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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 microcanonical ensemble is the (maximum entropy) probability distribution for a given specified total energy. What you've calculated is actually the maximum entropy distribution with no constraints on the energy, which is the same as the canonical distribution at infinite temperature ($\beta = 0$). To correctly calculate the microcanonical entropy, ...


3

Interesting and complicated question. The things to consider: "Black body radiation" assumes perfect absorption / radiation at all wavelengths. The greenhouse effect comes about from having absorption in the IR: the hot (short wavelength) radiation from the sun can penetrate the atmosphere, but the cooler earth radiates at a lower temperature - longer ...


2

You can combat that the same way we avoid getting too cold: apply insulation. The outer surface of the spacecraft may be very cold, but that doesn't mean the internal temperature is that cold. That is what the greenhouse effect does-it insulates the surface from space. Power dissipated inside the body all becomes heat. If you have solar arrays making ...


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

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

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 ...


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

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 ...


1

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 ...


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

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

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

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

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

It is more complicated than the physics of heat transfers. Our tactile sensations are pretty weird. One example would be that humans can taste "cold" and "cold" affects other tastes. There is not enough research done regarding the processes. In the multitude of cutaneous receptors, you have several that relate to temperature. One type of nocireceptors, ...


1

You may start from the density matrix $\rho$ of the grand canonical ensemble (assuming $k_BT=1$ as the unit of energy), $$\rho=Z^{-1}e^{-H+\mu n},$$ where $Z=\text{Tr}\,e^{-H+\mu n}$ is the partition function. Then $\langle n\rangle$ and $\langle n^2\rangle$ are defined by tracing with the density matrix $$\langle n\rangle=\text{Tr}\,n\rho\text{, and ...


1

The best way to understand the nature of intensive and extensive quantities in thermodynamics is like this: Take a system of your interest. Make it into two portions (one large portion and the other a small portion) by using a partition, for example. Then see the property of interest of the two samples. Density of the two portions will be the same as the ...


1

Was the container of boiling water still being heated when she put the ice cube in? If not, the water will stop boiling even without the ice cube as the container loses heat. If so, you could (with enough room around the ice cube) have boiling water near the container wall and non-boiling water near the ice cube. The water will not all be at the same ...


1

You are doing fine. You are correct to say the sun only shines on one side of the spacecraft, which the page you link to misses. Increasing the radiating area by a factor of six will decrease the temperature by a factor $\sqrt[4]6 \approx 1.565$ Dividing their $285$ by $\sqrt [4]6$ and multiplying by $\sqrt [4]{1.36}$ (to correct for your more accurate ...


1

It may be useful to have a look at the problem from a more quantitative perspective. Let $E$ denote the total internal energy of the system. For a thermodynamic system, it is defined as an ensemble average: $E = \sum_nE_np_n$, where $E_n$ and $p_n = e^{-E_n/k_BT}/Z$ are the energy and the realisation probability of the $n$th micro-state. Here $Z = ...


1

First of all, from a computation point of view, thermal equilibrium is much simpler than out-of-equilibrium. In equilibrium, the state of the system is (more of less) straightforwardly defined by it's Hamiltonian and a few thermodynamic parameter. Out-of-equilibrium contains everything that is not equilibrium and ranges from dropping a pencil to heavy ion ...


1

No. Your answer would be correct only for high vacuum - a density so low that collisions between molecules can be ignored. At atmospheric pressure the mean free path of an air molecule is only about a micron. If the heated volume is much larger than this we can ignore diffusion and thermal conduction in the short run. The constant collisions create what ...


1

Find the integration by applying the limits in the sixth equation. Multiply by c. Substitute $P=cV^{-n}$ twice.



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