Recently I saw a video on Youtube in which the guy takes some water in a blender and blends it at a very high speed for several minutes. Simultaneously he measures the temperature of water, which is at room conditions initially($23$ C or $73$ F ). Then, gradually the temperature of the water increases, and at the $100$ C mark, the water starts to boil. The explanation that he gives is that the heat is transferred due to the friction between the blender and the water.

First of all, if we are talking about heat, then the word friction to me is ambiguous. My understanding is that as we rotate the water at such a high pace, the kinetic energy of the water molecules increases, thereby increasing water's temperature. This transfer of energy should be called the transfer of heat. A very similar situation is with gases: a gas at a high temperature, which means at a high average kinetic energy has more heating effect because the average energy of the particles is higher, so more energy is transferred(and therefore more heat).

Now I am not sure if my intuition regarding heat is correct. I can't remember exactly but I read somewhere that the internal energy of a system can be increased without increasing its temperature.

So finally my question is: Transfer of heat is simply transfer of kinetic energy of one type of particle to the other: Isn't this statement true? Or is it something else?

  • $\begingroup$ Yes. Microscopically transferring heat is transferring kinetic energy. You can heat a sample by irradiation, too $\endgroup$ – Alchimista Dec 1 '17 at 13:49
  • $\begingroup$ The term friction they used is ok. They mean the collision between the blades and the water molecules. Is a firm of friction indeed, tough warming up the water was their goal. $\endgroup$ – Alchimista Dec 1 '17 at 13:52
  • $\begingroup$ 1st Law of Thermodynamics: $dU=\delta Q+\delta W$. So it's possible to increase $U$, even if $\delta Q$ (heat) is $0$. But that doesn't mean $dT=0$. $\endgroup$ – Gert Dec 1 '17 at 14:50
  • $\begingroup$ -1 Not clear what you are asking. During a change of state there is heat input but no change in temperature (=KE of particles). $\endgroup$ – sammy gerbil Dec 1 '17 at 15:46

Since temperature and entropy (its conjugate variable) are only well defined for large ensembles (i.e., many particles), it's unwise to define heat transfer in terms of the kinetic energy of single particles.

For example, if I were to reversibly compress a system, I might transfer kinetic energy from one type of particle to another—exactly your statement. Yet no heat transfer is occurring.

Heat transfer is better described as energy transfer driven by a temperature difference.


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