Does Copper's Lack of Dipole Movement Increase Thermal Radiation in Calorimetry?

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SUMMARY

The discussion centers on the impact of copper's lack of dipole movement on thermal radiation in calorimetry, specifically regarding heat loss from oil at 80°C in a copper calorimetry cup. Participants conclude that while copper is an excellent conductor of heat, its low heat capacity and absence of dipole movement do not significantly increase thermal radiation losses. The Stefan-Boltzmann Law is referenced to explain the relationship between temperature and radiation heat transport, emphasizing the importance of using absolute temperature in calculations.

PREREQUISITES
  • Understanding of thermal radiation principles
  • Familiarity with the Stefan-Boltzmann Law
  • Knowledge of heat capacity and its dependence on molecular structure
  • Basic principles of calorimetry
NEXT STEPS
  • Study the Stefan-Boltzmann Law in detail
  • Explore the concept of dipole movement in molecular chemistry
  • Investigate the thermal properties of different materials used in calorimetry
  • Learn about heat transfer mechanisms in liquids and solids
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This discussion is beneficial for physicists, chemists, and engineers involved in calorimetry, thermal analysis, and materials science, particularly those interested in the effects of material properties on thermal radiation and heat transfer.

nithin
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I have a question. Assuming I have oil at 80 degrees in a copper calorimetry cup , will the heat loss by thermal radiation be very high and will it affect my values?

I am thinking that it would not and that it would be negligible. But i do not know why. I heard somewhere that it would be high due to copper not having a dipole movement. Can someone please explain to me why?
 
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Your own experience with hot drinks surely tells you that a liquid at 80C cools pretty quickly whatever its in.

Copper is a very good conductor of heat and has a low heat capacity so it won't help much to keep the oil warm.

The heat capacity of a substance depends on its internal degrees of freedom for vibration, that's where the atomic dipole movement comes in.
 
nithin said:
I have a question. Assuming I have oil at 80 degrees in a copper calorimetry cup, will the heat loss by thermal radiation be very high and will it affect my values?

I am thinking that it would not and that it would be negligible. But i do not know why. I heard somewhere that it would be high due to copper not having a dipole movement. Can someone please explain to me why?
See the discussion of radiation heat transport and the Stefan-Boltzmann Law
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html

Let Thot = 80°C or 353 K and Tcold = 25° or 298 K. One must use absolute temperature in the S-B law.
 

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