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A body whose surface area is \(0.25 \mathrm{~m}^{2}\), emissivity is \(0.85\). and temperature is \(175^{\circ} \mathrm{C}\) is placed in a large, evacuated chamber whose walls are at \(27^{\circ} \mathrm{C}\). What is the rate at which radiation is emitted by the surface, in W? What is the net rate at which radiation is exchanged between the surface and the chamber walls, in \(W\) ?

Short Answer

Expert verified
Power emitted is approximately 128 W. Net rate of radiation exchange is approximately 111 W.

Step by step solution

01

- Convert Temperatures to Kelvins

First, convert the given temperatures from Celsius to Kelvin using the formula: \[ T(K) = T(^{\circ}C) + 273.15 \] For the surface: \[ T = 175^{\circ}C + 273.15 = 448.15 \ K \] For the chamber walls: \[ T = 27^{\circ}C + 273.15 = 300.15 \ K \]
02

- Formula for Radiation Emitted

Use the Stefan-Boltzmann law to calculate the power radiated by a surface: \[ P = e \sigma A T^4 \] where \( e \) is the emissivity, \( \sigma \) is the Stefan-Boltzmann constant \( 5.67 \times 10^{-8} \ \mathrm{W/m^2K^4} \), \( A \) is the surface area, and \( T \) is the temperature in Kelvin.
03

- Calculate Power Radiated by the Surface

Substitute the known values into the formula to find the power emitted by the surface: \[ P_{emitted} = 0.85 \times 5.67 \times 10^{-8} \times 0.25 \times (448.15)^4 \] Evaluate the expression carefully.
04

- Calculate the Net Radiation Exchange

To find the net rate of radiation exchange, subtract the power absorbed by the surface (from the chamber walls) from the power emitted by the surface. Use the formula: \[ P_{net} = e \sigma A (T_{surface}^4 - T_{walls}^4) \] Substitute the values: \[ P_{net} = 0.85 \times 5.67 \times 10^{-8} \times 0.25 \times ((448.15)^4 - (300.15)^4) \]
05

- Final Calculation

Perform the final calculations for the emitted and net power values using the provided formulas and numerical input.

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Key Concepts

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

Stefan-Boltzmann Law
The Stefan-Boltzmann law is pivotal when studying radiative heat transfer. It establishes the power radiated per unit area of a black body in terms of its temperature. The formula is expressed as \( P = e \sigma A T^4 \). Here, \( P \) is the power radiated, \( e \) is the 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. This law essentially tells us that the energy radiated increases dramatically with an increase in temperature due to the \( T^4 \) term.
Emissivity
Emissivity, denoted as \( e \), is a measure of how effectively a surface emits thermal radiation compared to a perfect black body. A perfect emitter (black body) has an emissivity of 1, while real surfaces have values between 0 and 1. Materials that are good reflectors, like metals, typically have low emissivity, whereas materials like black paint have high emissivity. In our problem, the emissivity is 0.85, indicating it is a fairly efficient emitter but not a perfect one. This is a critical factor in calculating the power radiated because the higher the emissivity, the more radiation the body emits. The formula \( P = e \sigma A T^4 \) incorporates this by multiplying the Stefan-Boltzmann constant \( \sigma \) with the emissivity \( e \).
Temperature Conversion
Temperature conversion from Celsius to Kelvin is essential in the application of the Stefan-Boltzmann law. The Kelvin scale is the standard unit of temperature in physics and provides a direct correlation to the energy radiated by a body. The conversion formula is straightforward: \( T(K) = T(^{\text{°}C}) + 273.15 \). This means if you have a temperature in Celsius, you simply add 273.15 to convert it to Kelvin. In our exercise, the temperatures were converted as follows:
  • For the surface: \( 175^{\text{°}C} + 273.15 = 448.15 \,K \)
  • For the chamber walls: \( 27^{\text{°}C} + 273.15 = 300.15 \,K \)
This conversion is vital because the Stefan-Boltzmann equation requires temperatures in Kelvin to accurately compute radiant energy.

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Most popular questions from this chapter

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