# Tag Info

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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 ...

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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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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 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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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 $$\text{meters}^3 \times \frac{\text{Newtons}}{\text{meters}^2}=\text{Newtons} \times ... 2 If a body has a positive temperature then it will emit radiation (over some very low limit perhaps). Imagine a hollow sphere at a uniform temperature T. radiation will be emitted from the walls inside the sphere and absorbed by the same walls. Consider a small part of this wall it loses energy by emission of radiation and gains energy by absorption of ... 2 The equilibrium mentioned in that quotation is between the radiation field and the walls of the container. The walls of the container are imagined to be held at some fixed temperature by some (unspecified) means. Under those conditions, the spectrum of the radiation in the cavity is that of a blackbody at the temperature of the walls. The radiation field ... 2 Any physical body has many degrees of freedom, not only mechanical, but also the field degrees of freedom. The energy is distributed amongst these degrees of freedom, so the radiation (as field excitations) is always present in a body. The energy exchange is always on and in the equilibrium conditions what is "radiated" is "absorbed" with the same rate. It ... 1 Can this scenario be considered as erasure of information as referred to in Landauer's principle? No. The reason is that the whole process is reversible and Landauer's principle specifically needs an irreversible step - otherwise no "erasure" can have taken place. Where's the problem? It's here: The state$$\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle) ...

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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 ...

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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.

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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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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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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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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 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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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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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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