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Chapter 13: Problem 23

A person is standing outdoors in the shade where the temperature is \(28^{\circ} \mathrm{C} .\) (a) What is the radiant energy absorbed per second by his head when it is covered with hair? The surface area of the hair (assumed to be flat) is \(160 \mathrm{cm}^{2}\) and its emissivity is \(0.85 .\) (b) What would be the radiant energy absorbed per second by the same person if he were bald and the emissivity of his head were \(0.65 ?\)

Short Answer

Expert verified
(a) 5.676 W, (b) 4.338 W.

Step by step solution

01

Convert Temperature

First, convert the temperature from Celsius to Kelvin, since the Stefan-Boltzmann law uses Kelvin. The conversion formula is \(T_K = T_C + 273.15\). Thus, the temperature is \(28^{\circ}C = 301.15\, K\).
02

Identify the Stefan-Boltzmann Law

The radiant energy absorbed per second can be calculated using the Stefan-Boltzmann Law: \(P = \epsilon \sigma A T^4\), where \(P\) is the power (radiant energy per second), \(\epsilon\) is emissivity, \(\sigma\) is the Stefan-Boltzmann constant \(5.67 \times 10^{-8} \text{ W/m}^2\text{K}^4\), \(A\) is the surface area, and \(T\) is the temperature in Kelvin.
03

Calculate Absorbed Energy with Hair

Convert the surface area from cm虏 to m虏: \(160 \text{ cm}^2 = 0.016 \text{ m}^2\). Now use the Stefan-Boltzmann Law: \(P = 0.85 \times 5.67 \times 10^{-8} \times 0.016 \times (301.15)^4\). This results in \(P \approx 5.676 \text{ W}\).
04

Calculate Absorbed Energy if Bald

For the bald condition, use the same formula with \(\epsilon = 0.65\): \(P = 0.65 \times 5.67 \times 10^{-8} \times 0.016 \times (301.15)^4\). This results in \(P \approx 4.338 \text{ W}\).

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Key Concepts

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

Emissivity
Emissivity is a measure of an object鈥檚 ability to emit or absorb thermal radiation compared to a perfect blackbody. It ranges from 0 to 1, where 1 represents a perfect emitter or absorber.
In the context of the given problem, the emissivity differs for hair and bald skin, affecting how much radiant energy is absorbed. Hair has an emissivity of 0.85, indicating that it absorbs 85% of all incident radiation, making it a more effective absorber than bald skin, which has an emissivity of 0.65.
Understanding emissivity is crucial in calculating radiant energy absorbed, as a higher emissivity means more energy absorbed and subsequently, more energy emitted as heat.
Radiant Energy
Radiant energy is the energy transported by electromagnetic waves. When the Sun鈥檚 energy reaches an object, it can be absorbed, reflected, or transmitted. The absorbed portion is what heats the object, allowing it to emit energy as thermal radiation.
In this exercise, radiant energy absorbed is calculated using the Stefan-Boltzmann law. It's expressed as power, which is energy per second, showing how much energy is absorbed by the head per second due to its exposure to the surrounding temperatures.
This calculation shows how factors like emissivity and surface area impact the amount of energy absorbed, directly influencing how warm an object feels.
Surface Area
Surface area is the measure of the exposed exterior of a three-dimensional object. It's crucial when calculating radiant energy absorption, as only the surface exposed to the environment can absorb energy.
In the exercise, the head's surface area is given as 160 cm虏. To perform calculations using the Stefan-Boltzmann law, this must be converted to square meters, resulting in 0.016 m虏.
A larger surface area would mean more area available for absorbing or emitting radiant energy, hence significantly affecting total energy calculations.
Temperature Conversion
Temperature conversion is often necessary when using scientific equations like the Stefan-Boltzmann law, which requires Kelvin as the temperature unit. To convert from Celsius to Kelvin, the formula is: \[ T_K = T_C + 273.15 \]
For our problem, converting 28掳C is straightforward: \[ 28掳C = 301.15 \, K \]
This conversion ensures calculations align with the Syst猫me International (SI) units, leading to consistent and accurate results. Understanding temperature conversion is vital, as improper or neglected conversions can result in significant errors in calculations involving thermal energy.

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Most popular questions from this chapter

A person's body is covered with \(1.6 \mathrm{m}^{2}\) of wool clothing. The thickness of the wool is \(2.0 \times 10^{-3} \mathrm{m} .\) The temperature at the outside surface of the wool is \(11^{\circ} \mathrm{C},\) and the skin temperature is \(36^{\circ} \mathrm{C} .\) How much heat per second does the person lose due to conduction?

A person eats a dessert that contains 260 Calories. (This "Calorie" unit, with a capital \(\mathrm{C}\), is the one used by nutritionists; 1 Calorie \(=\) 4186 J. See Section 12.7.) The skin temperature of this individual is \(36^{\circ} \mathrm{C}\) and that of her environment is \(21{ }^{\circ} \mathrm{C}\). The emissivity of her skin is 0.75 and its surface area is \(1.3 \mathrm{m}^{2} .\) How much time would it take for her to emit a net radiant energy from her body that is equal to the energy contained in this dessert?

A solid sphere has a temperature of \(773 \mathrm{K}\). The sphere is melted down and recast into a cube that has the same emissivity and emits the same radiant power as the sphere. What is the cube's temperature?

A closed box is filled with dry ice at a temperature of \(-78.5^{\circ} \mathrm{C}\), while the outside temperature is \(21.0^{\circ} \mathrm{C} .\) The box is cubical, measuring \(0.350 \mathrm{m}\) on a side, and the thickness of the walls is \(3.00 \times 10^{-2} \mathrm{m}\). In one day, \(3.10 \times 10^{6} \mathrm{J}\) of heat is conducted through the six walls. Find the thermal conductivity of the material from which the box is made.

A solar collector is placed in direct sunlight where it absorbs energy at the rate of \(880 \mathrm{J} / \mathrm{s}\) for each square meter of its surface. The emissivity of the solar collector is \(e=0.75 .\) What equilibrium temperature does the collector reach? Assume that the only energy loss is due to the emission of radiation.
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