Why does heat transfer occur at a faster rate at lower temperatures?

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Heat transfer occurs more rapidly at lower temperatures due to Newton's law of cooling, which indicates that the rate of heat transfer is proportional to the temperature difference between two bodies. In the case of heating copper from -30C to +7C, the temperature gradient is greater at lower temperatures, leading to faster heating rates. The specific heat capacity of copper remains relatively constant over small temperature ranges, but black-body radiation effects become significant at higher temperatures. To maintain a constant and linear rate of temperature change, an algorithm must account for these factors, including the surrounding environment's temperature. Understanding these principles is crucial for effective thermal management in engineering applications.
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Hi all,

First of all I'm a hardware engineer and therefore my phisycs knowladge is quite limited...
Secondly my English is not well also so ... please forgive me.

I have a heat body (copper) which I want him to heat from -30C to +7C.

I supply it a constant power of let's say 7W (it has no matter for the discussion).

The phenomenan I see is that (delta T)/(delta t) at low tempertures is much higher then in high tempertures. I mean that for example the time that takes from -30C to -20C is much faster than the time from 0C to 7C.

a. What is the reason?
b. I need to have some algorithm that will keep a constant and linear (delta T)/(delta t) so that the time from -30C to -29C will be the same time (lets say 1 minute) as from -3C to -4C. What are the formulas I need to know in order to do it?

Thanks
 
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a. this may not be very technical but say, the temperature of the environment may also play a part in the heating factor?

b. you could probably that your heat body is heated in an enclosed environment (don't know the thermodynamic terms), since i think the specific heat capacity of copper doesn't vary with temperature, at least not with such a small range (at very high temperatures you may experience difficulty due to black-body radiation)
 
You need Newton's 'law of cooling' which states that the amount of heat transferred between system A and system B is proportional to their temperature difference.

\frac{dE}{dt} = K(T_a - T_b)

this accounts for the rapid transfer at low temperatures. It solves to an exponential for E(t).
 
yes actually the answer should take account both reasons..deltaT and the radiation emmissions which vary propotional to T^4 and that's a lot of difference with a small temperature change. and the surrounding temperature of the environment.
 
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