# Tag Info

6

There are two parts to your question, and I'll address them individually, assuming that the room+ball system is an isolated system, i.e. no work is being done on the system, and no energy is being added / removed. Part I: How cold will it get? This depends on the size of the room, the materials that everything in the room is made of, the size and material ...

0

I would like to repeat the comment by Rossetto: During the move, the piston acquires some kinetic energy. ... If the gases are viscous, some of the energy will be converted into heat, warming up the gases. Depending on how this energy will be distributed among the gases, the final states will be different and determined by the dynamics of evolution. ...

1

Heat capacity is the ability of a material to store heat, higher the heat capacity higher the amount to heat stored by the material. Heat transfer usually varies inversely with heat capacity, i.e heat transfer will decrease with increase in heat capacity and vice versa. Thermal diffusivity is the ratio of thermal conductivity to the heat capacity, it says ...

1

Heat Capacity is well defined: $C = \frac{\Delta Q}{\Delta T}$ see: http://en.wikipedia.org/wiki/Heat_capacity Heat Transfer by Holman. It is a property that refers to the amount of energy required to increase the temperature of an object. This is slightly different than specific heat $C_p = \frac{\Delta Q}{\Delta T \times m}$ see: ...

0

If I understood well, your metal plate acts as a mirror. Seeing your own image in it is not surprising because what we call "IR" are actually photons. These photons have less energy than visible photons (IR means infra red, i.e. "energy below the energy of red photons") so you can't see them with your eyes. They are emitted by any hot object (that explains ...

1

The main thing you want the air conditioner to do is to remove the heat that enters the apartment and eject it to outside. The rate that the heat enters depends on the thermal conductivity of the apartment (which we assume to be fixed) and the temperature difference between the inside and the outside. The greater the heat difference, the greater the heat ...

0

For visibility I will post this as an answer as well. The problem in my text book wanted me to fit the numerical/dimensionless approximation to the graph in the text book, to estimate at what value of $T$ the graph would reach half its maximum value (in Figure 1.14 in Schroeder's Introduction to Thermal Physics that's $C_V=3R$). The estimation of ...

2

A number of factors go into determining the optimum daytime temperature settings for cooling: one of them is humidity. An air conditioner doesn't 'cool', it removes the heat, through the heat of the liquid refrigerant vented to the atmosphere. More humidity means in a residential setting without a lot of tonnage of air being moved is more 'wattage' is ...

0

If you're asking whether atmospheric pressure affects carburettors then the answer is that yes it does. In my wilder days I used to ride motorcycles that used carburettors (this was in the late 70s). While I have never noticed an affect of pressure variations due to the weather, on a trip over the Gross Glockner pass it was quite noticable that the engine ...

2

[As requested, I convert my comment into an answer, as it might also be useful for other people.] There is a very interesting series of works by Lieb and Yngvason on entropy and the second law of thermodynamics, based on the kind of axiomatic approach you seem to be interested in. You can start with this introductory paper, or this, this or this more ...

0

Erik, nothing will happen (as you already know!). Water and ice are in equilibrium at the triple point. No heat can flow at constant temperature. The water and ice have the same free energy per mole, so no spontaneous change can occur, and the total entropy cannot rise as a consequence of heat transfer. It is much simpler to just consider putting ice ...

1

I'm not sure how one can know that the half maximum corresponds to $kT/\epsilon \approx 1/3$ without resorting to the formula for the heat capacity. Still, notice that there are only two energy scales, i.e., $\epsilon$ and $kT$, in the problem. Then, whatever (dimensionless) number that determines whether the equipartition holds or fails has to be the ratio ...

0

Non-equilibrium thermodynamics is an emerging field of study. Asking whether the new concepts introduced by that emerging field will prove to be of practical value is a bit premature. The concepts are only now just starting to make their way out of academia. The concept certainly has merit. On a grand scale, a thermodynamic system that truly is in ...

1

Now we have the second law of thermodynamics, that says that entropy always increases. Second law does not say exactly that. It has more formulations, some of which use the concept of entropy. One such formulation is When thermally insulated system changes its state from one equilibrium state to another, its entropy cannot decrease. This statement ...

2

The heat capacity of an Einstein solid is given by $$C = Nk \left(\frac{\epsilon}{kT}\right)^{2} \frac{e^{\epsilon/kT}}{(e^{\epsilon/kT}-1)^{2}},$$ where $N$ is the number of degrees of freedom. So the value of the energy quantum $\epsilon$, or more precisely the ratio $x\equiv\epsilon/kT$ matters! The above equation tends to the ...

1

You have not specified how the pressure is controlled in the two systems. If they are each at the triple point pressure of 611.73 Pa there is no reason for heat to exchange and all will stay constant. If the pressures are different from this (and not on the freezing curve) energy can be released if there is heat flow by transferring heat between the ...

-1

The original state is not in thermodynamic equilibrium. The liquid side has a lot more heat than the solid side due to the heat of fusion.

1

Are not we simply saying that things more likely to occur, occur more times? Isn't it then, that the second law is simply an inmense tautology? No, this argument doesn't suffice to prove the second law. This argument only proves that thermal fluctuations away from equilibrum should be rare and short-lived. That's a statement that doesn't have anything ...

2

There are (at least) two things going on. Perhaps the easiest place to start is with the temperature as estimated from the radiation in the universe - possibly what you are referring to when you say the temperature is approaching 0K? The radiation in the universe takes the form of thermal blackbody radiation. It is emitted by material in thermal equilibrium ...

-2

I believe that you are safe in assuming it is cooler because it is spread out more. Wikipedia, on the subject

2

Ultimate physical motivation Strictly in the sense of physics, the entropy is less free than it might seem. It always has to provide a measure of energy released from a system not graspable by macroscopic parameters. I.e. it has to be subject to the relation $${\rm d}U = {\rm d}E_{macro} + T {\rm d} S$$ It has to carry all the forms of energy that cannot be ...

1

Thermodynamically entropy is defined by $$\mathrm{d}S = \frac{\mathrm{d}Q_{rev}}{T} \, ,$$ where $\mathrm{d}Q_{rev}$ is the heat, transferred reversibly. As you point out it can be shown that this quantity is a function of state. This implies that the entropy of any thermodynamic system has, up to a constant, a well defined ...

-1

Coefficient of performance (COP) talks about the amount of heat transferred for the amount of power used. When two points are close together in temperature, you can move a lot of heat for not a lot of energy expended. This is the principle behind using heat pumps for heating a house, for example. You take the heat from outside (which is "free"), and pump it ...

0

The method of modelling the chicken as a sphere, as mentioned on your other forum, might work something like this. Model the chicken as a sphere and use the heat equation, treating the surface as your boundary. The way to do this is discussed here. To obtain an approximate thermal diffusivity for a chicken you could use the equations you discovered above, ...

2

I think you will find more useful material by rather looking for "heat pumps" instead of refrigerators. If you teach them about the basic ideas of thermodynamics used in a heat pump first, understanding refrigerators becomes trivial. Three useful links that I just found: Physics of heat pump How heat pump works The basic physics of heat pumps

0

I think the key to the paradox is that you can't ignore the reflection coefficent. Lets say you have a lump of coal inside a shphere of polished steel. Yes, the coal emits much more heat than the steel; but that doesn't mean there is a net transfer of heat from the coal to the steel. Because the steel reflects back the excess heat. There is a fixed ...

1

Liquids evaporate at any temperature - not just at their boiling point. This is the reason, for example, why wearing a wet shirt on a windy day makes you so cold: the water evaporates, and in the process "takes some heat with it". The explanation for this is simple when you think about statistical thermodynamics. You have a lot of molecules whose energies ...

0

The curve represents the boiling point of water at different pressures and temperatures. If you are heating water at the pressure selected, it will be at the temperature indicated by the curve when the water boils. The latent heat of vapourization is always at the boiling point. The pressure and temperature determine where this point is.

1

No. As @Jim said, the heat would weaken the rock, which would cause a tunnel collapse before any sublimation could occur. Also, remember that the air in the tunnel would generally be at the same temperature as the rock (unless a large cooling system was put in), so thermal equilibrium would be maintained without any sublimation.

2

Yes, it's boiling. The clue is that you said some of the air above the liquid was sucked out, so the pressure is significantly lower than normal atmospheric. Boiling happens when the gas pressure at the surface of a liquid is lower than the liquid's vapor pressure. When boiling something in a open container, we heat it to make it boil. Water boils at ...

2

If you fill a bottle with hot water, the water in contact with the sides cools quickly. This sets up convection currents, which help to cool the whole bottle, and ensure its temperature is rather uniform. The syrup is viscous, so this doesn't happen. The syrup in the centre of the bottle remains hot - maybe 80 C after half an hour. The coldest part of ...

1

You can only measure a part of EU compliance. By using your utility meter and measuring the difference between the power consumed over say 10 minutes with the appliance on and off (with everything else in the house as off or steady as possible), you can measure consumed power. However, that's just one part of EU compliance. The other is power factor. ...

2

Your requirement that the measurement be made with equipment available in a kitchen is a severe constraint as I can't think of any way of measuring the electrical power supplied. If it's impossible to measure the electrical power in then the only other approach is to measure the thermal power out - i.e. measure the heat produced by the appliance. Given that ...

1

It has not be proven that The Second Law of Thermodynamics is physically derived from other basic physical principles. The H-Theorem is predicated upon some pretty serious, yet plausible, assumptions about how our universe works. To my knowledge, these assumptions have not themselves been explained using other principles and/or experimental verifications of ...

1

A certain volume of space with a uniform distribution of particles has maximum entropy. That is correct for non-interacting particles, but wrong for particles with the gravitational interaction. When gravity condenses these particles, it increases the entropy of the system, not decreases it, at least when the Jeans instability condition is satisfied. ...

0

I think you would need more information. Either how much energy per time interval is needed to maintain that room temperature, which would then already be your answer, or how thick your walls and windows are, what the material is exactly and what the temperature outside is.

0

There's no easy answer to your question as you don't define the nature of the gaz nor the nature of the cells. "Even if my understanding of the second law is incorrect, I don't understand why we can't extract heat energy from an object without a temperature gradient by placing it under certain conditions. For example, heat transfer via infrared radiation ...

0

Search AGWunveiled to see volcanic events on graph with measured average global temperatures. Any effect is lost in random scatter except Tambora in 1815 (approximately 10X the next largest) which was followed by 'the year without a summer', 1816.

1

The area of the manometer tube makes no difference. All that matters is the difference in the heights of the two ends (labelled $x$ in your diagram). That's why pressure units like the torr exist that are (or rather were) defined as the pressure difference when the difference in height of a mercury manometer is 1mm. All that matters is the height difference. ...

3

Ahh, I spent quite some time reading this problem, the problem with applying Dalton's Law of Partial Pressures is that we shouldn't be multiplying moles of $CO2$ with the total Pressure, rather we should multiply the mole fraction of $CO2$ with the total Pressure, in this case however, since the initial quantity/moles of oxygen is not known, it is not ...

1

It depends on the eruption, location, magnitude , timing etc. Have a look at the great Krakatoa eruption in 1883. In the year following the eruption, average Northern Hemisphere summer temperatures fell by as much as 1.2 °C (2.2 °F).[9] Weather patterns continued to be chaotic for years, and temperatures did not return to normal until 1888. Note, the ...

1

Volcanic eruptions release both stratospheric ash, which reduces insolation at the ground, and carbon dioxide, which has a long-term warming effect. The ash falls out of the atmosphere after a few years, but the extra carbon dioxide is brought from underground into the biological carbon cycle more or less in the same way as the carbon from fossil fuels. My ...

0

With a constant volume for the container, the sum of the volume of the liquid and vapor is constant. Let's start in an equilibrium position, temperature $T_1$. Volume of liquid is $V_l$ and vapor is $V_v$. Vapor pressure = $P_v$. Now we increase the temperature to $T_2$. The first thing that happens is that the pressure of vapor increases to ...

0

If you add heat to the system, it will be absorbed by liquid molecules changing to the gas phase. That reduces the number of liquid molecules (which reduces the liquid volume) and increases the number of gas molecules (which increases the pressure). The higher pressure will compress the liquid (probably a small amount) to a higher density. Both effects ...

1

The expansion takes place when the pressure is changed suddenly and drastically modifying the initial equilibrium state, remember that at difference of the reversible case the pressure is not changed infinitesimally. So in the formula $P \Delta V$ use the final pressure. The other questions you set up are very interesting, I recommend you this paper that ...

0

Floris's answer is of course very nice. I will just show very quickly how you can solve these scenarios in terms of exergy. Definition: In the case of a system coupled to a reservoir(environment), exergy is interpreted as the potential of a system to do work as it achieves equilibrium with the reservoir. Sometimes also referred to as "maximum shaft work" ...

3

Based on some "google research" I get the impression that the popularity of the perfume thought experiment stems from a 1975 Scientific American article written by David Layzer called The Arrow of Time. The article featured this figure visualizing the thought experiment: Of course, the notion that the second law of thermodynamics implies an asymmetry ...

2

The main assumption used here is that the direction of the velocity is distributed uniformly. This means that if I take a particle at random, the probability of its velocity pointing in any particular direction is the same for every direction; that is, no direction is more likely than the rest. Now suppose I take a whole bunch of particles and measure ...

2

The key step in this derivation is the assumption of isotropy of particle velocities. Think of the velocity vector of each particle as a randomly selected direction in space, in combination with a randomly selected speed from a specified distribution of speeds. The result of such a random sampling is that on average \$\langle v_x\rangle=\langle ...

1

As a rule, when the initial temperature difference is the same, the system that rejects heat at the lowest temperature will be the more efficient. That is because this involves the least amount of entropy. So without doing the math, I would say that case (b) will give rise to the greater amount of extracted work - which is also your conclusion. And yes, ...

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