How Long to Heat 100mg of Material Near a Copper Conductor?

AI Thread Summary
To determine how long it takes for a 100mg piece of material with a specific heat of 1 kJ/kg*K to heat to 200°C from a constant Joule loss of 0.25W, the necessary heat (Q) can be calculated using the formula Q = specific heat * mass * (200°C - ambient temperature). The time required to reach this temperature is then estimated as t = Q/0.25W, assuming an isolated system with negligible heat capacity of the conductor. However, in a real scenario, heat loss to the environment and the heat capacity of the copper conductor (0.385) must be considered, which complicates the calculation. To include the conductor's heat capacity, its heat contribution should be added to the total heat required. Accurate estimations may require additional data to account for temperature differences between the conductor and the material.
elcraft
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Hi.

Practical situation.

If I have 0.25W of constant Joule loss around a small piece of copper electric conductor, then I would like to find out how much time until a small piece of "Something" (let's say 100mg at 1 kJ/kg*K) around the conductor heats up to 200°C.

From my calculations, the necessary heat will be Q = Specific heat of the Something (?) * mass of "Something" * (200°C - ambient temp).

So the amount of time to heat this up is supposed to be Q/0.25W. Am I right?

Or is there any other delay or heat loss that won't go into heating the "Something" to 200C? :shy:
 
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If your system "conductor+something" is isolated from the environment and the heat capacity of the conductor is negligible, t=Q/0.25W. A real setup will heat the environment, too, and it will need longer to heat something.
 
Heat capacity for the conductor (copper) is 0.385. I would like to take it into account, but how?
 
Just add the heat capacity (specific heat capacity * mass) to the other one. If the heating process is very quick, your conductor will have a higher temperature than the "something", and your estimate will be a bit off - you would need more data to account for that.
 
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