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2

The cold air does flow down, but instead of flowing out of the fridge it is sucked into a channel, and pumped back out at the top of the fridge.


0

$T$ should be the actual temperature at which the water evaporates. That is, the temperature at the interface between the air and the water, not the boiling point. This is simply because $dU = TdS + pdV - \sum_i \mu_i dN$ (where $T$ is most definitely the temperature of the system), or by rearranging, $$ dS = \frac{1}{T}dU -\frac{p}{T}dV + ...


0

Specific entropies ($\mathrm{kJ\,kg^{-1}\,K^{-1}}$): \begin{array}{lrll} & \mathrm{^\circ C} & \text{liquid} & \text{vapor} \\ \text{Triple point} & 0.01 & 0 & 9.155 \\ \text{Normal boiling point} & 100. & 1.307 & 7.355 \\ \text{Critical point} & 374.15 & 4.430 & 4.430 \\ \end{array}


1

The inherent idea is, from that equation $$I = \exp (\frac{eU}{k_B T})$$ if you plot $(\ln I)$ versus $U$, that would be a straight line (of the form $y=mx$), with a slope $$\alpha = \frac{e}{k_B T}$$ That's all you have to do in the experiment, use least square fitting to find an accurate value of $\alpha$ and then find the Boltzmann constant using the ...


0

ΔS universe = ΔS system +ΔS surroundings . hotosystem I of plants has a quantum efficiency of 0.99 and an energy efficiency of >0.96. The system may be considered a canonical ensemble which consists of a Thermal bath in which the particles are suspended and monochromatic light emitted ny an incandescent lamp with a definite Tr. Please note that the lamp ...


0

Internal Energy = Heat - Work, or U = Q - W. By definition an adiabatic process has constant heat, so when finding the rate of change of each in a process by taking the time derivative: dU = dQ(0) - dW Q goes to zero, and U = W.


1

There's no violation of the second law here. You have a system that is out of thermal equilibrium. That black bodies absorb and radiate is the driving mechanism that tries to move this system toward thermal equilibrium. By way of analogy, suppose you are from a southern clime and take a trip at this time of year to a northern clime. You, as a southerner, ...


11

Not all the radiation from the outer shell reaches the inner shell. When you take into account the intensity distribution of radiation from the outer shell (Lambertian distribution, i.e. $\propto\cos\theta$) you will see that the amount of radiation for the inner to the outer shell is the same as in the other direction. No violation of the second law.


0

There was a time, when trains worked on coal. What sort of energy, do you think, it was? - Thermal. There's also solar energy, which is, if you think thoroughly, thermal energy. You turn heat (received by a solar panel) into an usable energy source. However, as I know, solar energy produces so little electricity, it's not worth all the effort. The way, I ...


2

To turn thermal energy into useful work completely one would need a thermal bath at the temperature of absolute zero. This is explicitly forbidden by the third law of thermodynamics. The best one can do is given by the efficiency of the (theoretical) Carnot cycle: http://en.wikipedia.org/wiki/Carnot_cycle. Th efficiency of the Carnot cycle only depends on ...


0

It is possible up to the Carnot limit, which is never 100%. In particular, the Carnot limit is higher if the temperature before extracting is higher. But you could only hit 100% if the temperature could reach infinity, which it can't.


1

To answer your second question, yes, condensation typically requires nucleation. Supersaturated vapours like the one you describe are the basis of the cloud chambers that used to be used as particle detectors. The energetic particles passing through the vapor would ionize molecules, and these ions would act as nucleation sites.


2

The concept of the storage heater is very common in Britain: they usually include a phase change medium which means that a relatively small mass of material can contain a lot of heat without becoming very hot (the latent heat of fusion provides a "thermal cushion" where the medium can give off a lot of heat at a constant temperature, keeping the rate of ...


1

UPDATED: I now think my previous answer was wrong, because the set up would be equivalent to the following question: Is a black body sphere inside a black body shell hotter than the shell? Just change the question to add a carefully crafted lens that focuses all the radiation into the sphere (you could make the shell as large as you want), which of course ...


1

I don't think you need to overthink this so much. Mechanical equilibrium in this context basically means that from a macroscopic point of view, all forces are balanced; this usually also means that the system's parts are at rest, though a system in uniform motion could be considered in mechanical equilibrium, I guess. The point that the authors are trying ...


0

Thanks for clarifying the question, but I'll restate it here in my words, since comments may disappear: Given a closed constant volume containing water and a head space, why is the thermodynamics of the water not affected by pressurizing the head space with an ideal gas? Lets simplify by stating I'll treat the water as an incompressible fluid. Adding ...


4

The heat that makes a filament lamp glow is derived from electrons bashing into the lattice of atoms in the filament and transferring energy to them. The kinetic energy of the electrons becomes vibrational energy of the lattice, and this is exactly what heat is. However the interaction of electrons with the lattice is also what resistance is, and ...


2

Water does not, in general, help extinguish a fire. Typical fires, however, can be successfully attacked using water alone, as it can cool the fuel at the base of the fire or generate a vapor barrier between atmospheric oxygen and the hot fuel. Water can accelerate liquid hydrocarbon fires by dispersing fuel. Water can generate explosive gaseous mixtures ...


14

To sustain fire, it is true that you need the tri-factor of oxygen,fuel, and heat. However extinguishing fire through the use of water, is different than one would think. Indeed, water "sucks" energy in order to change its phase, and thus reduces the heat factor, but the real crux lies in the water expansion properties. Water is heavier than hot air, and ...


0

This is valid for ideal gas whose molar number is constant $n$. Why? When a fluid changes volume, the equation $$ dU =dQ - pd V $$ is obeyed. Formally dividing by $dT$ we obtain $$ \frac{dU}{d T} = \frac{d Q}{dT} - p\frac{d V}{d T}. $$ If we now consider only processes where $V$ remains constant, the relation $$ \frac{dU}{d T}\bigg|_{V=const.} = \frac{d ...


41

To sustain a fire, you need three factors: fuel, oxygen, and heat. Take away one of the three and the fire goes out. Water removes heat. Most of this "removing heat" is the evaporation - roughly 540 calories / gram, so 7x more heat than is needed to get water from 20°C to boiling (with a tip of the hat to @Jasper for pointing out erroneous value in earlier ...


1

The Wien displacement law gives the maximum of a function, so the way to compute it is to start with the Planck function in frequency domain, $$ B(\nu,T)=\frac{8\pi\nu^2}{c^3}\frac{h\nu}{e^{h\nu/kT}-1} $$ Take the derivative with respect to $\nu$, set it equal to zero and solve for $\nu$. You'll likely have to use some numerical methods (e.g., iterative ...


0

First law of thermodynamics states that the change in internal energy, $dU$, of a system is equal to $\delta Q + \delta W$. The only way to separate the contributions from the two is to arrange the experiment so that one of the quantities is zero. If the change is adiabatic so that there is no heat exchange, $\delta Q = 0$ and hence $dU = \delta W$. On the ...


4

As mentioned in the other answers the etendue theorem rules this out for a system of mirrors and lenses. However, I think it's important to note that simple thermodynamic arguments are insufficient for the reasons given below. I will answer the question using mirrors rather than lenses as it makes the physics clearer. Suppose we have a massive mirrored ...


4

It is not possible because of conservation of etendue. This is based purely on geometry, not really a law of physics in that sense. No guarantees regarding quantum effects etc., but in the realm of ray optics it can't be done. Basically, given any source of light radiating from finite surface to half space, you can never concentrate the entire emitted ...


12

If you use complicated routes redefining "focusing" towards generating the temperature, yes. Physicists at CERN's Large Hadron Collider have broken a record by achieving the hottest man-made temperatures ever - 100,000 times hotter than the interior of the Sun. Scientists there collided lead ions to create a searingly hot sub-atomic soup known as ...


2

Because what you are doing is a flow process, with mass inflow and no mass outflow, you need to use the thermodynamic equation: $dU_{cv}={m_{in}d}{H}_{in}-{m_{out}d}H_{out}+\delta Q-\delta W_{shaft}$ If you insulate your air cylinder well enough, $\delta Q = 0$. Assuming that your air cylinder does not deform, $\delta W = 0$. Since you are filling your ...


1

A quick back-of-the-envelope calculation using the Stefan-Boltzman law for a sphere with radius $r$ $$ P = \sigma T^4 4\pi r^2 $$ lets me conclude that you would need to focus $1\,\textrm{m}^2$ of sunlight onto a sphere with a radius smaller than $ 2.8\,\textrm{mm}$ to get its temperature above $6000\,\textrm{K}$. How difficult is it to build such a ...


5

The thing is, even if you concentrate all the black-body radiation at temperature $T_0$ on a portion of matter $A$, this matter itself will, at least, radiate the same way all the energy when its temperature matches that of the black-body. So even assuming $A$ is a black-body, i.e. absorbs all radiation received and reflects none, it will itself emit energy ...


0

The energy change a gas is expressed by: $$dE = TdS - pdV +\mu dN$$ Admittedly, the process of filling a cylinder is rather complicated because several variables in this equation vary, and the way you fill it will influence their behaviour, but you can see that the result will give a production of heat. When you fill the cylinder you are connecting it to ...


0

Use this equation: $$\frac {P_1V_1}{T_1}=\frac {P_2V_2}{T_2}$$ Before, the pressure is 1 atm, the temperature is about 20°C and the volume is many cubic meters. After, the pressure is above 100 atm, the volume is about 1 cubit meter, and the temperature is significantly warmer, say 100°C. Filling in the equation: $$\frac {(1atm)V_1}{393K}=\frac ...


0

I would explain it by simply using the first law of thermodynamics: $$Q-W=\Delta U$$ $Q$ is heat added to the system, $W$ is work done by the system, $U$ is the internal energy. Keep in mind that internal energy is closely bound to temperature, so a change $\Delta U$ also results in a temperature change (which is what we are talking about). If you look at ...


45

Moonlight has a spectral peak around 650nm (the sun peaks at around 550nm). Ordinary solar cells will work just fine to convert it into electricity. The power of moonlight is about 500,000 times less than that of sunlight, which for a solar constant of 1000W/m^2 leaves us with about 2mW/m^2. After accounting for optical losses and a typical solar cell ...


20

At least one point in your favour is that the light we receive from the Moon has barely anything to do with its temperature. Instead it is mostly a secondary light source "reflecting" light from the Sun towards us. The second point in your favour (I think) is that the thermodynamic argument seems pretty weak. We are not trying to make Earth as hot as the ...


0

For what it's worth, it is claimed that the critical exponents differ above and below the critical point for some exactly solvable 2-dimensional model: http://www.ujp.bitp.kiev.ua/files/journals/49/11/491114p.pdf (Ukr. J. Phys., v.49, #11, p.1122 (2004)).


2

Critical exponents are properties of the RG fixed point that drives the phase transition. They are computed by linearising the RG flow equations close to the fixed point. The exponents are the derivatives of the beta functions evaluated at the fixed point. They know nothing of the way you approach the fixed point. In particular if you are flowing slightly ...


1

Starting with the definition of specific enthalpy $$h=u+pv,$$ we apply the differential operator to give $$dh=du+pdv+vdp.$$ Now, we know that the $du$ term can be expanded using fundamental thermodynamics. Namely, the change in the internal energy of the system $(du)$ is equal to the heat transfer to the system minus the work done by the system. In equation ...


1

Since the Helmholtz EOS is a function of density, you need to invert the function: $$ \rho=a^{-1}(a(\rho,T)) $$ For this particular, EOS, there is no analytic form for this, you must use iterative methods (e.g., Newton's method). Since you know the pressure and temperature, you can use the fact that $$ p=\rho RT\left[1+\delta\left(\frac{\partial ...


0

This is a very interesting question whose best answer is probably a very definite maybe... Defining a quasi-static process as one that is very very slow is not very useful as the simplest ferromagnetic hysteresis would show. If instead you define a quasi-static process as one that at any instant of time can be described by being on a path in the n+1 ...


0

Here's a simple, non-mathematical, answer. Although the pressure at the surface does depend on the velocity of air molecules that's not the whole picture. It is more precise to say that it depends on the rate of collisions. The collision rate depends on the velocity of the molecules, i.e. the temperature. But it also depends on density of molecules. Higher ...


0

The actual shock wave is quite short lived (I think it's visible for less than a second near 0:14 as a white sheet around the smoke/dust cloud) and doesn't propagate very far in this case. When the shock dissipates what's left is a pressure wave, the "bang" or sonic boom, and that propagates at the speed of sound. So I guess your video uses a reasonable ...


0

Your definition is a bit funky. I'll explain. Explanation As much as I like that you explicitly express internal energy $U$ as a function of state, for clarity I'm going to drop $\boldsymbol{R}$ for now. We can't directly measure the absolute internal energy of a system, we can only infer its change by understanding some defined process. Consider the ...


2

Why does the air pressure at the surface of the earth (resulting from collisions of molecules on the surface of the earth which has to do with the velocity of the particles) exactly equal the weight of the entire air column above it (which just has to do with the number and mass of the molecules in the air column)? That's not exactly true. Deviations ...


3

Suppose the pressure at the Earth's surface is $P$. Consider an air column of cross-sectional area $A$. The upward force on the column is $F_{\text{up}}=PA$. Denote the weight of the column as $W$. By definition of "weight", the downward force on the column is $F_{\text{down}}=W$. Suppose the pressure is too low, such that ...


0

I think your question is really: Why is Gibbs-Duhem equation not valid for inhomogeneous and/or small systems? This is related to why Thermodynamics does not apply to these systems. Examples, discussion and a Thermodynamical theory for these systems can be found in Terrell Hill's works link. But in short: both inhomogeneous and few-body systems have more ...


1

Let us first consider the trace of the density matrix: $$ \text{Tr}\; \exp\left\{-\beta H\right\}=\sum_{n_1,n_2,\ldots}\langle n_1,n_2,\ldots|\exp\left\{-\beta\sum_k\left(a_k^\dagger a_k+\frac{1}{2}\right)\hbar\omega_n\right\}| n_1,n_2,\ldots\rangle\\ =\prod_k \sum_{n_1,n_2,\ldots}\langle n_1,n_2,\ldots|\exp\left\{-\beta\left(a_k^\dagger ...


0

You have a mistake in the line $$ΔV= (.0314m^2)(.002m)$$ Note that 2 cm = 0.02 m, not 0.002 m...


1

Your basic intuition that anything that uses power will heat the room and reduce the consumption of the space heater is a good one. To first order it is correct. One can quibble that the heat is released in a different place, so may not heat the thermostat as effectively, causing a (slightly) higher room temperature and more total consumption, but that is ...


0

Incorrect. The heat produced by lighting device is very low (also depend on device type: LEDs, bulbs, etc) because it is made for produce light and heat is a sign of inefficiency. For example, the most inefficient lighting device is light bulb with 75% efficiency. It means that only 75% of the power converted into visible light, and another 25% converted ...


1

the cooling process depends on many factors: - convection, radiation and conduction inside the body. Exponential decay is good only if the lumped capacitance model is appropiate and if the relation with radiation is linear. I would say that in your case you also had convection inside the fluid. And ontop of that the relation with radiation is not linear it ...



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