Question

An aluminum soda can \(150 \mathrm{~mm}\) in length and \(60 \mathrm{~mm}\) in diameter is placed horizontally inside a refrigerator compartment that maintains a temperature of \(4^{\circ} \mathrm{C}\). If the surface temperature of the can is \(36^{\circ} \mathrm{C}\), estimate the heat transfer rate from the can. Neglect the heat transfer from the ends of the can.

Short answer

Neglect heat transfer from the ends of the can. Answer: The heat transfer rate from the aluminum soda can in the refrigerator is approximately 143.9 W.

Step-by-step solution

Identify the given values

We are provided with the following information: - Length of the soda can (L): \(150 \mathrm{~mm}\) - Diameter of the soda can (D): \(60 \mathrm{~mm}\) - Temperature of the refrigerator (T1): \(4^{\circ} \mathrm{C}\) - Surface temperature of the can (T2): \(36^{\circ} \mathrm{C}\) - Thermal conductivity of aluminum (k): approximately \(237 \mathrm{\frac{W}{mK}}\)

Calculate the temperature difference

The temperature difference (ΔT) can be calculated as ΔT = T2 - T1 ΔT = \(36^{\circ} \mathrm{C}\) - \(4^{\circ} \mathrm{C}\) = \(32^{\circ} \mathrm{C}\) Note that in this exercise, we don't need to convert the temperature to Kelvin since the difference between two temperatures remains the same.

Calculate the surface area of the can's side

We are asked to neglect the heat transfer from the ends of the can, so we only need to calculate the surface area of the side (cylindrical part) of the soda can. We can do that using: A = π x D x L First, we need to convert the diameter and length given in millimeters to meters: D = 0.06 m L = 0.15 m Now we calculate the surface area: A = π x (0.06 m) x (0.15 m) ≈ 0.0283 \(\mathrm{m^2}\)

Calculate the heat transfer rate

Now, we will use the formula for heat transfer by conduction: Q = kA(ΔT)/L Here, Q is the heat transfer rate, k is the thermal conductivity of aluminum, A is the surface area of the can's side, ΔT is the temperature difference, and L is the length of the can. Plugging in the values: Q = (\(237 \mathrm{\frac{W}{mK}}\)) x (0.0283 \(\mathrm{m^2}\)) x (\((32^{\circ} \mathrm{C})\)) / (0.15 m) ≈ 143.9 W Therefore, the heat transfer rate from the can is approximately 143.9 W.