Question

Determine the ratio of the heat flow into a six-pack of aluminum soda cans to the heat flow into a 2.00 - 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 \(0.100 \mathrm{~W} /(\mathrm{m} \mathrm{K})\) as the thermal conductivity of plastic.

Short answer

Answer: The ratio of heat flow into the six-pack of aluminum soda cans to the heat flow into the 2.00-L plastic bottle of soda is 22.167:1.

Step-by-step solution

1. Calculate the area of the soda can surface

To calculate the area of the soda can surface, we will use the equation for the surface area of a cylinder: \(A_{can} = 2 \pi r_{can} h_{can}\) where \(r_{can}\) is the radius of the soda can and \(h_{can}\) is the height of the soda can. Given the diameter of the soda can of \(6.00 \mathrm{~cm}\), we need to first find the radius, which is half the diameter: \(r_{can} = \dfrac{6.00 \mathrm{~cm}}{2} = 3.00 \mathrm{~cm}\) Now we can calculate the area of the soda can surface: \(A_{can} = 2 * \pi * (3.00 \mathrm{~cm}) * (12.0 \mathrm{~cm}) = 226.19 \mathrm{~cm^2}\) Remember that there are six cans in a six-pack, so the total area for the six cans would be: \(A_{total} = 6 * A_{can} = 6 * 226.19 \mathrm{~cm^2} = 1357.16 \mathrm{~cm^2}\)

2. Calculate the area of the plastic bottle surface

Similarly, we will calculate the area of the plastic bottle surface using the equation for the surface area of a cylinder: \(A_{bottle} = 2 \pi r_{bottle} h_{bottle}\) where \(r_{bottle}\) is the radius of the plastic bottle and \(h_{bottle}\) is the height of the plastic bottle. Given the diameter of the plastic bottle of \(10.0 \mathrm{~cm}\), we need to first find the radius: \(r_{bottle} = \dfrac{10.0 \mathrm{~cm}}{2} = 5.00 \mathrm{~cm}\) Now we can calculate the area of the plastic bottle surface: \(A_{bottle} = 2 * \pi * (5.00 \mathrm{~cm}) * (25.0 \mathrm{~cm}) = 785.4 \mathrm{~cm^2}\)

3. Calculate the heat flow of the soda cans

We will now calculate the heat flow of the soda cans using the formula for heat conduction through a cylindrical wall: \(Q_{can} = \dfrac{k_{Al} * A_{total} * \Delta T}{d_{can}}\) where \(k_{Al}\) is the thermal conductivity of aluminum, \(\Delta T\) is the initial temperature difference between the cans and the air, and \(d_{can}\) is the thickness of the soda can wall.

4. Calculate the heat flow of the plastic bottle

Similarly, we will calculate the heat flow of the plastic bottle using the formula for heat conduction through a cylindrical wall: \(Q_{bottle} = \dfrac{k_{plastic} * A_{bottle} * \Delta T}{d_{bottle}}\) where \(k_{plastic}\) is the thermal conductivity of plastic and \(d_{bottle}\) is the thickness of the plastic bottle wall.

5. Determine the ratio of heat flow

Finally, we can determine the ratio of heat flow from the soda cans to the plastic bottle by dividing their respective heat flow values: \(\dfrac{Q_{can}}{Q_{bottle}} = \dfrac{\dfrac{k_{Al} * A_{total} * \Delta T}{d_{can}}}{\dfrac{k_{plastic} * A_{bottle} * \Delta T}{d_{bottle}}}\) Simplifying the equation, we get: \(\dfrac{Q_{can}}{Q_{bottle}} = \dfrac{k_{Al} * A_{total} * d_{bottle}}{k_{plastic} * A_{bottle} * d_{can}}\) Plug in the given values for \(k_{Al}, k_{plastic}, d_{can}, d_{bottle}, A_{total}\), and \(A_{bottle}\): \(\dfrac{Q_{can}}{Q_{bottle}} = \dfrac{(205 \mathrm{~W/mK})* (1357.16 \mathrm{~cm^2}) * (0.100 \mathrm{~cm})}{(0.100 \mathrm{~W/mK}) * (785.4 \mathrm{~cm^2}) * (0.100 \mathrm{~cm})}\) Converting cm² to m², the areas become: \(A_{total} = 0.135716 \mathrm{~m^2}\) and \(A_{bottle} = 0.07854 \mathrm{~m^2}\) And converting the thicknesses from cm to m, they become: \(d_{can} = 0.001 \mathrm{~m}\) and \(d_{bottle} = 0.001 \mathrm{~m}\) Now, substitute the values again and compute the ratio: \(\dfrac{Q_{can}}{Q_{bottle}} = \dfrac{(205 \mathrm{~W/mK})* (0.135716 \mathrm{~m^2}) * (0.001 \mathrm{~m})}{(0.100 \mathrm{~W/mK}) * (0.07854 \mathrm{~m^2}) * (0.001 \mathrm{~m})} = 22.167\) So, the ratio of heat flow into the six-pack of aluminum soda cans to the heat flow into the 2.00-L plastic bottle of soda is 22.167:1.