67

This is more a question of chemistry and biology than physics. Solid objects don't burn (try dropping a lit match on a piece of structural lumber sometime -- it'll just go out). Instead, they release or decompose into flammable gasses on heating, and it's those gasses that burn. Some plants (eg. pines) produce volatile resins that provide an easy ignition ...


23

Water evaporates faster at higher temperatures, so plants will lose water faster in a heat wave than in milder conditions. Drier plants will catch fire easier and burn faster and hotter than moist plants.


8

So the infrared absorption process is already fully saturated, and thus further increases in atmospheric CO2 will not lead to any additional absorption. There is no such thing as a fully saturated absorption process of the $\mathrm{CO}_2$ line. Let $I_{\nu}(\nu)$ be the spectral distribution of the near-infrared emission that an observer would measure ...


7

EDIT – Due to the interest, while my previous answer was OK, I’ve summarised official Australian ABC (public broadcaster) Fact Checks on this subject to provide full and correct detail. The three essentials on how bushfire behaves are: a) weather, b) fuel and c) topography. In Australia, the Bureau of Meteorology produces ‘fire danger ratings’ in ...


5

Some reasons that I think that might be the cause: Higher temperature leads to higher rate of evaporation of from sources of water causing them to become dry and easily ignitable (reason being that wet things require more heat to burn than dry ones as wet things have water in them) Higher temperature lead to the loss in the efficiency of many vital enzymes ...


3

If you can convert all of the heat to work, you're reducing entropy by definition ($\Delta S = \frac{Q}{T}$ , If $Q<0$ then $\Delta S < 0$). If you allow yourself to let some heat flow into somewhere cold (heating something up instead of using all of the heat to work) you raise the entropy in the cold substance enough to let you not defy the second ...


3

The density being constant means you assume incompressibility, which is an approximation. In reality the density of the fluid at the bottom will be greater than at the top. In the case of a liquid this seems confusing, but remember that liquids have a low compressibility (hence the assumption of incompressibility), meaning that $\kappa = \dfrac{\partial p}{\...


3

Yes and yes to your first two questions. Fermions are massless at sufficiently high temperature when the VEV is zero and the symmetry unbroken. They acquire mass when the universe expands and cools, the VEV becomes nonzero, and the symmetry breaks.


3

Just change Boltzmann's constant. The equation you're referring to is probably a version of the equipartition theorem: $$\langle K\rangle =\frac{n}{2} k_BT$$ where $\langle K\rangle$ is the expectation value of kinetic energy of a particle, and $n$ is the number of degrees of freedom of the system in question. Since $K$ is expressed in units of ergs, and ...


2

Is temperature a measure of average kinetic energy (both components) or specifically the translational kinetic energy component? Vibration of atoms in a solid, vibration of atoms in a molecule of a liquid or a gas, and translational movement of molecules in liquid and gases are all translational kinetic energy in my opinion. What change is the mean free ...


2

As you say, this is only true in the classical approximation, when quantum effects are negligible. It is not true for water and ice at $0 ^\circ$C, where quantum effects are significant for the proton. But in melting lead, the classic approximation should be good. Then the atoms in the solid have the same kinetic energy as atoms in the liquid. Similar for ...


2

Fluid dynamics is an effective theory for the long wavelength, long time behavior of classical or quantum many body systems. In this limit the system reaches approximate local (but not global) equilibrium. This is the case because local equilibration takes place on a microscopic time scale (the collision rate between atoms, or leptons, quarks, etc), whereas ...


1

First of all, what was said in the question and assumed wrong: Constant heat capacity means $$\partial_{\rho}\partial_{T}\hat{u}=\partial_\rho \big(\partial_T \hat{f}(\rho, T) - \hat{f}(\rho, T) - T\partial_T\hat{f}(\rho, T)\big) = \partial_\rho \big(\partial_T f_2(T) - \hat{f}(\rho, T) - T\partial_T f_2(T)\big) = \partial_{\rho} \hat{f}(\rho, T) = \...


1

For a body of mass $m_A$, specific heat capacity $c_A$ at initial temperature $T_A$, placed in contact with a body of mass $m_B$, specific heat capacity $c_B$ at initial temperature $T_B$, the equilibrium temperature, $T_{eq}$ reached will be given by $$T_{eq}=\frac{(m_Ac_AT_A+m_Bc_BT_B)}{(m_Ac_A+m_Bc_B)}.$$ This would give an equilibrium temperature ...


1

What I can't understand is how this increase of pressure is not accompanied (or caused) by matching increase of density or temperature. There is a small increase in density. It's just that the pressure increase seems out of proportion when you think about how a gas reacts. Unlike an ideal gas where we assume the particles have insignificant volume, in a ...


1

Adding heat when air drying corn drastically shortens drying times. While the temperature range discussed in the article is lower than in Australia's current weather the general rule is surely applicable, and most likely to other plant parts as well. Not only is the terminal humidity of the plant material lower in hotter air, it is also reached earlier, ...


1

There’s a great explanation for this here. Essentially, you can model rubber as a mess of long, tangled molecules. When the rubber is unstretched, the molecules are more tangled, so in a higher entropy state. Stretching the rubber untangles the molecules, lowering the entropy. At higher temperature, the change in entropy results in more heating of the ...


1

Addressing directly you three inter-related questions: In a classical system, temperature, which is an intensive quantity, does not measure directly the average kinetic energy (translational or not) which is an extensive quantity. The correct statement is that each kinetic energy term of the Hamiltonian contributes by $\frac12 k_B T$ to the average energy ...


1

The fact that a heat engine cannot be 100% efficient is a consequence of the Kelvin-Plank statement of the second law, which can be summarized as Kelvin-Plank Statement of Second Law "No heat engine can operate in a cycle while transferring heat with a single heat reservoir" (my emphasis on cycle) COROLLARY to Kelvin-PLank: No heat engine can have a higher ...


1

From facts about particle physics and thermodynamics, we get an equation of state, which describes the stress-energy tensor. The stress-energy determines the pressure and density, which show up in the Friedmann equations. You solve the Friedmann equations, and you get predictions for, e.g., when the CMB will have had a certain temperature.


1

Global warming potential is given per mole, but you plot the atmospheric column-integrated effect. Methane concentrations are about 2 ppm in the atmosphere, that of $\rm CO_2$ is 400 ppm, still methane manages to make a visible blip in the total atmospheric absorption plot. As to any molecules "why is X a GHG, but Y is not?", the answer is in short "the ...


1

Although pressure increases with temperature, not all pressure is related to it. In fact only for an ideal gas pressure is proportional to T. You can exert pressure on a crystal or, in principle, on liquid helium, at $T=0$ by compression. The returning force is due to electronic Pauli repulsion. At finite T thermal motion adds to the pressure and causes ...


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