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.

Step by step solution

01

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: \(Q_\mathrm{aluminum} = 205\pi(0.06)(0.12)\frac{\Delta T}{0.001}\)

02

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: \(Q_\mathrm{six\ pack} = 6Q_\mathrm{aluminum} = 6 \times 205\pi(0.06)(0.12)\frac{\Delta T}{0.001}\)

03

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: \(Q_\mathrm{plastic} = 0.1\pi (0.10)(0.25)\frac{\Delta T}{0.001}\)

04

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.

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