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There are several ways to approach this problem. If we can estimate the power density achieved in $W/m^2$, then the temperature that can be reached follows from the Stefan-Boltzmann law. First method: 1) Take the total power collected, and see the size it got focused down to. You state the area of the mirror array is 0.6 m$^2$ (roughly), and with power ...


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Since at the infinitesimal singularity there was less space, there were less microstates. If you consider a cup of cold water in a pressurised, insulated vessel at absolute zero, the system has zero entropy because the atoms cannot move, there is no energy and thus less possible states the system can be in. Here we have the opposite, but the same situation: ...


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Let me show you that there is no contradiction by pointing out e.g. that for ordinary expansion periods (that is away from first order phase transitions, decouplings...) the total entropy is actually constant in time while the universe is getting bigger and cooler. Or, going back in time, the universe is getting hotter while S is kept constant. How is this ...


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Entropy is not the existence of heat or energy, but is more accurately described as the spread of energy. A universe with high heat and low matter density has very low entropy, the same way that a cup of hot water has low energy distribution when compared to a cold pool. If you throw the hot water into the cold pool the heat will spread throughout the pool ...


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It's an example of adiabatic expansion. If you have a container full of gas and you expand the container, the gas cools. Entropy is preserved. Adiabatic processes preserve entropy. Any decrease in entropy due to lowered energy, and correspondingly fewer possible velocities for the particles, is offset by an increase in entropy due to the expanding volume, ...


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In usual laboratory situations temperature will not drop in isobaric expansion -- i.e. expansion with constant pressure. However, you will have to add heat to cause the expansion and push back the piston. Usually the temperature will also increase, it is reflected by the $c_p$ quantity (specific heat capacity at constant pressure), $c_p=\frac{Q}{\Delta T}$. ...


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When a gas expands, it has do pressure-volume work against the piston, atmosphere etc. to make its room to occupy. Now, as the system spends energy to work, part of its kinetic energy decreases. Since, temperature is directly proportional to kinetic energy, the system's temperature decreases. An example: Much of chemistry takes place in vessels that are ...


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Intuitively, For a gas,if you apply heat to the container of gas the kinetic energies of the molecules or atoms increase,means heat added is used in increasing the kinetic energies of the molecules. As we know ,temperature of a gas depends on how fast the molecules of gas moving or vibrating ,so on heating temperature of the gas increases. Now these ...


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It means that the heat supplied to the gas is used in its expansion. From the text you quoted, it seems that part of the heat energy supplied to the gas is used for its expansion. This is what you usually see around you: when you heat a gas, that gas can expand a little but this does not mean that you don't manage to heat it.


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This is a difficult question for many reasons. One reason is likely because most of the introductory thermodynamics textbook problems that we are familiar with from childhood do not involve gravity. To illustrate this difficulty with gravity consider, for example, this snippet from an article in the New York Times Review of Books by physicist/mathematician ...


2

Your breath is the same temperature either way. The difference is how much ambient air is brought along with the breath by the time it reaches the object. Emitting a thin and fast stream of air will cause a lot of other air to follow along with it. When you are blowing on the soup to cool it, what you're really doing is using your breath to move a lot of ...


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The Molar Specific Heat at constant volume $C_v$ is: $$\frac{fR}{2}$$ and the Molar Specific Heat at constant pressure $C_p$ is:$$\frac{(f+2)R}{2}$$ ($f$ is the degrees of freedom of a molecule of the gas in concern, and $R$ has its usual meaning.) If you verify, the difference (i.e. $C_p-C_v$) is always $R$ regardless of the value of $f$.


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Simply because the Second Law of Thermodynamics states: Entropy > 0 (always) (think of taking a basketball and squeezing it into a golf ball) so that the density gets larger as the mass gets smaller. therefore, Entropy will always increase in a black hole, gaining Entropy [heat] as it gets smaller.The stars reaching critical mass(limiting mass) ~ ...


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The energy of any infalling mass is absorbed by the black hole. Classically, the temperature of a black hole is absolute zero, since it is a perfect absorber. If you include quantum mechanical effects, as Stephen Hawking did, you can show that black hole horizons will emit radiation in such a way that is consistent with the horizon being a hot body with ...


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The expansion of the universe is related to its cooling. Temperature is defined as the inverse of the partial derivative of entropy with respect to energy: $$ \frac{1}{T}=\frac{\partial S}{\partial E} $$ The derivative is taken at constant volume. This is the definition of temperature. This definition only makes sense for macroscopic bodies. I.e., ...


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Yes, all of them. You have reduced amount due to aeration (bubbles), cooling via convection and cooling via conduction into the side of the cup/bowl.


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Great answer found here: Ductile and Brittle Failure of Materials When a stress is applied to any object, it deforms, i.e., changes shape and/or size. This deformation is called elastic if the object returns to its original shape after the applied stress has been removed. Deformation that is permanent is called plastic deformation. All ...


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Olly, the first part of your thinking is correct, as the atoms receive more energy, the electrons do collide more energetically, but they also move "away" from the atom's center. The further they are from the center, the easier it is for an electric field to "move" them. This means that for the same effort (voltage), more electrons are moved (larger ...


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Entropy is defined in my book as ΔS=Q/T. Already, this is not correct. In general, $$ \delta S\ge\delta Q/T\;. $$ In the specific case where the system being heated is at always in thermal equilibrium, then $$ \delta S=\delta Q/T $$ So, clearly, $\delta S=\delta Q/T$ can not be taken as the definition of entropy. The definition of entropy is [See ...


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Consider each molecule. As you say, masses $m$ are constant but volumes usually grow with higher temperature. That hot air molecules will float up is not do to any changes in mass or so - but to changes in density $\rho$: $$\rho=\frac{m}{V}$$ The fluid of highest density will seek the bottom, and this will be the colder air molecules: ...


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In my statistical mechanics course last year, we derived that $S \propto \frac{1}{T}$ from the following considerations: Consider two boxes (each with $N_i$ particles, $V_i$ volume, $U_i$ internal energy, and $T_i$ temperature, where $i = 1,2$) separated by a wall through which they can exchange energy (heat). Clearly $\Omega_i \propto U_i$ and $\Omega_i ...


3

You asked for intuitive sense and I'll try to provide it. The formula is: $$\Delta S = \frac{\Delta Q}{T}$$ So, you can have $\Delta S_1=\frac{\Delta Q}{T_{lower}}$ and $\Delta S_2=\frac{\Delta Q}{T_{higher}}$ Assume the $\Delta Q$ is the same in each case. The denominator controls the "largeness" of the $\Delta S$. Therefore, $\Delta S_1 > \Delta ...


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I find that the question here relates directly to the definition of temperature, and I'll give a short version of it. For simplicity let us consider a system generated by two sub-systems, A and B, in thermal contact (meaning they only exchange energy in the form of heat). Let me state that, for $A$ and $B$ in thermal equilibrium, the entropy $S_{A}$ and ...


2

Heat added to a system at a lower temperature causes higher entropy increase than heat added to the same system at a higher temperature. How does this make intuitive sense? The formula defines entropy change. Since it defines new word not conceived before and thus devoid of sense, it is hard to imagine as having "intuitive sense". If your question ...


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The notion of temperature is all about how the equilibrium an otherwise isolated system shifts when the system's internal energy changes. So you do not need to worry about whether this internal energy is kinetic, potential, whatever. Actually the temperature is not quite the ensemble average kinetic energy. Your statement is true for an ideal gas and also ...


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Temperature is a quantity that determines how heat flows into and out of a system of particles when it is placed in contact with other systems. By this definition, measuring the temperature of a system with no mass inside is nonsensical; it's not absolute zero, it's undefined. Temperature can equivalently be defined as being proportional to the average ...


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A cubic metre void of anything cannot be described with a temperature. Spacetime itself does not have the property of temperature, so it would be incorrect to say such a void is at absolute zero. However, it is not necessary that any volume not at absolute zero has mass. The property of temperature could be held by photons or other massless particles. For ...


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As mentioned in the comment above, temperature is defined to be a measure of the average kinetic energy of the particles in a system. So with that definition, the answers to your questions should fall out naturally: If there is no mass in a volume, you could say the temperature is absolute zero. I would say it isn't defined because you cannot take the ...


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I have contributed this issue with this arXiv:1411.2425 and previous works. I stress that Gibbs entropy is not "defined" but rather "constructed". It is constructed on the expression of thermodynamic forces in the microcanonical ensemble, and is constructed in such a way that it reproduces them always and exactly. The construction is actually unique. ...


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Simple climate models use the fact that the total amount of water is conserved, and assume that there are three processes that affect it: evaporation, precipitation, and moisture flux (i.e. how the moisture already in the atmosphere moves around). We can write the moisture flux vector simply as the product of the velocity vector and specific humidity: ...


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My interpretation (based on reading the paragraph - no deep background in the physics of these devices so keep that in mind as you take this in): What this is saying is that if you store information as "heat", then the process of measuring the temperature of a "bit" requires some heat to flow - and this will in turn lower the temperature of the element you ...



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