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Chapter 1: Problem 121

An infrared heater for a sauna has a surface area of \(0.050 \mathrm{m}^{2}\) and an emissivity of 0.84. What temperature must it run at if the required power is 360 W? Neglect the temperature of the environment.

Short Answer

Expert verified
The infrared heater must run at a temperature of approximately \(386.29 \mathrm{K}\) to emit the required power of 360 W.

Step by step solution

01

Write down the Stefan-Boltzmann Law

The relationship between the power emitted by an object, its temperature, surface area, and emissivity is given by the Stefan-Boltzmann law: \(P = \varepsilon A \sigma T^4 \) where: - \(P\) is the power emitted (W) - \(\varepsilon\) is the emissivity (dimensionless) - \(A\) is the surface area (\(\mathrm{m}^2\)) - \(\sigma\) is the Stefan-Boltzmann constant \((5.67 \times 10^{-8} \mathrm{ Wm^{-2}K^{-4}})\) - \(T\) is the temperature of the object (K)
02

Plug in the given values

Now we can plug in the given values for the required power, surface area, and emissivity: \(360 \mathrm{W} = 0.84 \times 0.050 \mathrm{m}^2 \times (5.67 \times 10^{-8} \mathrm{ Wm^{-2}K^{-4}}) \times T^4 \)
03

Isolate the temperature term

Next, we will isolate the temperature term to solve for it: \(T^4 = \frac{360 \mathrm{W}}{0.84 \times 0.050 \mathrm{m}^2 \times (5.67 \times 10^{-8} \mathrm{ Wm^{-2}K^{-4}})} \)
04

Calculate the temperature

Now we can calculate the temperature by taking the fourth root of both sides: \(T = \sqrt[4]{\frac{360 \mathrm{W}}{0.84 \times 0.050 \mathrm{m}^2 \times (5.67 \times 10^{-8} \mathrm{ Wm^{-2}K^{-4}})}} \) \(T \approx 386.29 \mathrm{K} \) #Conclusion# The infrared heater must run at a temperature of approximately 386.29 K to emit the required power of 360 W.

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Key Concepts

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

Understanding Emissivity
Emissivity is a measure of how efficiently a surface emits energy as thermal radiation. This concept plays a crucial role in calculations involving thermal energy. Emissivity has a value between 0 and 1. A value of 1 indicates a perfect blackbody, which emits the maximum possible radiation. A value closer to 0 means the surface emits much less.
  • A high emissivity (e.g., 0.84) means the surface is good at emitting thermal radiation.
  • Materials with different surface qualities and colors can significantly affect emissivity.
In our problem, the emissivity of 0.84 tells us that the sauna's infrared heater is quite efficient at emitting energy.
The Role of Surface Area
Surface area is another key factor when calculating thermal radiation. It's the measure of how much exposed area a solid object has. A larger area can emit more radiation.
  • In calculations, surface area directly impacts the amount of energy emitted.
  • The unit is typically square meters (\(\mathrm{m}^{2}\)).
For our infrared heater, the surface area is \0.050\ \(\mathrm{m}^{2}\). This means the heater can emit a certain amount of power based on its size and emissivity.
Infrared Heaters in Focus
Infrared heaters are devices that use thermal radiation to transfer energy. They are especially common in saunas and heating systems. Infrared radiation is part of the electromagnetic spectrum, sitting between visible light and microwaves.
  • Infrared heaters are efficient because they directly heat objects without warming the air first.
  • They are often used for targeted heating, such as in saunas, to provide uniform and consistent warmth.
The use of infrared heaters allows for precise control over the environment's temperature, making them an ideal choice for saunas.
Temperature Calculation Using the Stefan-Boltzmann Law
Calculating temperature with the Stefan-Boltzmann law involves understanding the relationship between power, emissivity, surface area, and temperature. The Stefan-Boltzmann law is represented by: \[P = \varepsilon A \sigma T^4\] where:
  • \(P\) is the power emitted, in watts.
  • \(\varepsilon\) is the emissivity.
  • \(A\) is the surface area, in \(\mathrm{m}^{2}\).
  • \(\sigma\) is the Stefan-Boltzmann constant, \(5.67 \times 10^{-8} \, \mathrm{Wm^{-2}K^{-4}}\).
  • \(T\) is the temperature, in Kelvin.
By rearranging the formula to solve for temperature, you can find \(T\) using: \[T = \sqrt[4]{\frac{P}{\varepsilon A \sigma}}\] For our infrared heater, the computed temperature is approximately 386.29 K, ensuring it meets the required power of 360 W.

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Most popular questions from this chapter

In Miami, Florida, which has a very humid climate and numerous bodies of water nearby, it is unusual for temperatures to rise above about \(38^{\circ} \mathrm{C}\left(100^{\circ} \mathrm{F}\right) .\) In the desert climate of Phoenix, Arizona, however, temperatures rise above that almost every day in July and August. Explain how the evaporation of water helps limit high temperatures in humid climates.

Following vigorous exercise, the body temperature of an \(80.0 \mathrm{kg}\) person is \(40.0^{\circ} \mathrm{C} .\) At what rate in watts must the person transfer thermal energy to reduce the body temperature to \(37.0^{\circ} \mathrm{C}\) in \(30.0 \mathrm{min}\), assuming the body continues to produce energy at the rate of \(150 \mathrm{W}\) ? (1 watt \(=1\) joule/second or \(1 \mathrm{W}=1 \mathrm{J} / \mathrm{s}\) )

A person taking a reading of the temperature in a freezer in Celsius makes two mistakes: first omitting the negative sign and then thinking the temperature is Fahrenheit. That is, the person reads \(-x^{\circ} \mathrm{C}\) as \(x^{\circ} \mathrm{F}\). Oddly enough, the result is the correct Fahrenheit temperature. What is the original Celsius reading? Round your answer to three significant figures.

Broiling is a method of cooking by radiation, which produces somewhat different results from cooking by conduction or convection. A gas flame or electric heating element produces a very high temperature close to the food and above it. Why is radiation the dominant heat-transfer method in this situation?

(a) Suppose a cold front blows into your locale and drops the temperature by 40.0 Fahrenheit degrees. How many degrees Celsius does the temperature decrease when it decreases by \(40.0^{\circ} \mathrm{F}\) ? (b) Show that any change in temperature in Fahrenheit degrees is nine-fifths the change in Celsius degrees
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