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Thermo-electric coolers use electricity to cool things. Not sure if that's what you're looking for, as it requires physical contact. But in general, it's a lot easier to make heat than to remove it.


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Convection requires the hot air just above your soup that has been heated up to rise as its density reduces after being heated, allowing the colder air around the rising hotter air to replace the hotter air and the same thing will happen again. As the convection currents (the rising of the hot air and the sinking of the cold air) keep taking place just above ...


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When you touch something, you don't feel how hot/cold the thing is; you feel how hot/cold it makes your hand. Metal conducts heat more easily than wood. So if wood and metal are hot, the heat will flow more easily from the metal to your hand. If wood and metal are cold, the heat will flow more easily from your hand to the metal.


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The two materials possess different thermal conductivities. The metal appears to have a higher temperature because the heat escapes it into your fingers quicker than the wood. http://en.wikipedia.org/wiki/Thermal_conductivity italicized part edited to not confuse future readers.


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I managed to figure this out! It was done using the Clausius Clapeyron equation. $\frac{dP}{dT}=\frac{L}{T(V_2-V_1)}$ By letting $L=\frac{h_f}{m}$ and $V_1=\frac{1}{\rho_1}$,$V_2=\frac{1}{\rho_2}$ gives: $\frac{dP}{dT}=\frac{\frac{h_f}{m}}{T(\frac{1}{\rho_2}-\frac{1}{\rho_1})}$ By solving this differential equation it is seen that: ...


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There are several ways how the canonical ensemble can be derived (or maybe better say justified) in classical and quantum mechanics. Possible ways include: From the micro-canonical ensemble: This is essentially the approach outlined the answer by Sebastian Riese and addresses most directly your question. The short answer is yes, this is possible. However, ...


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I think that the main effect is that because humidity is increased, your sweat does not evaporate much when entering the bathroom. Since the evaporation of sweat makes you feel colder (that is what sweat is for!), you would feel warm in the bathroom. Using hot water may increase the temperature of the room as well.


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The answer depends on the nature of the "huge oxygen bottles"... You often see large ( around two stories high) tanks outside a hospital. They are distinguished by the name of the chemical firm, Union Carbide or Linde among others, painted on the side, and by the thick layer of frost found on the attached plumbing. These tanks hold liquid oxygen at low ...


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Using the diameter of the muzzle $d$ and the initial distance between the bullet and the end of the muzzle $l$ we can calculate the volume of the gas to be given by: \begin{equation} V\left(l\right)=\frac{\pi d^2\left(L-l\right)}{4} \end{equation} where $L$ is the total length of the muzzle. We therefore find from the ideal gas law that: \begin{equation} ...


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In some high-demand medical facilities, liquid oxygen at cryogenic temperatures (in tanks) is used (http://208.76.246.34/~ava/Oxygen-source.pdf). More often, gaseous oxygen in metal cylinders is used. The pressure may vary, but the above source gives 13700kPa (for UK).


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There is a method I like. It appears in the book H. Schulz: "Statistische Mechanik" (Harry Deutsch, 2005) (to my knowledge only available in German). It runs as follows: Consider a bath weakly coupled to a system, together they are prepared micro-canonically at energy $E$. The Hamiltonian of the system have only non-degenerate eigenenergies (otherwise we ...


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The sign is governed by the convention - whether the volume is of your system in consideration or not. If you decrease the volume of your system - you increase the energy of your system, so you require for total energy change to be positive. If the volume is describing your system then $ dV <0 $ and so $dE=-PdV>0$ is the correct expression If the ...


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You do not need any of those formulae. Visualize an ideal fluid element (no viscosity) in its rest frame. If it is incompressible, its volume does not change and all work done on it by pressure forces in any time interval equals change in its kinetic energy. If the flow is also adiabatic, no heat is transferred from one element to another. Since no work and ...


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Firstly, your are correct that in the second order phase transition the correlation time diverges, and there is a dynamical critical exponent related to it. By definition, the correlation time is defined by the time interval that two states are not correlated (similar to the definition of the correlation length), say they have quite different configurations ...


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The equation: $$ dS = \frac{dQ}{T} $$ only applies to reversible processes. For an irreversible process $dS \gt dQ/T$. To see this start with the expression for the change in internal energy: $$ dU = dQ - dW $$ The internal energy is a state function, so this equation always applies whether the process is reversible or irreversible. So for a reversible ...


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The change of entropy is Q/T only if the heat transfer is reversible if the process is irreversible you can't obtain the change of entropy through the formula Q/T After the heat transfer the change of entropy o the whole system is $\Delta S>Q/T-Q/T_0>0$


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Yes. Just for the other system, which is gaining heat. Entropy is lowered for the system loosing heat (negative $Q$), but is increased (even more than what was lowered) for the external system gaining heat (positive $Q$). $$\Delta S \geq \int \frac{dQ}{T}$$ Usually I have seen it denoted as $$\Delta S = \int \frac{dQ}{T}+\sigma_{gen}$$ where the ...


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I'd say it all comes down to your physical setup. When studying the Ising model one is usually interested in its behaviour at certain values of temperature and volume. On the other hand, the liquid-gas transition is usually described in a pressure-temperature diagram, i.e., at different but fixed pressures and temperatures. Bottom line: the choice of free ...


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There are some excellent recommendations above, but the best introduction to classical thermodynamics I've found is the first 70 pages of J. S. Dugdale's Entropy and its Physical Meaning. He follows the historical development of the subject, including page long quotes from the likes of Fahrenheit, Joule and Carnot. He develops the idea of the Carnot engine, ...


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It is true that there is no correct a priori way of deriving the "correct" entropy as the notion of "correct" will depend upon what we mean by this word. However, if statistical mechanics introduces a quantity that has the same name as the most important quantity of thermodynamics, I suspect it has, at the very least, to carry a meaning very close to the ...


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The heat flow (per unit area) through some thin layer, e.g. a boundary layer of water, is given by: $$ \frac{dQ}{dt} = \frac{K\Delta T}{d} $$ where $K$ is the thermal conductivity, $d$ is the thickness of the layer and $\Delta T$ is the temperature difference between the two sides of the layer. So a high thermal conductivity does indeed mean a high heat ...


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From an engineering perspective, there are many different fan designs including axial and centrifugal configurations, along with various blade designs including forward curved, backward curved, and radial. There is no formula to calculate the required fan speed, but if a specific fan configuration is known then fan similarity laws can be used to calculate ...


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The incoming waves will terminate against each other in the corners of the square pipe, meaning there is a deficit of heat in the faces of the square. If that's true, we should add more corners, and waste less material. A square has 4 corners, with not a lot of heat on the center of the faces. An octagon has 8 corners, with $ 1 \over 2 $ of the ...


1

the shape and area of the cross section of the pipe can change flow property along the stream. The fluid velocity in a pipe changes from zero at the surface because of the no-slip condition to a maximum at the pipe center. In fluid flow, it is convenient to work with an average velocity which remains constant in incompressible flow. Laminar flow in a round ...


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Using a dimensional analysis indeed can help you find dimensionless numbers as The Dark Side suggested Floris might help with, but beyond that there is no direct analytical method, no closed form solutions that can relate fan speed, shaft torque, flow rate and delta pressure. The issue is that at best the flow behavior is two dimensional, but more likely ...


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updated calculations - based on neutrino energy escaping and vapor inhalation risk Your math is close but not quite right. First - the number of tritium atoms. There are 1000/(16+3+3) = 45 moles (as you said) This means there are 45*2*$N_A$ = $5.5 \cdot 10^{25}$ atoms of Tritium Now the half life is 12.3 years or 4500 days, that is $3.9\cdot 10^8 $s. ...


1

Assuming we can treat the air in the room as an ideal gas, it will obey the ideal gas equation of state: $$ PV = nRT \tag{1} $$ where $n$ is the number of moles of the gas. The question tells us that the pressure is constant, and obviously the volume of the room is constant, so the only things that can vary are $T$ and $n$. The question tells us that $T$ ...


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The growth of internal energy or change in internal energy can be determined by the following equations $\Delta U = W + \Delta Q$ or $U = nc\Delta T$, with c being specific heat at constant volume Now, W is zero since the volume hasn't changed ($W = \int P dV $), so all that is needed is to determine change in heat energy. Heat energy added is equal to ...


1

You can pump heat from cold objects to hot objects if you pay some more energy (that's what your refrigerator is doing) and that doesn't violate second law of thermodynamics. You should note as you heat object, its thermal radiation will increase. Intensity (that is power per unit surface area) of thermal radiation is proportional to $T^4$ so when the ...


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A state is thermodynamically stable when its Gibbs free energy is at a minimum. $$G=U-TS+PV$$ Holding all variables but $P$ and $V$ fixed, it means that: $$dG=dP V+PdV=0 \implies {dP\over dV}=-{P\over V}$$ Since neither of $P$ nor $V$ could be negative ${dP\over dV}$ must be negative.


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From a purely thermodynamical point of view, why does that entropy have to be a maximum at equilibrium? Say there is equilibrium, i.e. no net heat flow, why can the entropy not be sitting at a non-maximal value? The statement that entropy has maximum possible value in equilibrium means this: given imposed constraints (volume, total energy, molar ...


1

In your case, heat conduction is considered to be quasi-static for the subsystems. Let $\Delta Q$ be positive number, energy transferred as heat. Then, change of entropy $S_2$ of the subsystem 2 is $\Delta Q/T_2$, change of entropy $S_1$ of the subsystem 1 is $-\Delta Q/T_1$. For the whole system containing both subsystems, the change of entropy is $$ ...


1

"typical" household freezers don't do minus 20. A premium dedicated freezer (no fridge part) might, but I don't think you want to buy one just for this. For a low-budget solution, get a styrofoam cooler, put your regular-temperature ice in it, and then add 2x mass dry ice. You will eventually get colder ice. Do cut a small drain hole in the bottom of the ...


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Don't quite know what you mean by "water at -20 degrees", as I would expect ice at these temperatures. But anyways you can reach such temperatures with a cooling bath


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Some domestic, commercially marketed deep freezers are able to reach that range of temperature. Here is one for example that advertises -20deg F, so almost -30 deg C. It really depends also on the amount of mass you want to lower the temperature to. Whatever device to be considered must provide outward heat flow to compete with whatever heat influx the ...


0

Entropy of a system can increase either by entropy transfer or due to entropy generation. If we live in a world where every process is reversible then entropy will never be generated only transfered from one system to other. Entropy generation is due to irreversibilities in a process.Lets consider water taken inside an adiabatic container we do some work on ...


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Irreversible heat loss has the implication there are two systems in contact with heat being transferred irreversibly from one to the other. In this case the total entropy of both systems should increase: the entropy of the heat donor system will decrease, and the entropy of the heat-receiving system increases by more than the loss in the first system.


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I think I understand your confusion. The answer would be: You've been misguided. There is no special link between heat and infrared radiation, except for the fact that most bodies radiate most of their heat in the infrared spectrum because they don't have enough energy (heat) to radiate at a higher frequency. See the graphs in this thread. So one could ...


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Yes there is, using electrostatic forces. See the details here: Electrostatic Fluid Accelerator I should add that it can't lead to a high compression ratio like a mechanical compressor, but technically it does compress the gas in the vicinity of the charging source. And that leads to flow. A perhaps more interesting followup question: can you conceive some ...


0

It might be possible to submerge a small enclosed canister under water at a very low depth so that the water pressurizes the canister and therefor pressurizes the air inside. or it might just explode. if it explodes try a more flexible material for the canister. if this is not the answer you were looking for then im sorry.


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You are correct in your proof. However the only situation in which both inequalities are satisfied is when the latter one is equal to zero. In this way you are saying that it is only zero for the reversible path. If the Clausius statement is violated then the limit of the composite system of Carnot engines is violated. This then comes back to why is the ...


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You will get most likely get uneven heating that is hard to reproduce - so I would say "no, that is not a good approach". Using a thermal bath like @BySymmetry suggested is much better - or wrap some resistive wire around it and run a known current through it for a known time. The key to good experiments is control and repeatability - your open fire solution ...


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The effect of salt etc. on the heat capacity is probably negligible, and the question in the header may be misleading. Cooling down water in a beaker is usually driven by convection. Liquid on top of the solution in the beaker cools quickly, and then has a higher density than the hot liquid at the bottom. The cold liquid falls down, hot comes up, cools, ...


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Try using Stephan-Boltzmann law to answer this question, here's a link that might help you: http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html


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You are confusing time with the flow of time. Time is just a coordinate like the spatial coordinates, that is we label spacetime points with four coordinates $(t, x, y, z)$. Indeed, in relativity (both flavours) the time and spatial coordinates get mixed up so different observers will disagree about what is time and what is space. But the obvious thing ...


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Entropy is a state variable. That means for every state you have a single value for entropy of that state. So the entropy difference of two states does not depend on the process that takes you from one state to the other. Therefore you can use this formula if the initial state is $(T_0,V_0)$ and final state is $(T,V)$. However the change in the Entropy of ...


1

Water is a complicated phase. I would expect a simplistic analysis to break down because water is really a quasi-crystal at loser temperatures. It tends to form cage-like structures because of hydrogen bonding. Like @SebastianRiese I think the argument that "there is little work done by the system in solid and liquid phase. The heat capacity must be ...


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Spilling supercooled liquid on your skin would actually be less painful than spilling a similar liquid that wasn't supercooled. This is because when a supercooled liquid freezes it gets warmer due to the latent heat of crystallisation released by the liquid to solid transition. So suppose we had some supercooled water and some other hypothetical fluid at ...


3

The rapidity of heat loss from black body radiation depends only on the temperature of the body and the difference to the temperature of the environment according to the Stefan-Boltzmann law. which describes the power radiated. This will happen in any case for any body immersed (first version of question) in a supercooled liquid if its temperature is higher ...


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Today I heard a thermodynamic argument for about this. Since there is little work done by the system in solid and liquid phase. The heat capacity must be (roughly) same for the solid and liquid phase. This one does not convince me at all. Especially, because no work is done by the system in the case one considers the constant volume heat capacity ...



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