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Temperature is fixed in the grand canonical ensemble. …

is reached, the state of $s$ will be $(N_s^f=N_s^i, V_s^f=V_s^i , T_s^f = T_S \neq T_s^i)$ while the state of $S$ will be unchanged (we are here assuming that $S$ is so large compared to $s$ that its temperature …

Temperature is proportional to the average kinetic energy, thus it is proportional to the average molecular speed squared. … Such a mass of air can be colder than the surrounding air, so the presence of wind can actually be linked to a decrease in temperature. …

Quoting from the article: When the fluid flows over the heated curved surface in proximity of the curved surface as the temperature of the curved surface increases, dynamic viscosity of the fluid at the … Increment of the temperature of curved surface induced the earlier detachment of the jet. …

So, the lower the temperature, the slower the reaction. …

However, the answer to shouldn't fixing temperature cause the energy to remain constant? is yes, if the number of particles $N$ is large enough. … equiv \frac{\delta E}{\langle E \rangle} \propto \frac 1 {\sqrt N}$$ which means that $$\lim_{N \to \infty} \frac{\delta E}{\langle E \rangle}= 0$$ i.e., in the limit of large system size fixing the temperature …

There are two ways to solve this "paradox". The first one is to say, like it was suggested by probably_someone in the comments, that a realistic system will never be perfectly isolated from magnetic …

The expression $$p(s) = e^{-E(s)/k_B T}$$ is valid in the canonical ensemble, where the fixed macroscopic variables are particle number $N$, volume $V$ and temperature $T$. … This means that for a macroscopic system, where $N\approx10^{24}$, the energy fluctuations will become vanishingly small, and fixing the temperature becomes equivalent to fixing the energy. …

But if the environment is at a non-zero temperature, it is going to emit radiation too! … It states that no reversible adiabatic process starting at nonzero temperature can possibly bring a system to zero temperature. …

increases and if we increase the energy the temperature decreases. … Since A has gained some energy, its temperature will decrease. …

for extremely idealized system like a Lennard-Jones fluid there is no analytical equation of state available, and only (very complicated) empirical expressions which are valid in limited pressure and temperature …

The solubility of NaCl increases with increasing temperature (though not very strongly), as you can see from this plot. … This means that at higher temperature more ions will be formed and therefore the resistivity will decrease (i.e. the conductivity will increase). …

The average kinetic energy is nothing else than the temperature of the ideal gas. … Experimentally, it is observed that the temperature has not changed. You conclude therefore that $E$ must be a function of $T$ only. You can find this argument in E. …

The average speed of an air molecule can be approximated by the following equation, which is exact only in the case of an ideal gas: $$\langle v \rangle = \sqrt{\frac{2RT}{M}}$$ This means at $25$°C …

The first thing that must be noticed is that the pressure axis is logarithmic while the temperature axis is linear. …