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4

If you add mass keeping the volume constant, the pressure would change. When a source says that pressure is independent of mass, it means that pressure is independent of mass if you keep the density constant, not the volume. The idea is that if you take two identical copies of a system, then combine them to make a system twice as large, extensive ...


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The proof behind Carnot's upper limit posed on the efficiency of heat engines is more robust than this. The quotes you've pasted are among the various statements of the second law of thermodynamics. Here I'll sketch for you some of the ideas of the proof, mainly to show where these formulations (related to Carnot) of the second principle come from. ...


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The simple answer, which is what I think you're hoping for is the following: At constant volume, the system (by definition) is not able to do work on the surroundings because work involves a change in volume. All the heat you put in is spent raising the temperature (internal energy). At constant pressure, some of the energy you put in goes into raising the ...


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I will not answer your question directly; only give you some tools that should help you answer the question (in practice) yourself. To focus the attention, find below a typical heating/cooling diagram for a frozen pure substance. The vertical axis marked $T$ represents temperature (in degrees Celsius). Three significant temperatures are indicated on the ...


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First law of thermodynamics $$\delta Q=\delta E+p\delta V $$ $$\therefore\frac {\delta Q}{\delta T} = \frac{\delta E}{\delta T}+nR$$ As $p\delta V$ can be written as $nR\delta T$ When volume is constant $$\delta Q= \delta E $$ $$\therefore\frac {\delta Q}{\delta T} = \frac{\delta E}{\delta T} $$ $\frac {\delta Q}{n\delta T}$ is Molar heat capacity. ...


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The above statement is not correct. First of all, you need to work against the force of friction while climbing stairs.So the energy is not entirely converted to PE.Rather a portion of it is dissipated. Secondly, even if we leave out friction, the basic flaw of the statement lies in the part: " the energy is converted to PE within system only. The person ...


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The connection of heat to entropy in thermodynamics is through: where S is entropy Q is heat T is temperature, and it is through differential changes. This in no way means that heat is entropy . The easiest way to acquire an intuition of entropy is to read up on the statistical definition which can be proven to be the same as the thermodynamic ...


1

First let us cut away some of the meat and the bones of your question. Let us forget about thermodynamics for a minute and classical physics; however we will need GRAVITY (not Newton’s version). Instead of calling it Gravity let us call it General Relativity or (GR). We can still use the term Gravity, but when discussing Quantum Mechanics it’s better to ...


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It is always true that, for an ideal gas, $\Delta U = C_V \Delta T$, regardless of the process. Remeber, we define $C_V=(\delta Q/dT)_V$. Since this is happening at constant volume (aka $\delta W=0$), we have $C_V=(\delta Q/dT)_V=(dU/dT)_V$. Then, since $U$ doesn't depend on volume for an ideal gas, we have that $C_V=dU/dT$ even if volume is changing. So ...


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Part of your problem comes from thinking that the potential energy is somehow located in or a property of the person alone. And the way the subject is usually introduced could easily lead you to think that, but it's not right. The potential energy is a property of the person-Earth system. In fact all potential energies are properties of systems of ...


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To start with the law of increasing entropy applies to isolated systems. The system you describe is isolated if one considers the total entropy of both the paramagnetic material and the permanent magnet, including any radiation. The order introduced in the paramagnetic material is balanced by a disorder in the permanent magnet plus any radiation from ...


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I haven't given this enough thought yet, but at a first glance, I would say no, a potential having $\mu, V, E$ as natural variables would not be a valid one. One possible attempt to obtain such a thermodynamic potential $Q$ that is a natural function of $\mu,V,E$, would be a Legendre transform of the entropy $S(E,V,N).$ We have: \begin{align} dS = ...


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Resistance changes with temperature The temperature coefficient of resistance, or TCR, is one of the main used parameters to characterize a resistor. The TCR defines the change in resistance as a function of the ambient temperature. The common way to express the TCR is in ppm/°C, which stands for parts per million per centigrade degree. The temperature ...


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Assuming you fix the temperature at the cell boundary, you can solve for the steady state temperature profile within the cell using (see any elementary heat transfer book) $$\nabla^2 T = -\frac{\dot q}{k}$$ where $k$ is the thermal conductivity on the inside of the cell and $\dot q$ is the heat generation rate per unit volume. The temperature only varies ...


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Warm air supports greater humidity (absorbs more moisture) than cool air. At night and in the early morning, when the air is cool, evaporation from river water at Niagara Falls hits cooler air which does not as readily absorb water, so is capable of less humidity, than warmer air during the day, so the river evaporation becomes visible as water vapor ...


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An open cup of hot water loses a lot of its heat through evaporation. You can easily test this: prepare two cups of hot water, and then float a thin layer of cooking oil on top of one. Come back in a half hour: you'll find that the cup without oil is a lot cooler, and the water level has also significantly gone down. Edit: I did the experiment: I poured ...


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When energy flows from a hotter subsystem to a colder subsystem the entropy of the combined system increases. $$\frac{dS_1}{dE_1}=\frac{1}{k_BT_1}<\frac{1}{k_BT_2}=\frac{dS_2}{dE_2}.$$ So if they exchange energy in a way that conserves energy then we get $-\Delta S_1<\Delta S_2$ and the total entropy increases. This does require that the systems ...


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If you touch the outside wall, does it feel like touching a heater? I guess not, so therefore we know that you are not losing 50% of the energy through the wall, but maybe something like 5%. The outside wall is cool, because the air gap and the wall are insulators. Adding more insulation reduces the conduction of heat through the insulators.


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It sounds like you'd enjoy reading some classic books on thermodynamics; one especially good one is the one by Fermi. It has lots of nice arguments; short and simple, but logically airtight. For your first question: an irreversible process increases the entropy of the universe. If an irreversible engine had the same efficiency as a reversible one, we could ...


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Introduction: Entropy Defined The popular literature is littered with articles, papers, books, and various & sundry other sources, filled to overflowing with prosaic explanations of entropy. But it should be remembered that entropy, an idea born from classical thermodynamics, is a quantitative entity, and not a qualitative one. That means that entropy ...


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There is no problem defining intensive quantities in the grand canonical ensemble, where the intensive quantities appear as the parameters defining the ensemble. Indeed, this is the ensemble closest to the application. For a treatment of (classical or quantum) thermodynamics solely in terms of the grand canonical ensemble see Part II of my book Classical and ...


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Short answer: The major difficulty lies with the definitions themselves, and none of the possibilities given has a real physical meaning which can be univoquely related to stress in non extensive systems in its conventional mechanical original meaning. The long one: What kind of systems does this apply to? This is not answered by referring to systems with ...



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