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Chapter 4: Problem 106

A barefooted person whose feet are at \(32^{\circ} \mathrm{C}\) steps on a large aluminum block at \(20^{\circ} \mathrm{C}\). Treating both the feet and the aluminum block as semi-infinite solids, determine the contact surface temperature. What would your answer be if the person stepped on a wood block instead? At room temperature, the \(\sqrt{k \rho c_{p}}\) value is $24 \mathrm{~kJ} / \mathrm{m}^{2},{ }^{\circ} \mathrm{C}\( for aluminum, \)0.38 \mathrm{~kJ} / \mathrm{m}^{2},{ }^{\circ} \mathrm{C}\( for wood, and \)1.1 \mathrm{~kJ} / \mathrm{m}^{2},{ }^{\circ} \mathrm{C}$ for human flesh.

Short Answer

Expert verified
Answer: The contact surface temperature when the person steps on the aluminum block is approximately \(30.6^{\circ}C\), and when steps on the wood block, the temperature is approximately \(26.9^{\circ}C\).

Step by step solution

01

Identify the initial temperatures of each material

The initial temperature of the foot is given as \(32^{\circ} \mathrm{C}\) and the initial temperature of the aluminum block is given as \(20^{\circ} \mathrm{C}\).
02

Apply the formula for interface temperature of semi-infinite solids

The formula for interface temperature of two semi-infinite solids in contact is given by: $$T_{interface} = T_1 \cdot \frac{\sqrt{k_2 \rho_2 c_{p_2}}}{\sqrt{k_1 \rho_1 c_{p_1}}+ \sqrt{k_2 \rho_2 c_{p_2}}}+T_2 \cdot \frac{\sqrt{k_1 \rho_1 c_{p_1}}}{\sqrt{k_1 \rho_1 c_{p_1}}+ \sqrt{k_2 \rho_2 c_{p_2}}}$$ Here, \(T_1\) is the initial temperature of the foot, \(T_2\) is the initial temperature of the material (aluminum or wood), and \(\sqrt{k \rho c_{p}}\) values for foot, aluminum, and wood are given in the problem.
03

Calculate the contact surface temperature for aluminum

Using the formula in Step 2, we can calculate the contact surface temperature when the person steps on the aluminum block: $$T_{interface_{aluminum}} = 32^{\circ} C \cdot \frac{24}{24+1.1} + 20^{\circ} C \cdot \frac{1.1}{24+1.1} \approx 30.6^{\circ}C$$
04

Calculate the contact surface temperature for wood

Similarly, we can calculate the contact surface temperature when the person steps on the wood block: $$T_{interface_{wood}} = 32^{\circ} C \cdot \frac{0.38}{1.1+0.38} + 20^{\circ} C \cdot \frac{1.1}{1.1+0.38} \approx 26.9^{\circ}C$$
05

Present the answers

The contact surface temperature when the person steps on the aluminum block is approximately \(30.6^{\circ}C\), and when steps on the wood block, the temperature is approximately \(26.9^{\circ}C\).

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Most popular questions from this chapter

Oranges of \(2.5\)-in-diameter $\left(k=0.26 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\right.\( and \)\alpha=1.4 \times 10^{-6} \mathrm{ft}^{2} / \mathrm{s}$ ) initially at a uniform temperature of \(78^{\circ} \mathrm{F}\) are to be cooled by refrigerated air at $25^{\circ} \mathrm{F}\( flowing at a velocity of \)1 \mathrm{ft} / \mathrm{s}$. The average heat transfer coefficient between the oranges and the air is experimentally determined to be $4.6 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F}$. Determine how long it will take for the center temperature of the oranges to drop to \(40^{\circ} \mathrm{F}\). Also, determine if any part of the oranges will freeze during this process. Solve this problem using the analytical one-term approximation method.

A steel casting cools to 90 percent of the original temperature difference in \(30 \mathrm{~min}\) in still air. The time it takes to cool this same casting to 90 percent of the original temperature difference in a moving air stream whose convective heat transfer coefficient is 5 times that of still air is (a) \(3 \mathrm{~min}\) (b) \(6 \mathrm{~min}\) (c) \(9 \mathrm{~min}\) (d) \(12 \mathrm{~min}\) (e) \(15 \mathrm{~min}\)

In Betty Crocker's Cookbook, it is stated that it takes $2 \mathrm{~h} \mathrm{} 45 \mathrm{~min}\( to roast a \)3.2-\mathrm{kg}$ rib initially at \(4.5^{\circ} \mathrm{C}\) to "rare" in an oven maintained at $163^{\circ} \mathrm{C}$. It is recommended that a meat thermometer be used to monitor the cooking, and the rib is considered rare when the thermometer inserted into the center of the thickest part of the meat registers \(60^{\circ} \mathrm{C}\). The rib can be treated as a homogeneous spherical object with the properties $\rho=1200 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=4.1 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}, k=0.45 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$, and \(\alpha=0.91 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\). Determine \((a)\) the heat transfer coefficient at the surface of the rib; \((b)\) the temperature of the outer surface of the rib when it is done; and \((c)\) the amount of heat transferred to the rib. \((d)\) Using the values obtained, predict how long it will take to roast this rib to "medium" level, which occurs when the innermost temperature of the rib reaches \(71^{\circ} \mathrm{C}\). Compare your result to the listed value of \(3 \mathrm{~h} \mathrm{} 20 \mathrm{~min} .\) If the roast rib is to be set on the counter for about \(15 \mathrm{~min}\) before it is sliced, it is recommended that the rib be taken out of the oven when the thermometer registers about \(4^{\circ} \mathrm{C}\) below the indicated value because the rib will continue cooking even after it is taken out of the oven. Do you agree with this recommendation? Solve this problem using the analytical one-term approximation method. Answers: (a) $156.9 \mathrm{~W} / \mathrm{m}^{2}-\mathrm{K}\(, (b) \)159.5^{\circ} \mathrm{C}$, (c) \(1629 \mathrm{~kJ}\), (d) \(3.0 \mathrm{~h}\)

During a picnic on a hot summer day, the only available drinks were those at the ambient temperature of \(90^{\circ} \mathrm{F}\). In an effort to cool a 12 -fluid-oz drink in a can, which is 5 in high and has a diameter of $2.5 \mathrm{in}$, a person grabs the can and starts shaking it in the iced water of the chest at \(32^{\circ} \mathrm{F}\). The temperature of the drink can be assumed to be uniform at all times, and the heat transfer coefficient between the iced water and the aluminum can is $30 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F}$. Using the properties of water for the drink, estimate how long it will take for the canned drink to cool to \(40^{\circ} \mathrm{F}\). Solve this problem using lumped system analysis. Is the lumped system analysis applicable to this problem? Why?

A 6-cm-high rectangular ice block $(k=2.22 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\( and \)\alpha=0.124 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}$ ) initially at \(-18^{\circ} \mathrm{C}\) is placed on a table on its square base \(4 \mathrm{~cm} \times 4 \mathrm{~cm}\) in size in a room at $18^{\circ} \mathrm{C}$. The heat transfer coefficient on the exposed surfaces of the ice block is \(12 \mathrm{~W} / \mathrm{m}^{2}\). \(\mathrm{K}\). Disregarding any heat transfer from the base to the table, determine how long it will be before the ice block starts melting. Where on the ice block will the first liquid droplets appear? Solve this problem using the analytical one-term approximation method.
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