# Tag Info

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Another possibility is that a fan would "de-stratify" the air in the room. If the temp sensor was fairly high up a fan would mix the hot and cooler layers of the air in the room making the sensor a bit cooler than it would normally be.

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A fan moves air around. It makes people feel cooler, by causing evaporation of skin moisture (sweat). A fan's motor also gets hotter. Air moving over a thermostat would have no affect (thermostats don't sweat), but the increase in temperature of the fan's motor, would increase the air temperature slightly, causing the air conditioning to work harder. If ...

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The microwave answer given above is good, especially if you have only one paper wrapped in foil because it would transfer a large fraction of the energy produced to the sample. If you have many of these (for example as a step on an assembly line) then immerse it in a hot medium. This would provide really efficient transfer of heat energy for each sample ...

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Heat a batch of high temperature silicone to 250°C, then drop in your foil wrapped paper.

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Put it in a microwave at the right power, and it will quickly heat up withouth burning (and heat whatever is inside). But do not try this at home. Too much power and you will get the microwave oven on flames.

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Not sure how hot you want it, but a hairdryer or heat gun would be quick, but not very energy efficient while it was blowing - but it turns on and off quickly so might be efficient by not having to be on for long. edit - ok for 250 degrees C - heat gun probably not hot enough. I would put an oven on at close to max temp and put the foil in, but even then ...

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However, isn't any closed loop on a PV diagram reversible? The arrows can simply be drawn in the reverse way to create a refrigerator. If any closed loop is reversible then why does the specific Carnot engine (a specific loop) have the highest efficiency? This was exactly the question I asked myself ten years ago :-) The problem is that often students ...

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The problem you are trying to solve is different from the standard textbook situation where $A$ and $B$ are two heat reservoirs. When considering equilibration of temperatures, you have to account for the finite heat capacity of the two objects. This allows you to consider a series of quasi-static reversible heat exchanges with a series of heat reservoirs, ...

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Here is a link to a study comparing heating water in a microwave to heating water in a conventional oven. Depending on the power of the microwave, the volume of the water, and time it's placed inside, the temperature will vary approximately linearly with time until either the system reaches equilibrium (for low power microwaves and large volumes of water) or ...

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If it is adiabatic then $\Delta Q_i$ will be always zero. The fact that it is irreversible doesnt matter. Any path that thakes you from A to B will result in the same change of entropy, as both initial and final states are in equilibrium. If you choose what is called a quasistatic path, which is idealized as a tranformation that occurrs slow enough so that ...

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In an ideal Stirling cycle, the isochoric steps have heat exchanged across an infinitesimal temperature difference, which is maintained by the regenerator having a continuous gradient of temperature between the hot and cold reservoirs. The gas can then cool or heat in alignment with that gradient. This is the very ideal part of the design that enables zero ...

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Of course I wasn't asking this question in order to answer it myself but because I discussed about it with some other "érudits" that were puzzled. But eventually, one pointed me the figure (sorry it's in french but you get the point) http://upload.wikimedia.org/wikipedia/commons/b/ba/P-V-T_Diagram_%28Water%29.fr.svg where the mixed phase zone are parallel ...

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Actually, that is not the correct formula for gases. In the usual model of ideal gases, the volume is given as a function of both temperature and pressure by the equation: $$V=\frac{n\bar RT}{p}$$ Both formulas you wrote are valid for solids and liquids (but not for gases). In those phases, volume is much more affected by temperature than by pressure, so ...

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$V_1/V_1$ units will get cancelled. Substituting the dimensions of remaining units, $$\frac{\text{length}^3\times \text{Force}}{\text{length}^2}=Force\times length$$ The formula for work too is $Force \times displacement~$. Since the unit of Work is Joules, The unit of $Force\times length$ will be Joules too

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Yes, the question is theoretical and so the response. Under enough pressure water will become a solid, regardless of temperature. That is, as far as it is still water. If pressure is high enough, the atoms will collapse and form neutron-degenerate matter (theorized to exist in the cores of neutron stars). I am not sure if there could be an intermediate mixed ...

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I would not write $dS=0$ for a cyclic process but rather its closed contour integral $\oint dS = 0$, then you also have $\oint \frac {\delta Q}{T} \le 0$ and from $T>0$ you will get that in a cyclic process you must have somewhere, sometime, at some stage of the cycle $\delta Q < 0$, that is some amount of heat must be rejected to complete the cycle. ...

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It should be in joules. The term $\ln \frac{V_2}{V_1}$ is unitless, because the units on the top and bottom of the fraction cancel. Pressure, in pascals, is defined as Newtons per square meter. You've already put the volume into cubic meters, and the cancellation goes $$\text{meters}^3 \times \frac{\text{Newtons}}{\text{meters}^2}=\text{Newtons} \times ... 0 For a reversible path between two states (1 and 2), entropy change of a system is NOT zero. It is$$\Delta S = \int_a^b \frac{dQ}{T}$$For reversible path between two states, entropy of the universe (Or any isolated system) is zero.$$\Delta S + \Delta S_\text{surroundings} = 0$$So You cannot just take any system and say that entropy change between two ... 3 I was told a long time ago that the sound is from "twinning" - this is where a metal under large stress experiences a reorientation of grains to relieve stress. However I am not convinced this is the case - typically when the engine parts (catalytic converted being probably the hottest) cools down, it will shrink - and there is some "give" in the mountings ... 1 As you drive, parts of the engine and exhaust system slowly heat up. They probably make cracking noises too, but you can't hear them over the sound of the engine and driving, and the noises-insulated driving cabin. When you stop driving, they cool down, and the hot metal contracts causing a cracking sound. 0 If I hold a ruler from one end and let it swing it will com to rest with one end, call it end A, pointing vertically down. If I start to move the point I'm holding down the ruler it will continue to hang down until suddenly, when my fingers pass the centre of mass, it will flip round and come to rest with the other end pointing down and end A pointing ... 0 I'm not sure whether this is an answer or a stretched out comment. Thermal energy is not really related to displacements, but instantaneous velocities; The diffusion behaviour of the different states is most certainly different. This is to say that in the solid, the particles can move at a great speeds, but immediately collide with particles next to them ... 0 What is important here is that the entropy is also very low today. So, since the Big Bang the entropy has been increasing but we're still quite far removed from the heat death state. So, the reason why at the macroscopic level time reversal inavariance appears to be broken is simply because the entropy today is low compared to the maximum entropy, and the ... 1 Yes, it looks like it's integrated (and sometimes it is even called integrating), but really it is due to the Euler theorem. Suppose X is a first order homogeneous function of n_1, n_2, \dots (this is equivalent to saying it is extensive) and that \mathrm{d}X(\mathbf{n}) = \sum_i \frac{\partial X(\mathbf{n})}{\partial n_i}\textrm{d}n_i. This is to ... 0 Question 1:$$\frac{dS}{dV}=\left(\frac{\partial S}{\partial T}\right)_V\frac{dT}{dV}+\left(\frac{\partial S}{\partial V}\right)_T\frac{dV}{dV}$$This doesn't make much sense, because is not a well defined expression. The differential$$ \tag{A} dS=\left(\frac{\partial S}{\partial T}\right)_VdT+\left(\frac{\partial S}{\partial V}\right)_TdV$$is just ... 1 Can this scenario be considered as erasure of information as referred to in Landauer's principle? No. The reason is that the whole process is reversible and Landauer's principle specifically needs an irreversible step - otherwise no "erasure" can have taken place. Where's the problem? It's here: The state$$\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle) ...

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Any physical body has many degrees of freedom, not only mechanical, but also the field degrees of freedom. The energy is distributed amongst these degrees of freedom, so the radiation (as field excitations) is always present in a body. The energy exchange is always on and in the equilibrium conditions what is "radiated" is "absorbed" with the same rate. It ...

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The equilibrium mentioned in that quotation is between the radiation field and the walls of the container. The walls of the container are imagined to be held at some fixed temperature by some (unspecified) means. Under those conditions, the spectrum of the radiation in the cavity is that of a blackbody at the temperature of the walls. The radiation field ...

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If a body has a positive temperature then it will emit radiation (over some very low limit perhaps). Imagine a hollow sphere at a uniform temperature $T$. radiation will be emitted from the walls inside the sphere and absorbed by the same walls. Consider a small part of this wall it loses energy by emission of radiation and gains energy by absorption of ...

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from the phase transition from vapor to liquid. Even if at the same temperature, you require energy to tranform water at 100C into vapor at 100C. And the reverse is also true.

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I understand that there was zero entropy or at least low entropy that became self agitated to bring about the big bang. The closest example to zero or very low entropy is a solution of Dropleons with quasiparicles structures in a liquid state that are self propagating. This is based on the initial conditions of the solution containing super conducting and ...

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The gas is expanded adiabatically and then isothermally. Thus the temperature it has at the end of adiabatic expansion stays the same even after the isothermal process. Ideal Gas equation after adiabatic expansion: $p_aV_a=nRT_a$, where index "a" shows after. You do not have $V_a, T_a$ in this equation. However, another equation you can write down is the ...

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Use a large and thick copper spoon. Put the spoon in the coffee for 3 seconds. Remove it and insert the spoon in your mouth. (beware it will be hot) Use the tongue and palate to cool spoon pushing hard. When the spoon is close to the temperature of your mouth (feels tepid) remove it and reinsert the spoon in the coffee by another 3 secondos plus 0.5 ...

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You don't have to discretize your problem (XY model). For each step, just take some value as the new $\theta$, and calculate the transition rate accordingly. Of course, when choosing the new value of $\theta$, better don't do it in a completely random way, otherwise your transition rate might be usually too small and you are just wasting time. Having said ...

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The rate of temperature change will be the power per unit mass times the specific heat. So if you have a certain mass of water $M$ flowing per second, at a velocity $v$, losing $\Delta P$ pressure per second, then work done is $v\Delta P A$ and $A = \frac{M}{\rho v}$ . Then with a heat capacity $c$ (about 4.2 kJ/kg/K for water), and the relationship between ...

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The key here is the air curtain. You can be certain that if it didn't save Costco money, they wouldn't bother with it! It takes a bit of power to push air that much. Two very helpful diagrams are in the youtube video Powered Aire - Cold Storage Air Curtain: In the first case, the air can mix and change temperature through convection, and it does so in a ...

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How can I calculate the gas pressure given particles per cubic centimeter, and its temperature in Kelvin? as pointed out in comment by KyleKanos $PV=Nk_BT$ where $P$ is pressure, $V$ is volume (in $m^3$), $N$ is the number of particles, $k_B$ is Botzmann's constant and $T$ is temperature in Kelvin. If you rearrange it $P= {N \over V}~k_BT$ so ...

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While I agree in principle with David Lynch's answer, I think it's good to take a closer look at the phase diagram (adapted from http://upload.wikimedia.org/wikipedia/commons/4/46/Carbon_basic_phase_diagram.png): I added the arrows to show possible paths you might follow. Red path: diamond would become graphite before melting; the molten carbon becomes ...

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I don't think I'd call the energy you extract this way an energy "loss". But imagine that you burn the coal and use that to boil water to steam. The steam is made up of very energetic particles. Given enough time, the particles will heat the materials inside the dome as they come to thermal equilibrium. But in the power plant, we don't let them do ...

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What follows is certainly not a comprehensive answer addressing all of your concerns. It is an answer to the question is there a way to see something clearly pathological like superluminal signals in the heat equation? I would argue that yes, there is. The general solution to the initial value problem $T(x,0) = T_0(x)$ for the heat equation on the ...

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Some corrosion always takes place (pure gold is not used for transmission lines, AFAIK:-) ), so the conductivity decreases with time, although for some materials this effect can be very small ...

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The line itself does not change much over years. What changes and therefore needs maintenance on power transmission lines is insulators, connectors and spacers. Insulators get dirty or simply break, connectors work loose due to thermal expansion and contraction, mechanical stresses and oxidation, and spacers can be damaged by wear due to these same ...

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It may be worth pointing out that blankets also (surprisingly) act as (thermal) radiation shields. This is the reason that "emergency blankets" can sometimes be found in survival kits that appear to be nothing more than thin shiny plastic. But they really make a difference in the amount of heat lost by a warm (37 °C) body on a cold night (cloudless sky - ...

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The thermal conductivities of a range of materials is given here. I can't find figures for the thermal conductivity of solid wool or cotton (i.e. a solid block with no air voids in) but the thermal conductivities of organic materials seem to be around $0.25$ Wm$^{-1}$K$^{-1}$. By contrast the thermal conductivity of air is $0.024$ Wm$^{-1}$K$^{-1}$, so given ...

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Normally your body heat would dissipate in the air, so when it's cold, your outer body cools down, because you are losing your body heat to the air near you. So when you cover yourself in a blanket, you stop your body heat from escaping, and as it is trapped, and your body continues to produce heat, you feel warmer and warmer under the blanket. Overall, the ...

1

In a Fermi-Dirac distribution, the relationship between temperature and the speed of particles is not intuitive. Even at cold temperatures, fermions can have high speeds simply because of degeneracy - the lower momentum states "fill up", leaving only states with large momentum available, and this is true even at very cold temperatures. However, the heat ...

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If you think about the infinite square well problem, the states with higher energy have higher momentum, (and also a higher velocity). However, it is better to think of the higher energy states as higher frequency standing waves. Because they have a higher frequency, the have to "travel faster", which is where the large velocity comes from in the Fermi ...

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Is the vacuum a required part of the problem? The available action time is increased if it were only oxygen suddenly removed, baring panic. The best case scenario with planning and specific conditions met is about 20 minutes, http://www.guinnessworldrecords.com/world-records/1000/longest-time-breath-held-voluntarily-(male) (Out of water the max time is ...

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At the microlevel, if you were to keep track of all the degrees of freedom of the system, then all of the internal energy change would be accounted for as work, heat exchange would always equal zero for any process. Whenever energy is exchanged, this is always due to molecules interacting with each other and thus performing work on each other. The whole ...

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