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Is this sentence always true? "The temperature rises as the kinetic energy rises"? If it's not, do we have any limitations for it?

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    $\begingroup$ Does throwing a ball increase its temperature? What's the difference in the motion of a ball and that of a gas of many particles $\endgroup$ – baker_man Jun 2 at 21:03
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    $\begingroup$ Some spin systems only have energy associated with spin up/down. They have temperature, but no kinetic energy. $\endgroup$ – mmesser314 Jun 2 at 21:20
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    $\begingroup$ Increase in KE doesn't always cause increase in Temperature, see en.wikipedia.org/wiki/Latent_heat $\endgroup$ – Negrawh Jun 2 at 23:49
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In Nitrogen, which has a boiling point lower than minus 100 degrees Celsius, when it passes its boiling point, it turns into gas, meaning that kinetic energy has increased.

Not quite sure what you mean by "passes the boiling point". But when it first reaches the boiling point, if it is at constant pressure, there will be no increase in temperature and no increase in molecular kinetic energy during a strictly phase change. Only after all the liquid turns into a gas, temperature increases. But before that happens, the addition of heat increases the distance between molecules and the internal molecular potential energy of the nitrogen.

Now, does this kinetic energy always cause our temperature to rise, or does it just happen to some extent?

After the phase change (liquid to gas) is complete the addition of more heat will increase the molecular kinetic energy and thus the temperature.

Hope this helps.

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Temperature and kinetic energy are manifestations of the same thing. In an ideal monoatomic gas, the average kinetic energy is $\frac{3}{2}k_b T$ ($k_b$ is Boltzmann constant). So the increase of the kinetic energy of your particles (not necessarily a gas) will definitely increase the temperature of your system. In terms of microstates, the temperature can be defined as $k_b T := \frac{1}{\beta}$ where $\beta := \frac{d \ln \Omega}{dE}$ ($\Omega$ being the total number of microstates) so this means that if the increase in the number of states is high when you increase the energy of your system (i.e. $\beta$ is big), then the temperature of the system you are studying is low. This makes sense if you think of it when $T \rightarrow 0$ since there is (generally) just one state at $0 K$, and adding energy into the system will increase a lot the number of your states. Contrary to the case in which you have a lot of energy, if you pump the same amount of energy you did before, if there are a big number of microstates (lots of kinetic energy) the system won't change much, so $\beta$ will be small.

For the case of the ball, it depends. Temperature is a macroscopic concept, which helps you describe (roughly) the evolution of your system in terms of few variables (without the need of knowing exactly the position and momentum of every particle). If what you want to study is the trajectory of a ball, it doesn't make sense to talk about the temperature of the ball (unless you want to study the relation of it interacting with an atmosphere, etc). If what you want is to study how it cools in a fridge, you could define its temperature as before. It all depends on what extent you want to study your system.

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  • $\begingroup$ What's the point of the discussion of a ball. I don't see the OP asking about a ball. $\endgroup$ – Bob D Jun 2 at 23:00
  • $\begingroup$ The post was edited and removed that part $\endgroup$ – BestQuark Jun 3 at 1:28
  • $\begingroup$ Gothcha. OP should not have done that. $\endgroup$ – Bob D Jun 3 at 1:45
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When there is a heat flow to the system, both energy $E$ and entropy $S$ increase.

What happens during the phase change is that the increase in energy and the increase in entropy are linearly related.

Temperature can be defined as: $$T = \frac {\partial E}{\partial S}$$

During the phase change: $\Delta E = T \Delta S$

Before and after that point, temperature increases as heat flows to the system, and that relation is not linear anymore.

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