How fast would a body reach thermal equilibrium ?

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SUMMARY

The discussion centers on the thermal equilibrium of a 1 kg steel cube placed in a room at 20°C. Participants emphasize the importance of Newton's Law of Cooling, represented by the formula \(\frac{dQ}{dt}=hA(T_{\text {room}}-T_{\text {cube}})\), where \(A\) is the surface area and \(h\) is the heat transfer coefficient. Factors such as increased surface area and air movement are acknowledged as significant in accelerating the heating process. However, precise figures and calculations for the rate of temperature change are not provided.

PREREQUISITES
  • Understanding of Newton's Law of Cooling
  • Basic knowledge of heat transfer coefficients
  • Familiarity with thermal equilibrium concepts
  • Ability to calculate surface area of geometric shapes
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  • Research detailed applications of Newton's Law of Cooling in thermal systems
  • Explore methods to calculate heat transfer coefficients for various materials
  • Investigate the impact of surface area on heat transfer rates
  • Learn about forced convection and its effects on thermal equilibrium
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Edi
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For example, I place a simple steel cube weighting 1 kg in a room [filled with air] at 20C [room temperature]. - how fast would the cube heat up to 20C (yes, actually equilibrium would be a bit lower than 20C, but lets.. imagine that is is a infinitely large room and .. yes, you get it ) ?
Yes, I know that increasing surface area would increase the rate of energy exchange and moving air [wind] would speed up the process too.. but I don't know the exact figures and formulas.. :(
 
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Edi said:
For example, I place a simple steel cube weighting 1 kg in a room [filled with air] at 20C [room temperature]. - how fast would the cube heat up to 20C (yes, actually equilibrium would be a bit lower than 20C, but lets.. imagine that is is a infinitely large room and .. yes, you get it ) ?
Yes, I know that increasing surface area would increase the rate of energy exchange and moving air [wind] would speed up the process too.. but I don't know the exact figures and formulas.. :(
I'd use Newton's law of cooling.
[itex]\frac{dQ}{dt}=hA(T_{\text {room}}-T_{\text {cube} })[/itex] where A is the surface area of the cube and h is the heat transfer coefficient (depending on the material of the cube).
 

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