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Consider Object A which has a temperature of 70°C and Object B which has a temperature of 20°C. The objects are placed right next to each other.

Now, as far as i know, the temperature is defined as the average kinetic energy of the particles in a system. Again, as far as i know, there are particles colliding inside, resulting in a range of different speed and hence kinetic energy.

So if we are talking about the average kinetic energy, that means there are some high kinetic energy particles in object B (which has a lower temp) and there are some low kinetic energy particles in object A (which has a higher temp) right? If so, couldn't the high temp particles in object B transfer some of its kinetic energy to object A, resulting a temperature change?

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Individual transfers of kinetic energy due to collisions can be the “wrong way round” but on average for many collisions there is a net transfer of kinetic energy from a hot body to the colder body ie there are more “hot to cold” transfers than there are “cold to hot” transfers.

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  • $\begingroup$ Thank you sir, so heat transfers that are from cold to hot are possible? (my teacher never mention this!) $\endgroup$ – Fred Weasley Jan 12 at 7:14
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    $\begingroup$ @FredWeasley When you talk about heat transfer you are referring to the result of many, many collisions which means that there is a net transfer of kinetic energy from a hot body to a cold body. However there will be some individual molecular transfers such that kinetic energy is transferred from a colder body molecule to a hotter body molecule. $\endgroup$ – Farcher Jan 12 at 7:24
  • $\begingroup$ @Fred Weasley just be careful. Referring to these individuals transfers as heat would be anyway uncorrect. Heat comes together with the collective picture in mind. $\endgroup$ – Alchimista Jan 12 at 10:12
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I agree with Farcher. The net affect is an average one.

But I also feel you shouldn't be thinking in terms of "high temperature particles". Temperature is a macroscopic measure of the average translational kinetic energy of all the particles. I could be wrong, but I don't think a single particle has a "temperature".

Hope this, along with Farcher's answer, helps.

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  • $\begingroup$ @PM2Ring The OP certainly did. Look at his last sentence. $\endgroup$ – Bob D Jan 12 at 20:19

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