How much heat a bar is putting out?

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SUMMARY

The discussion centers on determining the temperature at the center of a bar based on its heat output. It emphasizes that without additional information such as mass, emissivity, and thermal conductivity, the question lacks meaning. The participants agree that knowing the power output and emissivity can help calculate the surface temperature, which, combined with thermal conductivity, may allow for estimating the core temperature under steady-state conditions.

PREREQUISITES
  • Understanding of thermal conductivity
  • Knowledge of emissivity and its role in heat transfer
  • Familiarity with steady-state thermal analysis
  • Basic principles of heat transfer and thermodynamics
NEXT STEPS
  • Research methods for calculating surface temperature using emissivity
  • Learn about steady-state heat transfer analysis
  • Explore thermal conductivity in different materials
  • Study the relationship between mass, heat output, and temperature in thermodynamic systems
USEFUL FOR

Students in physics or engineering, thermal analysts, and professionals involved in heat transfer calculations will benefit from this discussion.

edwin.07
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If you know how much heat a bar is putting out, what is the temperature at the center of the bar?
 
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Meaningless question without more information. You could have a moderate-heat bar with a large mass putting out the same amount of heat as a hotter bar of lower mass, and they would have different temperatures.

If this is a homework problem, you need to show the exact question and show your own attempt at a solution.
 
Not really my field but I agree with phinds. More info needed.

If you know the power (aka heat) and the emissivity you might be able to use that to work out the surface temperature of the bar. Then if you know the thermal conductivity between the core and the surface you might be able to work out the core temperature. Some assumptions are required - such as it all being in steady state eg something is heating the core of the bar to compensate for the heat losses.
 

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