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 \(1400 \mathrm{~W}\). 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 102.27°C.

Step by step solution

01

Identify the formula for one-dimensional steady-state heat conduction

The formula we will use for this problem is given by Fourier's Law of Heat Conduction in the one-dimensional steady-state case: \(Q = kA\dfrac{T_1 - T_2}{L}\), where \(Q\) is the heat transfer rate, \(k\) is the thermal conductivity of the material, \(A\) is the area of the surface through which heat is being conducted, \(T_1\) and \(T_2\) are the temperatures of the two surfaces, and \(L\) is the thickness of the material.
02

Calculate the area of the bottom of the pan

To find the area of the bottom of the pan, we use the formula for the area of a circle: \(A = \pi r^2\), where \(r\) is the radius of the circle. Given the diameter of the bottom of the pan is \(15\,\text{cm}\), the radius is half of that: \(r = \dfrac{15}{2}\,\text{cm} = 7.5\,\text{cm}\). So, the area is: \(A = \pi (7.5)^2 \approx 176.71\,\text{cm}^2\).
03

Rearrange the heat conduction formula to solve for the unknown temperature

We know the temperature of the inner surface of the pan, so we need to find the temperature of the outer surface: \(T_2\). Rearrange Fourier's Law of Heat Conduction for \(T_2\): \(T_2 = T_1 - \dfrac{QL}{kA}\).
04

Substitute the given values into the formula and calculate the unknown temperature

Now we have the rearranged formula and the necessary values to solve for \(T_2\). Substitute them into the equation: \(T_2 = 105 - \dfrac{1400 \cdot 0.4}{237 \cdot 176.71}\). Calculating the temperature of the outer surface, we get: \(T_2 \approx 102.27^{\circ}\text{C}\). Hence, the temperature of the outer surface of the bottom of the pan is approximately \(102.27^{\circ}\text{C}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Fourier's Law of Heat Conduction
Imagine a chilly winter evening, and you're holding a steaming cup of cocoa. Ever notice how the heat from the cocoa seems to move into your hands through the cup? This movement of heat is guided by the Fourier's Law of Heat Conduction. Simply put, this law states that heat transfers from a region of higher temperature to a region of lower temperature within a material, and it does so at a rate that's proportional to the negative gradient of temperatures and the area through which the heat travels.

Mathematically speaking, if we take a solid object, like the bottom of an aluminum pan, the rate at which heat is traveling through it is represented as:
\[\begin{equation}Q = kA\dfrac{T_1 - T_2}{L}\end{equation}\],where Q is the heat transfer rate (how fast that cozy warmth moves), k stands for the material's thermal conductivity (a measure of how well the material transfers heat), A is the area through which heat is being transferred (like the bottom of your cup), and T_1 and T_2 are the temperatures on the two sides of the material, with L being the material's thickness.

This relationship is crucial when engineers design anything from cookware to spacecraft, as it helps predict how heat will behave and ensure your cocoa stays warm, but your hands do not get too hot!
Thermal Conductivity
Now, let's zoom in on one aspect of Fourier's law: thermal conductivity, represented by k. Think of thermal conductivity as a personality trait of materials that describes how well they can conduct heat. Different materials show off different abilities to transfer heat; metals, for instance, tend to be great at this, which is why they're often used in cookware and heat exchangers. It's why the aluminum pan from our example conducts heat so effectively to its contents.

The thermal conductivity of a material is measured in watts per meter Kelvin \((W/mK)\). This number tells you how much heat will pass through a material over a certain distance with a specific temperature difference between the two sides. For instance, with aluminum's high conductivity of \(237 W/mK\), it's a superstar at delivering heat quickly and evenly, which is why it's a popular choice for chefs and astronauts alike. Yet materials with low thermal conductivity, like wood or Styrofoam, are sought after for insulation because they keep the heat in (or out).
Steady-State Heat Transfer
You may wonder, what happens over time to that warmth in your hands or the heat from a pan on a stove? Well, after a while, the heat transfer reaches a sort of equilibrium called steady-state heat transfer. This doesn't mean that heat stops moving; rather, it implies that heat moves at a consistent rate. The temperatures at each point don't change anymore, even though the heat march continues across the material.

In this state, our trusty Fourier's Law equation works like a charm, allowing us to calculate temperatures at various points without worrying about changes over time. This concept helps in designing systems that require constant temperatures, like heating buildings or cooling electronic components. For our aluminum pan, this means that once the stove reaches a certain setting, the temperature at the bottom of the pan isn't going to suddenly spike or drop as long as the conditions stay the same - crucial for not burning dinner!
Thermal Resistance
Another way to think about heat movement is to consider its opposite: what slows it down. This is where thermal resistance comes into play. Just as electrical resistance slows down the flow of electricity, thermal resistance slows down the flow of heat. The thermal resistance of a material is a measure of how hard it is for heat to pass through it.

It's often expressed with the letter \(R\) and calculated as the ratio of the thickness of the material \(L\) to its thermal conductivity \(k\): \(R = \dfrac{L}{k}\). So, if you have two pans made of the same material but one has a thicker base, the thicker one would have greater thermal resistance, meaning it would be slower to heat up the food inside it.

Understanding thermal resistance helps not only in cooking but also in keeping our homes comfortable and energy-efficient. Insulation in your house walls, for instance, is all about high thermal resistance; keeping the toasty warm or cool air right where you want it – inside.

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Most popular questions from this chapter

The outer surface of a spacecraft in space has an emissivity of \(0.8\) and a solar absorptivity of \(0.3\). If solar radiation is incident on the spacecraft at a rate of \(950 \mathrm{~W} / \mathrm{m}^{2}\), determine the surface temperature of the spacecraft when the radiation emitted equals the solar energy absorbed.

A series of experiments were conducted by passing \(40^{\circ} \mathrm{C}\) air over a long \(25 \mathrm{~mm}\) diameter cylinder with an embedded electrical heater. The objective of these experiments was to determine the power per unit length required \((\dot{W} / L)\) to maintain the surface temperature of the cylinder at \(300^{\circ} \mathrm{C}\) for different air velocities \((V)\). The results of these experiments are given in the following table: $$ \begin{array}{lccccc} \hline V(\mathrm{~m} / \mathrm{s}) & 1 & 2 & 4 & 8 & 12 \\ \dot{W} / L(\mathrm{~W} / \mathrm{m}) & 450 & 658 & 983 & 1507 & 1963 \\ \hline \end{array} $$ (a) Assuming a uniform temperature over the cylinder, negligible radiation between the cylinder surface and surroundings, and steady state conditions, determine the convection heat transfer coefficient \((h)\) for each velocity \((V)\). Plot the results in terms of \(h\left(\mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\right)\) vs. \(V(\mathrm{~m} / \mathrm{s})\). Provide a computer generated graph for the display of your results and tabulate the data used for the graph. (b) Assume that the heat transfer coefficient and velocity can be expressed in the form of \(h=C V^{m}\). Determine the values of the constants \(C\) and \(n\) from the results of part (a) by plotting \(h\) vs. \(V\) on log-log coordinates and choosing a \(C\) value that assures a match at \(V=1 \mathrm{~m} / \mathrm{s}\) and then varying \(n\) to get the best fit.

Hot air at \(80^{\circ} \mathrm{C}\) is blown over a 2-m \(\times 4\) - \(\mathrm{m}\) flat surface at \(30^{\circ} \mathrm{C}\). If the average convection heat transfer coefficient is \(55 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the rate of heat transfer from the air to the plate, in \(\mathrm{kW}\).

Consider a sealed 20-cm-high electronic box whose base dimensions are \(50 \mathrm{~cm} \times 50 \mathrm{~cm}\) placed in a vacuum chamber. The emissivity of the outer surface of the box is \(0.95\). If the electronic components in the box dissipate a total of \(120 \mathrm{~W}\) of power and the outer surface temperature of the box is not to exceed \(55^{\circ} \mathrm{C}\), determine the temperature at which the surrounding surfaces must be kept if this box is to be cooled by radiation alone. Assume the heat transfer from the bottom surface of the box to the stand to be negligible.

Consider a 20-cm thick granite wall with a thermal conductivity of \(2.79 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The temperature of the left surface is held constant at \(50^{\circ} \mathrm{C}\), whereas the right face is exposed to a flow of \(22^{\circ} \mathrm{C}\) air with a convection heat transfer coefficient of \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Neglecting heat transfer by radiation, find the right wall surface temperature and the heat flux through the wall.

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