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Chapter 19: Problem 54

(a) Calculate the rate at which body heat is conducted through the clothing of a skier in a steady-state process, given the following data: the body surface area is \(1.8 \mathrm{~m}^{2}\) and the clothing is \(1.0 \mathrm{~cm}\) thick; the skin surface temperature is \(33^{\circ} \mathrm{C}\) and the outer surface of the clothing is at \(1.0^{\circ} \mathrm{C} ;\) the thermal conductivity of the clothing is \(0.040 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (b) How would the answer to (a) change if, after a fall, the skier's clothes became soaked with water of thermal conductivity \(0.60 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} ?\)

Short Answer

Expert verified
The rate of heat conduction through dry clothing is 230.4 W, and through wet clothing, it is 3456 W.

Step by step solution

01

Understand the heat conduction formula

The rate of heat conduction can be calculated using the formula: \( Q = \frac{k \cdot A \cdot \Delta T}{d} \), where \( Q \) is the heat conduction rate, \( k \) is the thermal conductivity, \( A \) is the surface area through which heat is conducted, \( \Delta T \) is the temperature difference, and \( d \) is the thickness of the material.
02

List the given data

The given data are: - Body surface area \( A = 1.8 \; m^2 \) - Thickness of clothing \( d = 1.0 \; cm = 0.01 \; m \) - Skin surface temperature \( T_1 = 33 \; ^\circ C \) - Outer surface temperature of clothing \( T_2 = 1.0 \; ^\circ C \) - Thermal conductivity of dry clothing \( k = 0.040 \; W/m \cdot K \)
03

Calculate the temperature difference

Subtract the outer surface temperature from the skin surface temperature to find \( \Delta T \): \( \Delta T = T_1 - T_2 = 33 \; ^\circ C - 1.0 \; ^\circ C = 32 \; K \)
04

Calculate the rate of heat conduction for dry clothing

Plug the values into the heat conduction formula: \( Q = \frac{0.040 \cdot 1.8 \cdot 32}{0.01} = \frac{2.304}{0.01} = 230.4 \; W \)
05

Calculate the rate of heat conduction for wet clothing

If the clothes are soaked with water, the thermal conductivity changes to \( k = 0.60 \; W/m \cdot K \). Using the same values for \( A \), \( \Delta T \), and \( d \), plug into the formula again: \( Q = \frac{0.60 \cdot 1.8 \cdot 32}{0.01} = \frac{34.56}{0.01} = 3456 \; W \)

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Key Concepts

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

Thermal Conductivity
Thermal conductivity is a material's ability to conduct heat. It quantifies how well a material can transfer heat from the hot side to the cold side. A higher thermal conductivity means the material is better at conducting heat. For instance, metals, with high thermal conductivities, quickly transfer heat, whereas materials like clothing or insulation have low thermal conductivities and slow the heat transfer process.
In the exercise, the skier's clothing has a thermal conductivity of 0.040 W/m路K. When the clothes become wet, this increases significantly to 0.60 W/m路K due to the presence of water, which is a better conductor of heat. This significant change dramatically affects the rate at which heat is conducted away from the skier's body, emphasizing the importance of keeping dry to stay warm.
Steady-State Heat Transfer
In steady-state heat transfer, the temperature distribution in a material does not change over time. The amount of heat entering a material is equal to the amount leaving it, resulting in a constant heat flow.
In the problem, steady-state heat transfer implies that the skier's body continuously loses heat at a constant rate through the clothing. The heat transfer rate doesn鈥檛 fluctuate because the temperatures of the skier's skin and outer clothing surface remain constant.
This simplification is useful for calculating the heat loss over time, which in turn helps in designing better clothing to maintain body temperature.
Surface Area and Heat Flow
The surface area through which heat is conducted significantly impacts the rate of heat transfer. More surface area means more space for heat to travel through, resulting in increased heat flow.
For the skier, the body surface area is given as 1.8 m虏. This large area allows for substantial heat transfer. Formulaically, it鈥檚 part of the numerator in the heat conduction equation, hence directly proportional to the heat flow rate. Therefore, if the surface area were smaller, the rate of heat loss would also decrease.
This concept helps explain why tight, well-fitted clothing is more effective at retaining warmth in colder environments.
Temperature Difference in Heat Transfer
Temperature difference, symbolized as \(\Delta T\), drives the heat transfer process. It is the difference between the hot and cold sides of the material.
In the skier example, the skin temperature is 33掳C, and the outer clothing is at 1掳C. The calculation yields a \(\Delta T\) of 32 K. The greater this temperature difference, the higher the rate of heat conduction. So, when the clothes are dry, this results in a controlled heat loss of 230.4 W. However, the presence of water and its higher conductivity amplifies this rate to 3456 W.
Understanding \(\Delta T\) is crucial for designing clothing and materials that can protect against heat loss in fluctuating environmental conditions.

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Most popular questions from this chapter

For each of the situations described below, the object considered is undergoing some changes. Among the possible changes you should consider are: (Q) The object is absorbing or giving off thermal energy. (T) The object's temperature is changing. \(\left(E^{\text {int }}\right)\) The object's internal energy is changing. (W) The object is doing mechanical work or having work done on it. For each of the situations described below, identify which of the four changes are taking place and write as many of the letters \(\mathrm{Q}\), \(\mathrm{T}, E^{\text {int }}, \mathrm{W}\), (or none) as are appropriate. (a) A cylinder with a piston on top contains a compressed gas and is sitting on a thermal reservoir (a large iron block). After everything has come to thermal equilibrium, the piston is moved upward somewhat (very slowly). The object to be considered is the gas in the cylinder. (b) Consider the same cylinder as in part (a), but it is wrapped in styrofoam, a very good thermal insulator, instead of sitting on a thermal reservoir. The piston is pressed downward (again, very slowly), compressing the gas. The object to be considered is the

At \(20^{\circ} \mathrm{C}\), a rod is exactly \(20.05 \mathrm{~cm}\) long on a steel ruler. Both the rod and the ruler are placed in an oven at \(270^{\circ} \mathrm{C}\), where the rod now measures \(20.11 \mathrm{~cm}\) on the same ruler. What is the coefficient of thermal expansion for the material of which the rod is made?

A room is lighted by four \(100 \mathrm{~W}\) incandescent lightbulbs. (The power of \(100 \mathrm{~W}\) is the rate at which a bulb converts electrical energy to thermal energy and the energy of visible light.) Assuming that \(90 \%\) of the energy is converted to thermal energy, how much thermal energy is transferred to the room in \(1.00 \mathrm{~h}\) ?

Converting kinetic energy into thermal energy produces small rises in temperature. This was in part responsible for the difficulty in discovering the law of conservation of energy. It also implies that hot objects contain a lot of energy. (This latter comment is largely responsible for the industrial revolution in the 19 th century.) To get some feel for these numbers, assume all mechanical energy is converted to thermal energy and carry out three estimates: (a) A steel ball is dropped from a height of \(3 \mathrm{~m}\) onto a concrete floor. It bounces a large number of times but eventually comes to rest. Estimate the ball's rise in temperature. (b) Suppose the steel ball you used in part (a) is at room temperature. If you converted all its thermal energy to translational kinetic energy, how fast would it be moving? (Give your answer in units of miles per hour. Also, ignore the fact that you would have to create momentum.) (c) Suppose a nickel-iron meteor falls to Earth from deep space. Estimate how much its temperature would rise on impact.

A small electric immersion heater is used to heat \(100 \mathrm{~g}\) of water for a cup of instant coffee. The heater is labeled "200 watts," so it converts electrical energy to thermal energy that is transferred to the water at this rate. Calculate the time required to bring the water from \(23^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}\) ignoring any thermal energy that transfers out of the cup.
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