# Tag Info

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What this demonstrates is that the Earth's crust must have an incredibly high insulating function.

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Foil is used to create a reflective surface, reducing heat loss by radiation. Also a covering which will reduce heat loss by convection. A heat sink transmits heat by conduction.

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The force balance on the piston at any time during the compression is given by: $$P_gA+F-mg-kx-P_{atm}A=0$$ where x is the upward displacement of the spring from its unextended length. If we multiply this by the differential (upward) displacement of the piston during the process $dx=\frac{1}{A}dV$, we obtain:$$P_gdV+Fdx-mgdx-kdx-P_{atm}dV=0$$Integrating ...

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This is not an answer, just a cool phase diagram of distinct topological phases in the Haldane model with extended hopping. I found it here https://arxiv.org/abs/1212.4046 This is not thermodynamics, so it probably doesn't answer your question.

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In general there are several ways in which heat can be transferred from two bodies. Some of them are: Advection is the transport mechanism of a fluid from one location to another, and is dependent on motion and momentum of that fluid. Conduction by contact Heat moves between bodies placed in physical contact one to the other. Aluminum has a high thermal ...

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Your system is the spring and the gas. When the first law is written in this form, $\Delta U= Q-W$, then $\Delta U$ is the change in internal energy of the system which includes the spring ($U_{\rm final} -U_{\rm initial})$, $Q$ is the heat input to the system (positive if into the system and negative if out of the system) and $W$ is the work done by the ...

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The equation of state for a thermodynamical system with a fixed number of particles takes the form $$f(p,V,T)=0$$ for some function $f$. For example, in the case of the ideal gas, one has that $$f(p,V,T) = pV-nRT = 0$$ As a result, $p,V,$ and $T$ are not all independent of one another, in the sense that you are not free to specify all three independently. ...

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This experiment (with tea and milk) seems to show that immediate addition of the milk gives the warmest drink: But the experiment is far from perfect, especially because of the lack of replication. And this simplified derivation, based on Newton's Law of Cooling finds the same. But it too is very open to well-reasoned critiques. And here's my own ...

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The specific heat at constant pressure is not a function of volume due to the fact that it is defined in terms of the change in enthalpy with temperature at constant pressure. The specific heat at constant pressure is defined by $$c_{p}=\biggl(\frac{\delta h}{\delta T}\biggr)_P$$ Specific enthalpy is $$dh=du+d(pV)=q-pdv+pdv+vdp$$ For a constant pressure ...

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Specific heats of solid are generally constant upto good ranges of temperature. At higher temperatures specific heats do become a function of temperature ( according to Einstein and Debye Model) However say if $C_p$ is a function of Pand T then the equation of state can help to calculate $C_p$ as a function of T and V. Similarly for $C_v$.

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In black hole physics the term “lukewarm black hole” refers to asymptotically de Sitter black hole solutions that have equal temperatures of the black hole and cosmological horizons ($T_b = T_c$) (generally, the temperatures are different). For example, for solutions from Reissner–Nordström–de Sitter family the black hole would be lukewarm when its mass and ...

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@J. Murray's answer gets the radiation part right. As @MSalters pointed out in a comment, assuming 100 W of heat generated means the outside of the clothes need to be approximatley -70˚C (with emissivity 0.74). With body temperature 37˚C that would mean a difference of 107 K. The heat conducted through the clothing layer $Q$ is given by: $$Q=\frac{kA}{d}\... 44 A super-thick sweater probably isn't the way to go - you may be better off wrapping yourself in aluminum foil. The body loses heat through a handful of mechanisms: During conduction, your body transfers heat to the surrounding air which is in contact with your skin. This raises the temperature of the air, which (if the air is still) decreases the rate of ... 1 I will adopt a different viewpoint than other answers. The flow of time can be used in relativity. Some answers above and this answer to Does it actually take infinite (observer) time for someone to fall into a black hole? point to time as in the Block Universe. Space-time is an unchanging 4D block. Time does not flow. According to this answer to ... 0 The question is an old conundrum and can be found in various guises on the Internet and in handbooks. One can summarise it as follows: If I add milk to my coffee and wait 5 minutes before drinking it and another person waits 5 minutes and then adds the milk to his/her coffee, who is drinking the hottest coffee? A theoretical derivation of the end ... 0 I will try to answer this restricting myself to quantum critical points. I'll be trying to connect a gap closing of the system with infinite correlation length. In general a system with a finite gap between the ground state and the excited states will have a finite length correlation lengths. This link, at least in my mind, can offer a way out of the ... 1 No, it doesn't, even in the case of a perfect wire with no heat loss. Here is why. To create a magnetic field requires the expenditure of work. In the case of a coil of wire, the work to create the field is done by the time integral of (voltage x current). Once the current has achieved steady state, that current is then required to sustain the field. When ... 1 To address something nobody yet has, though not answering the question (as I can't top the others): In fact, it ends up so cold that it feels much colder than I would expect given my relatively warm room (\sim 25^\circ). However, this could be an illusion caused by room temperature still being less than body temperature, or the mug being fairly cold. ... 0 In this problem all three modes of heat transfer occur: radiative conductive convective Firstly, we assume the furnace operates in steady state, so that all heat generated by the LPG burner dissipates away through the furnace's wall. In that case, all temperatures are constant in time. Also I'll approximate the furnace by a single wall with surface area ... 0 The second law is the probabilistic statement. Think that there are only two gas molecules inside the box. In the beginning, the two gas molecules will stay at the corner. You will pretty often see two molecules get back to the corner again and again if you have the small enough box. You better see the second law of thermodynamics as an asymptotic statement ... 1 As it's been asked about "fan", the fan can't cool anything by itself (imagine using a fan inside vacuum XD), it brings the fluid outside the container to motion hence helping to transfer the heat from container to the moving fluid(air in usual cases). If we use much more cooler fluid we must be able bring down the temperature significantly. 0 When there is a heat flow to the system, both energy E and entropy S increase. What happens during the phase change is that the increase in energy and the increase in entropy are linearly related. Temperature can be defined as:$$T = \frac {\partial E}{\partial S}$$During the phase change: \Delta E = T \Delta S Before and after that point, ... 2 In Nitrogen, which has a boiling point lower than minus 100 degrees Celsius, when it passes its boiling point, it turns into gas, meaning that kinetic energy has increased. Not quite sure what you mean by "passes the boiling point". But when it first reaches the boiling point, if it is at constant pressure, there will be no increase in temperature and ... 1 Temperature and kinetic energy are manifestations of the same thing. In an ideal monoatomic gas, the average kinetic energy is \frac{3}{2}k_b T (k_b is Boltzmann constant). So the increase of the kinetic energy of your particles (not necessarily a gas) will definitely increase the temperature of your system. In terms of microstates, the temperature can ... 3 A quick explanation of how evaporation cools your coffee. Let's assume that enough time has already passed so that the coffee is at room temperature. How can it get even cooler? Although the coffee may be at room temperature overall, not all of the molecules will be at the same temperature. Some will be hotter (have more energy) and some cooler (less energy)... 0 Almost 2 months later, I seem to have found a fitting answer to my own question. My question was aimed at how pressure is different from heat, if both involve vibration of paticles.   https://youtu.be/v3pYRn5j7oI And there he is, Dr. Richard P. Feynman, telling us how :) It turns out, that the extent of vibration is what differs across the two, heat ... 0 For general processes – reversible or irreversible – there are two complications with respect to thermostatics: 1 – The notion of "state" is generalized. A state at time t may be given not just by the values of some set of thermodynamics (or thermomechanics) variables at that time – say, \bigl(V(t), T(t)\bigr) – but by the full history of those values ... 2 Other answers here are interesting, but here is a simple version, without formulae. Consider the hypothetical situation that the drink has already cooled to reach equilibrium, ie it has already cooled to room temperature of 25C. The fan then facilitates evaporation of one more water molecule, which will be one of the more energetic (hotter) molecules. So ... 4 Take the Ising model in dimension d\ge 2 for concreteness. Let \langle\cdot\rangle_{\beta,+} denote the plus state at inverse temperature \beta and zero magnetic field. This is obtained by putting the interaction in a box with boundary condition given by +1 spins, and then taking the thermodynamic limit where the box fills the entire space. The OP's ... 0 I don't know if you have to memorize these terms for your studies, but the idea of "quasi-static" process is somewhat obsolete and misleading. The division between "reversible" and "irreversible", on the other hand, has a clear mathematical definition, involvig a time-reversal operation (which consists in replacing t by -t and making some so-called "... 1 In your flow chart in the quasistatic branch you need to ask is there friction or no friction. If has friction it’s irreversible. If no friction it’s reversible. So to be reversible it must be quasistatic with no friction. Hope this helps 5 I used my hand sanitizer(thanks to COVID) which is the alcohol of 70%v/v will surely an easy experiment to demonstrate it, and my result was yes the temperature of the liquid is below the room temperature, and I feel it depends upon volatility of the liquids which cause the process of evaporation , and cool down the liquid , in your example, the temperature ... 40 Yes, as other answers have stated, the temperature could drop below room temperature through evaporative cooling. In fact it could get as cold as the wet-bulb temperature of the air in the room. If you know the temperature and humidity of the air, you can figure out the wet-bulb temperature by using a psychrometric chart: Find your room temperature on the ... 3 Typically a convective boundary condition is of the form$$ Q = h \cdot A \cdot (T(t) - T_\text{env}) = h \cdot A \cdot \Delta T(t) $$(where h can be a function of fluid velocity if necessary) so it is not really possible that heat can be transferred past the environmental temperature(if T is bigger than Tenv we have cooling if the reverse we have heating).... 1 I don't see how it can be cooled below room temperature. But if the environment is dry and large enough, the fan can remove molecules from the liquid, increasing the water vapor content in the air. In that process, the air may cool, transfering energy to evaporate the water. Eventually the room can be cooled. As a consequence the drink can reach ... 5 There are two types of liquid i.e. volatile & non-volatile liquids. Consider two cases for these liquids Case 1: Let the liquid be volatile say water. The liquids are mostly volatile. The fan blows air over the liquid surface. This makes the liquid to evaporate & taken away in form of vapors by blowing air. In this process, the liquid molecules on ... 14 Yes, it can become cooler than room temperature. The fan blowing on the liquid surface acts like the evaporation stage in a refrigerator. There's no recycling of the vapour by condensation as in a refrigerator, but that's clearly not important here! It's no doubt possible even to freeze the liquid, but this is more easily demonstrated by blowing air at room ... 0 I think it's always good to keep in mind that these "accessible" microstates and their "increase" or "decrease" are not real physical properties or processes of the system. Strictly speaking the system has only one accessible microstate at time t+\mathrm{d}t: namely the one determined by its microstate at time t and the equations of motion. The ... 1 I think the main confusion here is that you use \Omega and the entropy interchangeably. \Omega is not directly proportion to entropy. Rather S, the entropy, is proportional to \log \Omega. Let’s call the two systems you have A and B. The entropy is additive S_{tot} = S_A + S_B. However the total number of states isn’t, rather we have \Omega_{... 1 Nothing is absurd here. When writing$$\Delta U+\Delta \text{KE} +\Delta \text{PE}=W+Q$$the work term W is work done by non-conservative forces as well as work done by any external conservative forces you haven't included in \Delta \text{PE}. This is also true in your "classical mechanics" expression$$\Delta \text{KE}+\Delta \text{PE}=W$$The W ... 0 I assume they take the heat capacity to be constant from 4 to 100 degrees. This sample of water can take up some heat \Delta Q = c_p m \Delta T . So now the question is: what does the heat do to the sample ? For an isobaric process it can increase its temperature and its volume. Both processes contribute to the uptake of the heat, which is described by ... 0 True. It's very easy to explain. Thermal conductivity ratio between water and air is about 23 times ! This means water transmits heat 23 times better than air - you can do nothing about about. So if you want to cool matter fast - you need to search for a maximum thermal conductivity materials. 0 Your question is interesting, but it is ill-posed because it contains some wrong definitions and some circular elements. First of all, the thermodynamic entropy of a system depends on the description of the system. For example, if you describe air as a gas, it has one thermodynamic entropy; if you describe it as a mixture of gases, it has a different ... 0 This is actually a really profound question. If I understand Boltzmann's answer to this question, it is that indeed, if you were on the downward slope of the fluctuation, you would see time running backwards. However, were you a creature that came into being during that process, it would seem natural to you that the "future" causes the "past". Since we are ... 2 To make water evaporate faster, you need some means by which you can increase the vapor pressure of water. In the case of salt, dissolving it in water decreases the vapor pressure of the water, by furnishing ions in solution whose company the water molecules enjoy (in a manner of speaking). Since the ions themselves cannot evaporate, this slows down the ... 6 The heat transfer from 'hot' air or water to 'cold' sausages is roughly determined by Newton's law of cooling/heating:$$\boxed{\frac{\text{d}Q}{\text{d}t}=hA[T_{\infty}-T(t)]}\tag{1}$$where, for heating: \frac{\text{d}Q}{\text{d}t} is the rate of heat transfer into of the body, which determines how quickly the body's temperature rises, h is the heat ... 27 I do exactly the same. It is a very effective way of defrosting food fast. Compared to air water has a much higher heat capacity and a much higher thermal conductivity. That means heat flows from the water into the sausages much faster than it would in air and the water cools less than air would as it heats the sausages. Aluminium foil has a much, much ... 2 I believe it has to do with the speed of heat exchange between sausage and air in one case and sausage and water in the other. Also, frozen sausage essentially means that the water in this sausage is turned into ice - you can test that a block of ice in water melts faster than in air. The most efficient technique is microwave in low power mode (if the ... 1 This is more a computational question than a physics one, and the only reason I'm writing this as an answer is because it's a bit too long for a comment: to me your problem seems to be in the ToVSolve function. def ToVSolve(pc,x_c): # Initialization of m(r), p(r) and r (dimensionless) dr=0.0024 r = pylab.arange(0.0012,2.4,dr) #1.2e-3,2.4,dr m, p = pylab.... 0 The chain rule result is incorrect. It should be:$$= \frac{NkTq^{N-1}}{q^N}\left (\frac{\partial q}{\partial V} \right)_T 

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