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The factors that most matter when you are near lava: The fractional solid angle of lava as subtended at the observer ("how much lava do you see") The temperature of the lava The reflectivity of the clothing you are wearing Any effect of air flow (wind blowing towards lava or away from it) Toxic fumes... In essence, if we treat lava as a black body ...


17

I got close enough to slowly flowing lava to stick a rock hammer in it, but you had to pull back quickly -- it felt like a bonfire. It was tolerable 8 feet away. The lava was about 6 inches thick, oozed less than an inch per second and showed orange-red on an advancing toe that was only about six inches in diameter. The rest of the flow was silvery black ...


6

There's actually not one simple answer to your question, which is why you are a bit confused. To specify your problem fully, you must specify exactly how and whether the gas swaps heat with its surroundings and how or even whether it is compressed. You should always refer to the full gas law $P\,V=n\,R\,T$ when reasoning. Common situations that are ...


3

The main problem with your approach is that you are using the wrong area for the radiation "window". The area over which the exchange of radiative energy can take place is just the window in the door - the walls "see each other" and that part of the radiation has no net effect. So you want to use just $H\cdot W$ for the area. Secondly the window is ...


3

The second law of thermodynamics forbids materials that conduct better in one (forward) direction than the reverse direction - such a material placed between two containers at thermal equilibrium would drive the temperature away from equilibrium, decreasing the entropy of the whole system and paving the way for a perpetuum mobile... Reflectance and ...


2

If we assume that both the aluminum foil and the paint have scalar heat conductivities then their composite will also be scalar, and thus symmetrical. Being both material either poly-crystalline or amorphous this is probably reasonable assumption. @Floris suggested that I expand on this, but not being my area I can only summarize a few ideas from ...


2

It's tempting to think that heat will flow more readily across a cavity from a high emissivity surface to a low emissivity surface than in the opposite direction when the surface temperatures are interchanged, but this is fallacious. The reason is that repeated inter-reflection between the surfaces restores symmetry. Call the radiant flux across the cavity ...


2

No. The parcel of gas expands adiabatically as it rises. This makes it cool. Here's one way of looking at what happens: No heat is transferred to or from the parcel of gas (so, adiabatic). At the same time, the parcel of gas expands as it rises. That means the parcel is doing work on the external environment. The temperature of the parcel must drop to ...


2

The definition of a physical concept can be a differential form but can’t be the difference of functions. $\Delta S=S_{final}-S_{initial}$ is an equation but not the definition of entropy. Thermodynamics itself now can hardly explain “what is the entropy really" , the reason please see bellow. 1.Clausius’ definition \begin{align}dS=\left(\frac{\delta ...


2

Experimental data is given in http://journals.aps.org/pr/abstract/10.1103/PhysRev.98.889 - unfortunately I only have access to the abstract. It may be worth taking a look. The shape of the cathode does not matter. The material does. Key to solving this problem is knowing the work function of the material - that is the minimum energy that an electron needs ...


2

Correct me if I'm wrong, but your line of thinking goes like this... Since quantum fields do not commute in general one can have finite variances for, e.g., particle number. Since the vacuum states defines a probability distribution we can find the corresponding entropy. However, here we are dealing with quantum physics. The entropy is in general $S ...


2

If there's hardly any ice at all, the water will cool as the ice warms to 0°C, then cool some more as the ice melts. The cooling stops when the ice melts. If there's hardly any water at all, the water will cool to 0°C. If the ice is cooler than 0°C, that tiny bit of water will freeze. If there's an intermediate amount of water, the water will cool to 0°C ...


2

To see that you are correct, look to radiative heat transfer amongst black bodies. Consider two black bodies, call them bodies A and B, arranged as flat plates facing one another. Suppose body A has a temperature $T_A$ and body B has a temperature $T_B$. Both plates are black bodies, so each radiates energy at a rate given by the Stefan-Boltzmann law: $dE/dt ...


1

I can't think of any source that would claim you need homogeneity to do thermodynamics. Textbooks might usually assume systems have a single constant pressure and temperature because it's easier, but it's not a requirement. Intensive variables can be functions of position. Any discussion of the buoyant force needs position-dependent pressure. Any discussion ...


1

Thermodynamics does deal with inhomogeneities and non-equilibrium conditions... ...but indeed it requires some "macroscopic smearing" of quantities. Problems like diffusion of particles, or heat are dealt with and very well explained by thermodynamics, specifically with Maxwell's Thermodynamic Potentials formalism. But the thing is, the thermodynamic ...


1

Dielectric heating is the priciple of a microwave oven. Water $H_2O$ has a strong dipole moment. Since the water molecule is not linear and the oxygen atom has a higher electronegativity than hydrogen atoms, the oxygen atom carries a slight negative charge, whereas the hydrogen atoms are slightly positive. As a result, water is a polar molecule with an ...


1

Temperature doesn't rise as an object falls. Temperature is just the average kinetic energy of the individual particles within an object, but their individual velocities is random. A water droplet falling is an ordered falling that you can consider as a form of work, and you feel it as the splattering on your feet.


1

The $Q=H$ refers to the system itself, the $2.3MJ$ latent heat rejected by the $1kg$ steam as it condenses to liquid at $100C$ is the same heat that would be measured if the surroundings were at $99.9999999999C$ that process being effectively reversible, and therefore then we would have $\Delta S=0$. But since $Q$ is rejected to the environment that is at ...


1

Here's a point of view from thermodynamics that might be useful. Typically, the intensive quantities (in the form they're usually defined) arise as derivatives of the total (internal) energy $U$ by some particular extensive quantity. Thus: Temperature $T=\frac{\partial U}{\partial S}$, the derivative with respect to the entropy Pressure $P=-\frac{\partial ...


1

If two systems $X$ and $Y$ are in contact such that they can exchange energy, then the statement $X$ and $Y$ are in equilibrium. is equivalent to the statement $X$ and $Y$ are at the same temperature. Here is a proof. The total entropy of the combined system is $$S = S_X + S_Y \qquad (1)$$ where $S_X$ is the entropy of system $X$, and ...


1

Regarding the first viewpoint, it has also been long emphasized that thermodynamics is about studying phenomena related to a property of equilibrium systems called temperature. The equilibrium is always important in all thermodynamics --- although you can make up some measure of temperature for a nonequilibrium system, none of the familiar thermodynamic ...


1

As @DavidVercauteren hinted in his comment and @user31748 pointed out in his answer (albeit with other words), the problem here is that the process undergone by the system is in fact not irreversible: the irreversibilities occur elsewhere. There is a missing piece to the puzzle, which, after some thought, I present below: The key here is that the ...


1

Yes, your analysis is correct. Water in equilibrium with ice is at a temperature of 0 degrees C. The reason that the water doesn't spontaneously turn to ice has to do with the latent heat of fusion of water: in order for water to turn to ice at zero degrees C, you need to remove quite a lot of heat from it. In the case of water / ice, the latent heat is ...


1

The Stirling cycle as you describe it is not reversible. The transfer of heat from thermal reservoirs along paths 4->1 and 2->3 is not a reversible process, because heat is being transferred between two objects at different temperatures. To reverse the process, you would need to spontaneously transfer heat from a colder to hotter reservoir, which violates ...


1

I'd look at it as an energy storage vs loss situation. Take a patch of earth (square slab) and neglect rotation of the earth around its axis (days) so that the patch always faces the sun. At any time it's receiving an incident solar flux (assume constant) and emitting due to its own temperature. The slab also has some thermal mass (capturing the ground, ...


1

More broadly, the free energy functions for phases of a number of elements, including carbon, are gathered in 'SGTE Data for Pure Elements', A. T. Dinsdale, CALPHAD 15(4), 317-425 (1991). This provides a self-consistent set of free energy functions to use in phase diagram calculations. For carbon, the source listed is P. Gustafson, Carbon 24, 169-176 ...


1

Essentially from the definition of thermodynamic equilibrium yes. The Zeroth Law of Thermodynamics states that if a body $A$ is in thermodynamic equilibrium with $B$ and $B$ is in thermodynamic equilibrium with $C$ then $A$ is in thermodynamic equilibrium with $C$ and we say that $A$, $B$ and $C$ have the same temperature. This is precisely the setup you ...


1

This is a very interesting, important and at the same time subtle matter which is useful not only in thermodynamics, but other areas as well, and not everyone quite understands it. First off, two statements that you'd better just memorize for now: In a reversible process, equilibrium is never left. Yes, that is a paradox. That's why all real processes are ...



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