of soda has a diameter of \(10.0 \mathrm{~cm}\), a height of \(25.0 \mathrm{~cm}\), and a thickness of \(0.100 \mathrm{~cm} .\) Use \(0.100 \mathrm{~W} /(\mathrm{mK})\) as the thermal conductivity of plastic.

Short answer

The ratio of heat flow into a six-pack of aluminum soda cans to the heat flow into a 2.00-L plastic bottle of soda is approximately 5.928.

Step-by-step solution

Calculate the heat flow for a single aluminum can

We will use the following formula for heat flow through a cylindrical surface: \(Q = kA\frac{\Delta T}{d} \) Where, \(Q = \) Heat flow \(k = \) Thermal conductivity \(\Delta T = \) Temperature difference \(A = \pi Dh = \pi(0.06\mathrm{m})(0.12\mathrm{m})\) (Surface area of the cylinder) \(d = \) Thickness Given the thermal conductivity of aluminum, \(k = 205 \mathrm{ ~W} /(\mathrm{m} \mathrm{K})\), and the thickness \(d = 0.100 \mathrm{ ~cm} = 0.001\mathrm{m}\), the formula for a single aluminum can becomes: \(Q_\mathrm{aluminum} = 205\pi(0.06)(0.12)\frac{\Delta T}{0.001}\)

Calculate the heat flow for the six-pack of aluminum cans

Since the cans are identical, we can multiply the heat flow from the single aluminum can by the number of cans in the six-pack: \(Q_\mathrm{six\ pack} = 6Q_\mathrm{aluminum} = 6 \times 205\pi(0.06)(0.12)\frac{\Delta T}{0.001}\)

Calculate the heat flow for the 2.00-L plastic bottle

We follow the same procedure as with the aluminum can, but for the plastic bottle: Given the thermal conductivity of plastic, \(k = 0.100 \mathrm{ ~W} /(\mathrm{m} \mathrm{K})\), diameter \(D = 0.10\mathrm{m}\), height \(h = 0.25\mathrm{m}\), and thickness \(d = 0.100 \mathrm{ ~cm} = 0.001\mathrm{m}\), the formula for the plastic bottle becomes: \(Q_\mathrm{plastic} = 0.1\pi (0.10)(0.25)\frac{\Delta T}{0.001}\)

Calculate the ratio of heat flow for the six-pack of aluminum cans to the plastic bottle

Now, we can calculate the ratio of heat flow for the six-pack of aluminum cans to the plastic bottle: \(Ratio = \frac{Q_\mathrm{six\ pack}}{Q_\mathrm{plastic}} = \frac{6 \times 205\pi(0.06)(0.12)\frac{\Delta T}{0.001}}{0.1\pi (0.10)(0.25)\frac{\Delta T}{0.001}}\) Notice that \(\Delta T\) and \(\pi\) cancel out: \(Ratio = \frac{6 \times 205(0.06)(0.12)}{0.1 (0.10)(0.25)}\) Now, plug in the values and calculate the ratio: \(Ratio \approx 5.928\) The ratio of heat flow into a six-pack of aluminum soda cans to the heat flow into a 2.00-L plastic bottle of soda is approximately 5.928.

Key concepts

Thermal Conductivity

Thermal conductivity is a measure of a material's ability to conduct heat. It's an essential property in understanding how heat flows through different objects. When you take an object from the refrigerator, like a soda can or bottle, the material’s thermal conductivity will determine how quickly the object's temperature can equalize with the surrounding air.

Metals, such as aluminum used in soda cans, typically have high thermal conductivities. For example, the thermal conductivity of aluminum is given as 205 W/(m K). This high value implies aluminum can conduct heat rapidly.

• Better conductor of heat means faster heat transfer.
• Leads to a quicker temperature rise towards room temperature.

Plastic, on the other hand, like the material in soda bottles, has a much lower thermal conductivity, around 0.100 W/(m K).

• Poor conductor of heat results in slower heat transfer.
• Takes longer to warm up to room temperature compared to aluminum.

The different thermal conductivities directly affect how quickly these objects heat up when removed from the refrigerator.

Cylinder Surface Area

When calculating heat transfer through an object, the surface area is a crucial factor. For cylindrical objects like soda cans and bottles, the surface area can be calculated using the formula for the lateral surface area of a cylinder: \[ A = \pi Dh \]where \(D\) is the diameter and \(h\) is the height of the cylinder.

The surface area directly influences the rate of heat transfer, as more surface area allows for more heat to flow. For example, each aluminum can's surface area can be calculated using their given dimensions (diameter = 0.06 m, height = 0.12 m), resulting in:

• Aluminum can: \( A = \pi(0.06)(0.12) \)

Similarly, for the plastic soda bottle with a diameter of 0.10 m and a height of 0.25 m:

• Plastic bottle: \( A = \pi(0.10)(0.25) \)

These computed areas are critical in the heat flow formula, as they multiply with the thermal conductivity and the temperature difference to find out the total heat transfer.

Temperature Difference

The temperature difference, denoted as \( \Delta T \), plays a pivotal role in determining the rate of heat flow between two objects. A larger difference typically signifies a greater potential for heat transfer.

When you pull a cold object like a soda from the refrigerator into a warmer room, this temperature difference dictates how fast the heat will flow into the object. The basic heat transfer equation, \[ Q = kA\frac{\Delta T}{d} \], indicates that \( Q \), the heat flow, is directly proportional to \( \Delta T \). Larger temperature difference means:

• Greater the force driving the heat flow.
• Faster temperature change in the object.

However, in the given exercise, the temperature difference \( \Delta T \) remains constant for both objects (the cans and the bottle). Therefore, it simplifies the comparative calculations, since it becomes a common factor that cancels out while calculating the ratio of heat flows.

Refrigeration

Refrigeration involves removing heat from an enclosed space to keep items inside cooler than the surrounding environment. It's the main reason why your soda stays cold when it's in the fridge.

Once removed from the refrigerator, heat will start to move from the warmer environment into the cold soda. The material properties, discussed earlier, such as thermal conductivity and surface area, will dictate how quickly this process occurs. In designing a refrigeration system or selecting materials for containers, engineers consider these factors to optimize performance and energy efficiency.

• Aluminum cans will allow for rapid cooling and warming due to high conductivity.
• Plastic bottles provide slower heat transfer, making them potentially advantageous for stage cooling.

Understanding and managing these dynamics is vital not only in the effective storage of goods but also in their transport and usage over time. By learning how heat transfer principles apply, we can design more effective cooling solutions and choose more suitable materials for packaging based on their thermal characteristics.