Boltzman Constant, Emissivity and Surface Area of a filament

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SUMMARY

The discussion centers on calculating the surface area of a tungsten filament heated to 2.30 x 103 K with an emissivity of 0.31, delivering 35.0 W of power. Using Stefan's Law, the equation P(net) = εσAT4 was applied, where σ is the Stefan-Boltzmann constant (5.670 x 10-8 W/m2K4). The calculated surface area of the filament is approximately 7.12 x 10-5 m2, confirming the solution's correctness.

PREREQUISITES
  • Understanding of Stefan's Law and its application in thermal physics
  • Knowledge of emissivity and its significance in heat transfer
  • Familiarity with the Stefan-Boltzmann constant and its value (5.670 x 10-8 W/m2K4)
  • Basic algebra skills for solving equations
NEXT STEPS
  • Study the derivation and applications of Stefan's Law in thermal radiation
  • Explore the concept of emissivity in different materials and its impact on thermal efficiency
  • Learn about the relationship between power, temperature, and surface area in thermal systems
  • Investigate the properties of tungsten as a filament material in incandescent lamps
USEFUL FOR

Students in physics or engineering, particularly those studying thermodynamics, as well as professionals involved in designing and analyzing thermal systems and lighting technologies.

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Homework Statement



A tungsten filament in a lamp is heated to a temperature of 2.30 x 10^3 K by an electric current. The tungsten has an emissivity of 0.31. What is the surface area of the filament if the lamp delivers 35.0 W of power?

Homework Equations

Stefan's Law

(a greek letter that looks like P?) P(net)=ekAT^4

The Attempt at a Solution



35.0=0.31(5.670x10^-8 w/m^2*K^4)(2.30*10^3)^4A

35.0 = 491876.5257A

35.0/491876.5257=A

7.1156069x10^-5


is this correct?

Thanks in advance for your help.
 
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