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imagine if I have hot metal piece I want to cool it as fast as possiple, should I put it in strong circulated water bath or just stick it between 2 thick copper plates ?

the calculated heat diffusivity of copper is : thermal conductivity x specific heat x density = 385 x 0.385 x 8960 = 1328096 m²/s

while for water it is just : 0.6 x 4.2 x 1000 = 2520 m²/s only !!

so, the copper can transfere heat 527 times faster than water. should I expect that copper will cool the metal piece faster ? or there is something I missed make water have higher cooling power than copper plates ?

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marked as duplicate by probably_someone, JMac, Gert, stafusa, Buzz Dec 11 '18 at 2:33

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  • $\begingroup$ Water can evaporate/boil, which transfers a large amount of heat away quite quickly, both from the latent heat of vaporization and from the increased mobility of the gas. $\endgroup$ – probably_someone Dec 10 '18 at 17:01
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    $\begingroup$ But the water vapor also creates an isolating barrier which slows down heat transfer again. That's why water drops dance in a hot frying pan and last longer than they should. en.wikipedia.org/wiki/Leidenfrost_effect $\endgroup$ – Hilmar Dec 10 '18 at 17:19
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    $\begingroup$ @probably_someone Not only that, but water can move freely in a liquid state as well. This is also supposedly circulated water, adding another, very significant factor. Madara, look up convection. You seem to be applying the conduction equations to both situations, but moving water allows convection, and thus the effective conductivity is greatly increased. $\endgroup$ – JMac Dec 10 '18 at 18:17
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As a practical matter, a circulated water bath will cool a hot metal piece faster than copper plates in most conceivable configurations. Your calculation of diffusivity is based on an assumption of an immovable conductor, metallic copper or liquid water. This is an unsupportable assumption for water, which will convect and likely undergo a phase change. In your question, you said the water was circulating.

As an engineering problem, in which you calculate heat flux for both situations and compare the results, there is insufficient information provided about the "hot metal piece," including its mass per area of interface, its initial temperature, the mass per area of the copper plate (thickness), its initial temperature, the temperature of the water (assuming a substantially large volume), and its flow rate. The discussion starts to fall apart if the "hot metal piece" is tungsten, in which case the copper would melt.

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The initial heat transfer rate to the thick copper plates will certainly be greater than the heat transfer rate to the circulating water bath, simply because the conductive heat transfer rate to the copper will be substantially higher than forced convective heat transfer rate to the water. But that does not necessarily mean the hot metal plate will reach some final lower temperature sooner with the thick copper plates than with the circulating bath. It will also depend on the heat capacity of the thick copper plates (a function of their mass) compared to the heat capacity of the circulating water bath (a function of the mass of the water bath).

Not only will the heat transfer rate to the copper plates be higher, but the rate of increase of the temperature of the plates will also be higher because the specific heat (heat capacity per unit mass) of the copper is much lower than the water. Once the copper plates reach the same temperature as the hot metal plate between them, heat transfer stops.

The heat transfer rate to the circulating water bath will be lower, but the rate of increase of the temperature of the water bath will also be lower because of the large heat capacity of water. Only when the temperature of the water bath equals the hot metal, will heat transfer cease.

So the answer to the question posed as the heading of your question (initial rate of cooling) is the copper plates. But if your question is which one, the copper plates or water bath, will reduce the temperature of the hot metal to a specific lower temperature faster, the answer may depend on the total mass of the copper plates vs. the total mass of the water bath, and perhaps some shape factors. It may become a matter of solving two calorimetry problems.

And, of course, all of this will depend on the initial temperatures of the thick copper plates and water bath.

Hope this helps.

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