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1

Certainly, you agree $dH=C_P dT$ if we're at constant pressure. If $H$ and $C_P$ don't actually depend on pressure, then you can use this equation regardless of whether pressure changes. However, to determine $C_P$ and $H$ you first need an equation of state (such as $PV=NkT$). Without knowing the specific equation of state (aka, if your gas is ideal or ...


0

Enthalpy is defined as $H=U+pV$ where $U$ is the internal energy. This definition has nothing to do with pressure being constant or not. If the system under consideration has only pressure $p$ and thermal $T$ internal variables and interaction with its environment (simple homogeneous system) then for constant external pressure $dp=0$, and we have $dH = dU + ...


0

The change in Helmholtz Free Energy (divided -$k_B T$) is the change in entropy of the system $A$ plus environment, assuming the entropy of the environment changes only by exchanging energy with $A$. The Gibbs Free Energy is the entropy of the system plus environment, assuming the entropy of the environment changes only by exchanging energy and volume with ...


0

According to quantum theory, expected average energy of harmonic oscillator when in contact with thermal reservoir of temperature $T$ is $$ \epsilon_{av} = \frac{\hbar \omega}{2}+ \frac{\hbar \omega}{e^{\frac{\hbar\omega}{k_B T}}-1}. $$ If $k_B T \gg \hbar \omega$, this is approximately equal to $$ k_B T, $$ else the average energy is lower. For other ...


1

Dr. Robitaille says that blackbody radiation is not universal, even inside cavities where the surfaces are all at thermal equilibrium. That is highly controversial since the electromagnetic fields in a cavity are usually considered as an additional substance, called a "photon gas" which is also at thermal equilibrium and hence has a temperature. This ...


0

Friction is caused when a Normal force ($L$ in the diagram above) pushes one surface onto another, causing a friction force of magnitude $F_f=\mu L$, with $\mu$ the coefficient of friction. When a force $F > F_f$ drags the surface (of the bloc in the diagram) over the other surface (of the 'floor' in the diagram) work is being done: $W = \int F_f ds = ...


0

Conservation of energy. It takes work to overcome friction and that work has to go some where. The work goes to heat.


1

Thermal diffusivity is the ratio of heat conduction ability to heat storage ability of a material at constant pressure. A high ratio indicates that the material will conduct heat away more readily than it can retain the heat. A low ratio indicates that the material has a great capacity to store heat, and will not be able to conduct heat away as rapidly. ...


0

The thermal conductivity of steel is significantly less than that of aluminum. Heat generated by the friction of asbestos on the surfaces of both discs is carried away much faster by aluminum than by steel. Therefore the steel disc in the viscinity of contact with the pin, as well as the asbestos pin, will retain heat and will measure a higher temperature ...


1

I can’t write this computer simulation for you (at least not based on the data provided) but will instead explain a few basic relationships that govern the heating and cooling of objects. I hope this helps. Consider a building an object that is composed of $n$ objects of masses $m_i$ with specific heat capacities $c_{p,i}$, then the building has an overall ...


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Firstly its the potential energy of the earth-object system as the two answers have said. "Potential energy of the object" is a loosely spoken phrase for the same for things happening on the surface of the earth where $g$ is taken to be a constant.Secondly work is not done by a person. Work is done by a force.There is an important thing you need to know. ...


4

If you have a glass of water, every molecule of water is under the influence of the gravitational potential field. Each molecule pushes down on the molecules below. The molecules of water near the bottom of the glass experience the downward force from all molecules above them (each molecule transfers the force downward). If you made a hole near the bottom of ...


2

The Maximum Entropy principle, principally popularized by Jaynes, is known by most people having studied statistical physics. The way I see it, although Jaynes considered it as crucial in the foundations of equilibrium statistical mechanics and other people in the field still do (like Roger Balian for instance), it is more taught and thought of as a useful ...


1

A vacuum will offer you no advantage in cooling your aluminum block, ceteris paribus. In a vacuum, the only method of heat transfer is radiation. You will lose the benefit of any of the other three methods listed below. There are four ways for an object to lose heat: (1) Conduction - Thermodynamic energy is transferred through physical contact with ...


1

A force does not require a constant input of energy to exist. Energy is only required to perform work, which is exerting a force over a distance. $$ W = \mathbf F \centerdot \Delta \mathbf x $$ That distance is key. In your example, if the size of the container does not change, no energy is expended no matter how long the force lasts. If the force is used ...


3

A recent work that uses nanothermodynamics and includes a computational investigation of the kind you are asking about for an Ising lattice: R.V. Chamberlin, The Big World of Nanothermodynamics Sec.5 of the following paper makes a reference to another paper that appears to have tested the limits of usual thermodynamics in single polymer stretching ...


0

Disclaimer : I'm not sure that the following is the exact/complete answer to the question but maybe these elements could help. Equilibrium of a system under an external field Let say that you have an open macroscopic system $\Sigma$ (composed of identical particles) which is under the influence of an external time-independant but space-dependant ...


0

The linear expansion coefficient of a material $\alpha_L$ gives the change in the length $\Delta L$ of a bar of that material after a change in temperature $\Delta T$ given the original length of the bar $L$: $$\Delta L= L\ \alpha_L \Delta T $$ Modeling your spring as a 2 cm long steel bar and using steel's $\alpha_L$ of $1.2 \times 10^{-5}$ (m/m°C) with a ...


0

The temperature does not depend linearly on the radius because the flow lines of heat flux are not parallel. This means that the heat flux, and therefore the temperature gradient, decrease as the flux lines diverge, and increase as the converge. Here the flow lines are radial. By conservation of energy, the rate of heat passing through any cylindrical ...


3

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 ...


1

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 = ...


0

A cooler that has ice and water in it will be held at 32 degrees Fahrenheit until all the ice is melted. The rate at which the ice melts depends on the rate at which heat can enter the container. The rate at which heat crosses any thermal boundary can be modeled as: $$\dot Q=\frac{\Delta T}{\sum 1/h_i}$$ Where $h_i$ represent the thermal conductivity of ...


1

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 ...


1

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 ...


2

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 ...


0

Yes it is possible to have water coming out be colder than the water going in. Imagine a big pipe connecting a big container of ice water to a big container of boiling water. Then one end starts out colder than the other. Now raise the container of boiling water higher up so that water starts to flow from the boiling water to the freezing water. Heat ...


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 ...


0

The closest existing solution to what you're looking for is probably Peltier cooling. Of course Peltier coolers don't break the Second Law of Thermodynamics either: they are heat pumps, with a cold sink and a hot sink.


0

As long as you're not trying to violate the second law of thermodynamics, yes. There are thermo-electric effects where a voltage can be generated from a temperature difference. This mechanism isn't really considered efficient though. Thermodynamically, the situation isn't too different from a regular power plant anyways, so it's never used for large scale ...


2

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 ...


0

There are 3 forms of heat transfer: radiative, convective, and conductive heat transfer. All 3 forms of heat transfer are normally operating at the same time.


0

For the PV diagram, the saturation line is the horizontal red dotted line in your diagram. At the far left on the horizontal red dotted line, the curve is at the "bubble" point of water, where saturated liquid with a "speck" of a bubble exist (100%- of liquid water). At the far right of the horizontal red dotted line, the curve is at the "dew" point of ...


1

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 ...


0

As a rough planning figure for building design etc, 1 human = 100 watt.


0

Entropy is fundamentally a subjective quantity, it is the amount of information needed to specify the exact physical state of a system given it's macroscopic specification. The fact that you don't know the exact physical state of a system, yet the laws of physics are such that information is never lost, implies the second law of thermodynamics (one also ...


1

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 ...


1

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 ...


0

Let's frame this question in terms of a heat engine (Carnot engine). Here is a diagram I made for a class when teaching this stuff. Heat flow $\dot{Q}$ has an associated entropy flow $\dot{Q}/T$. The job of a thermodynamic engine is extract/filter as much useful work as possible from a flow of energy and entropy. To answer your question in layman terms. ...


1

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 ...


0

Reversible engine is the most efficient and all reversible engines have the same efficiency. It can be shown like this: If there is a machine X that is more efficient then the R (reversible) engine then we could reverse the R to give warm reservoir its heat which X took but to do that we would have to take more heat from cold reservoir then the X because X ...


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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. ...


0

Your intuition is correct; if you use an ideal gas, the temperature will be unchanged. However, in a real gas there are extra attractive forces between the molecules, so that it costs energy for the gas to expand. The keyword to search for is the 'Joule-Thomson effect'. This cannot be used as a free refrigerator (which would violate the 2nd law) because the ...


0

To answer this, we need to clear up what 'heat' is. "Heat" is not really a thing; it means nothing more than "energy transfer in some disorganized way". The two main modes of heating in this case are conduction (atoms in your heater warm up adjacent air atoms) and radiation (atoms in your heater emit infrared light). When we say something is a thermal ...


-2

There's no work done for a person climbing upstairs because the energy is converted to PE within system only. The person is the system. How true is the above statement? I think it's true enough. You do work on a brick when you lift it up. You add energy to it, and we call this energy gravitational potential energy. Then when you drop the brick this ...


1

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 ...


1

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 ...


2

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 ...


0

At normal atmospheric pressure water is ice at 273 K (-0.15 Celsius). Applying heat causes the ice to melt isothermally (at 273 K) and the Latent Heat of Fusion (of ice) has to be applied to cause that state transition to happen. Once melted, more heat then has to be applied to heat the water from 273 K to 277 K, taking into account the Heat Capacity of ...


0

The thermal conductivity for water is k=0.56 W/m⋅K,all right,but this is for a lake,for exemple,when the heat transfer is between the air and the cold-water layer.The unit of measurement is W/mK ,not W/(m^2)K. So it's for the thickness ,not for the surface! If you want to calculate the heat production in the time unit ,H,start from: H=4π(r^2)ΔT α ,where ...


1

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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