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Chapter 14: Problem 55

A hiker is wearing wool clothing of \(0.50-\mathrm{cm}\) thickness to keep warm. Her skin temperature is \(35^{\circ} \mathrm{C}\) and the outside temperature is \(4.0^{\circ} \mathrm{C} .\) Her body surface area is \(1.2 \mathrm{m}^{2} .\) (a) If the thermal conductivity of wool is $0.040 \mathrm{W} /(\mathrm{m} \cdot \mathrm{K}),$ what is the rate of heat conduction through her clothing? (b) If the hiker is caught in a rainstorm, the thermal conductivity of the soaked wool increases to \(0.60 \mathrm{W} /(\mathrm{m} \cdot \mathrm{K})\) (that of water). Now what is the rate of heat conduction?

Short Answer

Expert verified
Answer: The rate of heat conduction through the hiker's dry wool clothing is 748.8 W, while it increases to 11,232 W when the wool is soaked. The significant increase in heat loss when the wool is soaked makes the hiker feel much colder in the rainstorm.

Step by step solution

01

Write down the formula for the rate of heat conduction

The rate of heat conduction (Q) can be found using Fourier's Law of heat conduction, which is given by the formula: \(Q = \frac{kA(T_1 - T_2)}{d}\) Here, \(k\) is the thermal conductivity \(A\) is the area \(T_1\) is the temperature of the warmer side (the hiker's body) \(T_2\) is the temperature of the colder side (the outside temperature) \(d\) is the thickness of the material
02

Determine the rate of heat conduction when the wool is dry (Q dry)

Given the thermal conductivity of dry wool, \(k_d = 0.040\,\text{W}\, /(\text{m}\cdot \text{K})\). The thickness of the wool in meters is \(0.50\,\text{cm}=0.005\,\text{m}\), and the body surface area is \(1.2\,\text{m}^2\). The skin temperature is \(35^{\circ} \mathrm{C}\) and the outside temperature is \(4.0^{\circ} \mathrm{C}\). We can now plug these values into the Fourier's Law formula to get the rate of heat conduction for dry wool: \(Q_{dry}=\frac{0.040 * 1.2\,(\,35 - 4\,)}{0.005} \mathrm{W}\) \(Q_{dry}=748.8 \mathrm{W}\)
03

Determine the rate of heat conduction when the wool is soaked (Q soaked)

When the wool is soaked, the thermal conductivity increases to \(k_s=0.60\,\text{W}/(\mathrm{m}\cdot\text{K})\). We can again use the Fourier's Law formula to find the rate of heat conduction: \(Q_{soaked}=\frac{0.60 * 1.2\,(\,35 - 4\,)}{0.005} \mathrm{W}\) \(Q_{soaked}=11\,232 \mathrm{W}\)
04

Compare the rates of heat conduction and summarize

The rate of heat conduction through the hiker's dry wool clothing is \(748.8\,\mathrm{W}\), and the rate of heat conduction through the soaked wool clothing is \(11\,232\,\mathrm{W}\). The rate of heat loss increases significantly when the wool is soaked, which will make the hiker feel much colder in the rainstorm.

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Most popular questions from this chapter

A blacksmith heats a 0.38 -kg piece of iron to \(498^{\circ} \mathrm{C}\) in his forge. After shaping it into a decorative design, he places it into a bucket of water to cool. If the available water is at \(20.0^{\circ} \mathrm{C},\) what minimum amount of water must be in the bucket to cool the iron to \(23.0^{\circ} \mathrm{C} ?\) The water in the bucket should remain in the liquid phase.
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