How Is Temperature Calculated in a Heated Copper Block?

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Discussion Overview

The discussion revolves around the calculation of temperature distribution within a heated copper block, specifically focusing on determining the temperature at a point a distance x from a heated end after a certain time. The context includes theoretical and mathematical reasoning related to heat transfer principles.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • Froskoy inquires about calculating the temperature at a specific point in a copper block heated at one end to 800K, considering the block's dimensions and time.
  • One participant references a standard treatment of the problem in a specific text, suggesting that it is a well-established topic in heat transfer literature.
  • Another participant proposes an assumption of no heat loss from the sides of the block and introduces the equation for the rate of heat flow, indicating that the temperature gradient is a key factor in the calculation.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the specific methods or equations to use for the calculation, and multiple approaches and assumptions are presented without resolution.

Contextual Notes

The discussion does not clarify the assumptions regarding boundary conditions or the specific dimensions of the copper block, which may affect the calculations.

Froskoy
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Hi,

Suppose a copper block is heated on one side so that one end is at 800K. Given the dimensions of the copper block, is there a way of calculating the temperature of a point in the block distance x from the heated end after a given time?

With many thanks,

Froskoy.
 
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This is a standard problem exhaustively treated in, for example, Mikhailov and Ozisik's "Unified analysis and solutions of heat and mass diffusion".
 
If it can be assumed that there are no heat losses from the sides of the block then the equation for rate of heat flow is:
ΔQ/Δt = k A Δθ/Δx Δθ/Δx is the temperature gradient
 
Thanks! Your replies were really useful.
 

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