# Tag Info

8

A mixed state represents a lack of knowledge on the system (maybe caused by the observer, maybe more fundamental). It is a notion in my opinion closely related to the Bayesian interpretation of quantum theories (seen as non-commutative probability theories). Quantum states are (non-commutative) probabilities, i.e. they encode all the information on ...

7

This answer has been updated and the response has changed! You've discovered a famous problem in thermodynamics. In our case the piston will not move. The mechanical argument is right, while the maximum entropy argument is inconclusive. To see that $P_1=P_2$ is an equilibrium position you can also apply conservation of energy. Since there is no heat ...

7

In physics, heat is energy that spontaneously passes between a system and its surroundings in some way other than through work or the transfer of matter. When a suitable physical pathway exists, heat flows spontaneously from a hotter to a colder body.The transfer can be by contact between the source and the destination body, as in conduction; or by radiation ...

7

Formally, the two entropies are the same thing. The Gibbs entropy, in thermodynamics, is $$S = -k_B \sum p_i \ln p_i$$ while the Shannon entropy of information theory is $$H = -\sum p_i \log_2 p_i.$$ These are equal up to some numerical factors. Given a statistical ensemble, you can calculate its (thermodynamic) entropy using the Shannon entropy, then ...

5

From the original assumptions: So far, my understanding of water evaporating is the following: The higher the temperature, the higher the vapor pressure, therefore the faster water vaporizes. The rate of water boiling, assuming a constant boiling temperature, is dependent on the rate of heat transfer to the water, not the vapor pressure. At sea level, ...

4

I will only make some qualitative and conceptual observations here that does not really deal with the specific problem, but shares some insight on why the two notions in the question are not conflicting. This may or may not answer the question. One way of treatment suggests that since $P_1=P_2$ and and since only mechanichal work (exchange) is allowed - ...

4

If the phase change occurs at the temperature of interest, then the system can give off a lot of heat without cooling down very much. Thus, melting ice is a great way to maintain something at a temperature around 0°C, and melting paraffin-18-Carbons is good if you are trying to maintain temperature around 20 °C. With a melting point of 28°C, the material ...

4

If you take a bottle of gas and carry it with you on a supersonic plane, then the molecules will go much faster without the temperature changing. If you let pressurized gas flow through a well-designed nozzle (De Laval nozzle), the gas will accelerate to supersonic velocity (i.e., faster than the original thermal speed of the molecules) while the ...

4

Conservation of energy relies on the symmetry of your system under time translation (see Noether Theorem). In a system that is not time translation invariant, eg expanding universe, energy doesn't have to be conserved.

4

Here is a simple experimentalist's answer: a) Tin huts in the sun get extremely hot, as anybody who has left a car in the sun will know, much hotter than the outside temperature. b) People in Bangladesh will have about the same IQ as people in other countries, therefore these tin huts will have at least two windows for a cross current in an effor to lower ...

4

The answer is the same reason why a glass of water left out at room temperature will evaporate. Even though most of the particles will be below the boiling point, the equilibrium one expects is not entirely in the liquid phase. The occasional particularly energetic water molecule will vaporize, just as the occasional neutral hydrogen atom will be struck by a ...

3

The simplest answer, if I did not misunderstood your question, is to adiabatically compress the gas, both pressure and temperature will raise.

3

This is certainly an interesting question. As the question is currently put: The entropy argument says that the piston will eventually move. To know how fast it will move, we have to look at the rate of transfer of heat across the piston. If you have a perfectly insulating piston, this rate is zero. Therefore the first answer has to be qualified with: "...

3

I would like to answer with the words of L.D. Landau, from his book Statistical Physics (first edition $1958$):

3

The relations between thermodynamics and quantum field theories is treated in books and papers about non-equilibrium statistical mechanics. Here macroscopic many-particle systems are considered, and the focus is either on equilibrium, or on a dynamical description at finite times. For a readable introduction see, e.g., J. Berges, Introduction to ...

3

We know that: irreversible+adiabatic = $\Delta S>0$, thus, if $\Delta S=0$ the process is either: irreversible+non-adiabatic, reversible+adiabatic, or reversible+non-adiabatic. Then you can conclude that: $\Delta S=0$+adiabatic=reversible. Regarding you example of an irreversible adiabatic cycle: It is impossible and the example is flawed. The ...

3

The hamiltonian of a perfect crystal can be approximated at low temperature as the sum of harmonic oscillator hamiltonians. In 1D we have $$H = \sum_{i=1}^N \frac{p_i^2}{2 m} + \frac 1 2 m \omega^2 \sum_{ij} ( r_i- r_j)^2$$ where the $ij$ sum is over nearest neighbors. It is possible to verify that the eigenvalues of this hamiltonian are $$E_n = \left( ... 3 Suppose you start with your (stationary) 1kg block of gold. If you raise its temperature you have to add energy to it, and that means it's different after you've raised its temperature. For example you could shine an infrared lamp on it, in which case you've added the energy from the IR lamp. The mass changes because if you add an energy E the mass goes up ... 3 For a candle flame, following processes occur. 1. heat transfer (from flame to surrounding and to candle) 2. material transfer (wax vapor diffused outwards and oxygen diffused inwards) 3. heat generation (chemical reaction at stoichiometric mixture location) with gravity, the above will shift due to free convection flow. This will accelerate heat transfer, ... 3 So Pratchett's quote seems to be about energy, rather than entropy. I supposed you could claim otherwise if you assume "entropy is knowledge," but I think that's exactly backwards: I think that knowledge is a special case of low entropy. But your question is still interesting. The entropy S in thermodynamics is related to the number of indistinguishable ... 3 So far there have been quite a few insightful answers about statistical mechanical entropy, but so far the only mention of thermodynamic entropy has been made by CuriousOne in the comments, so I thought it would be useful to give a short general reminder about the subtle difference between the notion of entropy in thermodynamics and the formulas that come up ... 3 Your intuition that the same amount of fluid goes down and then up by the same amount is incomplete, you are forgetting what happens inside the fluid. It is easier to see using solid blocks as in the figure below: Here you can see that the effect of moving block 1 down is to shift block 2 to the right, and moving block 3 back up the same amount that ... 3 The noise from a fan is mostly the vibrations caused by the fan motor, and the rush of air past the fan blades. You can buy a quieter fan. It will cost more. Some new fans have no moving parts, no blades : http://electronics.howstuffworks.com/gadgets/home/dyson-bladeless-fan.htm The obvious easy way to create an air flow (cross-breeze) without creating ... 3 As a rough estimate: we can approximate the shape of the can with a cylinder of radius R=3,2 cm and height h=15,5 cm (those are the dimensions of a Cola can). Its total area can be found by using$$A_T=A_L+2A_B= 2\pi R h + 2\pi R^2$$which gives us A_T=375,8 cm^2. Assuming that water vapor condensation forms a 1 mm thick, continuous sheet of ... 3 The answers boils down to yes, the larger the rate of of heat (assuming you can transfer it at any rate you want), the larger the vaporization rate. The rate of change in internal energy at constant volume is U=U_0+\dot Q\Delta t, where \Delta U is the total internal energy necessary to change the phase, and \dot Q is the rate of heat transfer. Thus \... 3 Toasting not only browns the surface, it dries out and cooks the interior. Any method that applies heat to the surface and waits for it to diffuse into the interior is limited by the rate of diffusion. You can speed it up by a higher surface temperature, but this leads to uneven cooking. Radiation and particles can penetrate to the interior, and can be ... 3 Is this taken to be an additional (and apparently implicit) assumption? You are correct. Take two arbitrary points A,B on the PV (or any other) plane, and draw an arbitrary curve connecting them: you have just defined a reversible transformation connecting A and B. This is because every point in the PV (or any other) plane represents an ... 3 Entropy Demystified (The Second Law Reduced to Plain Common Sense) by Arieh Ben-Naim. Authored discussed not only the thermodynamics origin of entropy but also the same notion in the context of information theory developed by Claude Shannon. 3 We have reached temperature as low as 0.00000001K in the lab and as high as millions of degrees in a nuclear bomb. We achieve such low temperatures by using different techniques. Laser cooling is one of them. In laser cooling, where we fire photons in a certain direction which are accepted and reemitted by atoms in such a way that they retain a component of ... 3 First get data lined up. It seems the container is not small :) volume of water: 4.80\times 10^{-4}~m^3 heat transfer surface area (ignore wood supporting face): 0.038m^2 mass of water: 0.48~kg heat capacity of water is 4.2kJ/kg-K calculate how much heat required for the water to reach room temperature.$$\bigtriangleup Q=m c (27-10)= 34.3kJ ...

Only top voted, non community-wiki answers of a minimum length are eligible