Question

Consider an 800-W iron whose base plate is made of \(0.5-\mathrm{cm}\)-thick aluminum alloy $2024-\mathrm{T} 6\left(\rho=2770 \mathrm{~kg} / \mathrm{m}^{3}\right.\(, \)c_{p}=875 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, \alpha=7.3 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}$ ). The base plate has a surface area of \(0.03 \mathrm{~m}^{2}\). Initially, the iron is in thermal equilibrium with the ambient air at \(22^{\circ} \mathrm{C}\). Taking the heat transfer coefficient at the surface of the base plate to be $12 \mathrm{~W} / \mathrm{m}^{2}, \mathrm{~K}$ and assuming 85 percent of the heat generated in the resistance wires is transferred to the plate, determine how long it will take for the plate temperature to reach \(140^{\circ} \mathrm{C}\). Is it realistic to assume the plate temperature to be uniform at all times?

Short answer

Answer: It takes approximately 57.6 seconds for the plate temperature to reach 140°C. The assumption of a uniform temperature is not entirely realistic, but it can be considered a reasonable approximation for this problem, as the rapid heat conduction in metals like aluminum minimizes temperature variations.

Step-by-step solution

Compute the power transmitted to the base plate

Calculate the power (in watts) that is transmitted to the base plate by multiplying the total power of the iron (800 W) by the percentage of heat transferred to the plate (85%). Since 85% of the generated heat is transferred to the base plate, we have: \(P_{transferred} = 0.85 \times 800 = 680 ~\text{W}\)

Calculate the volume and mass of the plate

In order to compute the change in temperature, we need to find the mass of the aluminum plate. To find that, first, we have to find the volume. Multiply the thickness of the plate (0.5 cm) with the surface area of the plate (0.03 m²): \(V = 0.005 \times 0.03 = 1.5 \times 10^{-4}~\text{m}^3\) Now, use the density of the material (\(\rho = 2770~\text{kg}/\text{m}^3\)) to find the mass of the plate: \(m = \rho \times V = 2770 \times 1.5 \times 10^{-4} = 0.4155 ~\text{kg}\)

Calculate the heat absorbed by the plate

To find the absorbed heat over time, multiply the power transmitted to the base plate (\(P_{transferred}\)) with the time (t): \(Q_{absorbed} = P_{transferred} \times t = 680 ~\text{W} \times t\)

Compute the change in temperature for the plate

The heat absorbed will cause the temperature of the plate to change. Using the heat capacity of aluminum (\(c_p = 875~\text{J}/\text{kg} \cdot \text{K}\)), the mass of the plate (m), and the change in temperature (\(\Delta T = T_{final} - T_{initial} = 140-22 = 118~\text{K}\)), we can compute the heat absorbed: \(Q_{absorbed} = m \times c_p \times \Delta T\)

Solve for time t

Now, we can equate the absorbed heat to the heat transfer equation and solve for time, t: \(680 ~\text{W} \times t = 0.4155 ~\text{kg} \times 875 ~\text{J}/\text{kg} \cdot \text{K} \times 118 ~\text{K}\) Divide both sides by 680 W: \(t = \frac{0.4155 \times 875 \times 118}{680} \approx 57.6 ~\text{seconds}\) So, it will take approximately 57.6 seconds for the plate temperature to reach 140°C.

Discuss the uniformity of plate temperature

The assumption of a uniform temperature throughout the plate at all times is not realistic because there will always be some small variations in temperature distribution. However, given the rapid heat conduction in metals, especially aluminum, these variations will be minimal and may not significantly affect the overall behavior of the iron. Overall, the assumption of uniform temperature can be considered a reasonable approximation for this problem.