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An aluminum pan whose thermal conductivity is $237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\( has a flat bottom with diameter \)15 \mathrm{~cm}$ and thickness \(0.4 \mathrm{~cm}\). Heat is transferred steadily to boiling water in the pan through its bottom at a rate of \(800 \mathrm{~W}\). If the inner surface of the bottom of the pan is at \(105^{\circ} \mathrm{C}\), determine the temperature of the outer surface of the bottom of the pan.

Short Answer

Expert verified
Answer: The temperature of the outer surface of the bottom of the pan is approximately 80.7°C.

Step by step solution

01

Identify the conduction heat transfer formula

In this problem, we will be dealing with heat transfer due to conduction. The general formula for heat transfer through conduction in a solid is given by Fourier's law of conduction: \(q = kA \dfrac{\Delta T}{l}\), where \(q\) is the heat transfer rate (W), \(k\) is the thermal conductivity (W/m⋅K), \(A\) is the surface area (m²), \(\Delta T\) is the temperature difference between the two surfaces (K), \(l\) is the thickness (m).
02

Convert the given dimensions

First, we need to convert the given dimensions to the SI units: Diameter of the pan, \(D = 15 \mathrm{~cm} = 0.15\ \mathrm{m}\), Thickness of the pan, \(L = 0.4 \mathrm{~cm} = 0.004\ \mathrm{m}\).
03

Calculate the surface area

The surface area of the bottom of the pan is given by the formula for the area of a circle. \(A = \pi (\dfrac{D}{2})^2\)
04

Calculate the temperature difference

Now we can solve the Fourier's law formula for the temperature difference between the two surfaces. \(\Delta T = \dfrac{ql}{kA}\)
05

Determine the temperature of the outer surface

We are given the temperature of the inner surface and we need to determine the temperature of the outer surface. We can use the calculated temperature difference in step 4 to find the temperature of the outer surface: \(T_{outer} = T_{inner} - \Delta T\) Note that subtraction is used since heat is transferred from the inner surface to the outer surface. Now, we can plug in all the given values and known constants to calculate the temperature of the outer surface of the bottom of the pan.

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Most popular questions from this chapter

A flat-plate solar collector is used to heat water by having water flow through tubes attached at the back of the thin solar absorber plate. The absorber plate has a surface area of \(2 \mathrm{~m}^{2}\) with emissivity and absorptivity of \(0.9\). The surface temperature of the absorber is $35^{\circ} \mathrm{C}\(, and solar radiation is incident on the absorber at \)500 \mathrm{~W} / \mathrm{m}^{2}\( with a surrounding temperature of \)0^{\circ} \mathrm{C}$. The convection heat transfer coefficient at the absorber surface is \(5 \mathrm{~W} / \mathrm{m}^{2}\). \(\mathrm{K}\), while the ambient temperature is \(25^{\circ} \mathrm{C}\). Net heat absorbed by the solar collector heats the water from an inlet temperature $\left(T_{\text {in }}\right)\( to an outlet temperature \)\left(T_{\text {out }}\right)$. If the water flow rate is \(5 \mathrm{~g} / \mathrm{s}\) with a specific heat of $4.2 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}$, determine the temperature rise of the water.

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How does forced convection differ from natural convection?

In the metal processing industry, heat treatment of metals is commonly done using electrically heated draw batch furnaces. Consider a furnace that is situated in a room with surrounding air temperature of \(30^{\circ} \mathrm{C}\) and an average convection heat transfer coefficient of $12 \mathrm{~W} / \mathrm{m}^{2}\(. \)\mathrm{K}$. The furnace front is made of a steel plate with thickness of \(20 \mathrm{~mm}\) and a thermal conductivity of $25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$. The outer furnace front surface has an emissivity of \(0.23\), and the inside surface is subjected to a heat flux of $8 \mathrm{~kW} / \mathrm{m}^{2}$. Determine the outside surface temperature of the furnace front.

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