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41

Candle wax expands considerably when hot and molten. So while burning the candle the level in the glass rises. But when the candle is extinguished the outer region (nearest the glass) cools down quicker (candle wax doesn't conduct heat very well) and solidifies first, becoming immobile. The molten remainder then shrinks before solidifying. So it's the ...


12

Yeah, I would guess that the alternate heating/cooling of the wax in the sun pushes it up the side of the glass. Presumably the surface tension between the wax and the glass is quite strong and holds the wax up once it's been pushed up. Subsequent cycles cause wax to "backfill" the wax that's been pushed up. It would be interesting to design an experiment ...


10

You seem to make the implicit assumption that your vessel is placed in an environment that does not emit any thermal radiation, i.e. is already at 0 K temperature. The temperature of your container will asymptotically decrease to 0 K but will never actually reach it. Assuming black-body radiation, fixed heat capacity $c$, and sufficient thermal ...


7

The energy comes from the sun, that much is certain. One possible mechanism is capillary action resulting in a meniscus, assuming the sun heats the wax to a (near) liquid state. The other possible mechanism is a vaporization/redeposit cycle. During the day, heat from the sun creates wax vapor, which is heavier than air [citation needed] and therefore ...


6

It is entirely possible that the entire candle has melted in the sun. If this happens the molten wax has a significantly greater volume than the solid wax candle and so the whole level will rise up the glass ie liquid wax takes up more volume than solid wax. As it cools again the level will fall as it solidifies. Typically it will cool from the top and ...


5

Concerning the meaning of "temperature coefficient of [measurement]". Generally, when it appears without additional adjectives, the "temperature coefficient" of anything refers to the fractional slope of a linear fit to that quantity as a function of temperature. In the introductory class we often introduce A linear coefficient of thermal expansion, in ...


4

Looking around, the root mean square speed of air at $20$ C is about $500 m/s$, and given that you have $\langle v^2 \rangle \propto \, T$ so that $v_{rms}(T) = \sqrt{\langle v^2\rangle}$ varies with $\sqrt{T}$ then have $$v_{rms}(15) = v_{rms}(20)\times \frac{\sqrt{15+273}}{\sqrt{20+273}} \approx 496 m/s$$ and $$v_{rms}(25) = v_{rms}(20)\times ...


4

There are some issues with the experimental setup you proposed (apart from the fact that when its temperature is lowered the gas would become a liquid and then a solid - if it's not $^4$He: in that case it will stay a liquid). Let's see why. In the picture above, I've sketched your experimental setup. The black box must be impermeable to matter in order ...


4

It won't work because your perfect vacuum is permeated by the cosmic background radiation, which itself is only asymptotically reducing to zero with the expansion of the universe. Trying to exclude the cosmic background radiation backs you into the infinite steps that forms the basis of the third law again. Also, using a container results in quantum ...


4

If you take a bottle of gas and carry it with you on a supersonic plane, then the molecules will go much faster without the temperature changing. If you let pressurized gas flow through a well-designed nozzle (De Laval nozzle), the gas will accelerate to supersonic velocity (i.e., faster than the original thermal speed of the molecules) while the ...


3

Temperature can be thought of as the vibration or oscillation of individual particles. More the vibration, more the temperature. The frames velocity is just the velocity of its mean position, as the vibration is independent of the frame velocity, so is the temperature.


3

The "buoyant force" does not arise from a principle, but is what remains of the gravity force when you subtract from it an average hydrostatic pressure. Let's assume that you have some reference density $\rho_0$. Then if $\rho=\rho_0$ everywhere and the fluid is at rest, $p = \rho_0 g z$. Now let's call this baseline pressure $p_0(z)$, without loss of ...


3

We're talking about the Coanda Effect, right? Then I think this article could provide some useful insights. Quoting from the article: When the fluid flows over the heated curved surface in proximity of the curved surface as the temperature of the curved surface increases, dynamic viscosity of the fluid at the vicinity of the wall is increasing with ...


3

Solid and cold are two distinct concepts. As you can change the state (solid, liquid) with temperature (cold, warm), there must be a relation to it. For each degree of freedom you have a thermal energy of $k_\mathrm B T/2$. Atoms in a crystal have a certain binding energy. If the thermal energy is way larger than the binding energy, no crystal can form. If ...


3

The quantity $\left(\frac{\partial S}{\partial T} \right)_{\{\alpha\}} = TC_v$ is essentially proportional to the heat capacity of the thermodynamic system under study. As far as I know, there is no principle of thermodynamics that forbids such a quantity to be negative. Considerations such as "yes otherwise matter would not be stable" lie outside the ...


3

The simplest answer, if I did not misunderstood your question, is to adiabatically compress the gas, both pressure and temperature will raise.


2

One really, really energetic one might do it. But it would need to be very close to the speed of light, and I'm not sure you could capture its energy to boil the water. Assuming you want to convert the mass of a non-relativistic neutron to energy, it's about 940 MeV per neutron. Boiling a gallon of water (starting at room temperature, 20°C) requires raising ...


2

Your question is: assuming a rod of radius $R=1$ cm with a fixed heat production per unit volume, totaling $Q=100$ W/m, surface temperature $T_0$ and thermal conductivity $\lambda$, what is the temperature $T(r)$ as a function of $r$? First, observe that within a cylindrical shell of radius $r$, the total heat production is $q=Q(r/R)^2$. The temperature ...


2

For the hypothetical case of a thermally perfectly insulated system, I'm sure you can work out yourself from the specific heat and the enthalpy of fusion for water. Given that the enthalpy of fusion (330 kJ/kg) and the specific heat of ice (2 kJ/kg-K) have a ratio of 165 K, and you need the entire ice bucket to stay below the melting point, and your 20:1 ...


2

Your scenario is actually a classic transient conduction problem tackled in undergraduate engineering heat transfer, so we can handle this scenario easily. I took the figure below and adapted a derivation from a popular heat transfer textbook by Incropera and DeWitt: In this figure a warm object is placed in a tank filled with a known liquid (the ...


2

The temperature of a true vacuum would be a measure for the energy distribution of the photon gas in that vacuum. You can derive the occupation of the electromagnetic modes in a volume with Bose-Einstein statistics, which is essentially what Planck did to describe the emission spectrum of a black body. However, you don't need to do understand the details of ...


1

The Ideal Gas Law PV=nRT explains what happens. It can be written as PV/T=nR. For our purposes nR can be considered a constant, PV&T are all expressed in absolute units. Thus if the container volume is constant, P/T is a constant.


1

Your question seems to be about heat transfer through convection. The formula that describes this phenomenon is: $dQ/dt=h*A*(To-Tenv)$ where $dQ/dt$ is the heat transferred per unit time, $A$ is the area of the object, $h$ is the heat transfer coefficient, $Ta$ is the object's surface temperature and $Tenv$ is the fluid temperature (temperature of air ...


1

Whether something is solid or not, generally depends on two things: the attractive forces between molecules/atoms and the temperature. The attractive forces will try to lock the molecules together and make it harder for the material to change shape (or being solid), but this can only happen when the molecules aren't moving fast. If they are moving too fast, ...


1

Molecules of solid object are still able to move and can even move in high speed if the object in high temperature . Being solid only mean that those molecules is hard to separate or make room ( like liquids does ) ,it doesn't mean those molecules have to be standing still . That what I think . Sorry for my bad English


1

My hypothesis is this. When the Sun heats up the wax, the wax gets soften or melt. The surface tension between the glass wall and the wax then drag the wax up the wall.


1

If you are comfortable with electric circuits this method is useful: The temperature difference between 2 parts of a conducting body acts the same way as when you have a potential difference between 2 parts of a conducting wire. The heat conductivity is equivalent to the electric conductivity of the wire which is inverse of its resistance. And the last ...


1

Assuming the water can't cool, then after about 30 minutes, all the neutrons have decayed to protons and electrons, the resultant antineutrinos escape, but an average of around 0.5 MeV per decay will get thermalised in the water (the range of the electrons will be of order 1cm). As Floris points out, it takes 15.8kJ to heat a gallon of water by 1K, so this ...


1

In an exam, Alice scored $50$, Bob scored $40$, Eve scored $10$. Now to raise the class mean, the teacher decided to add $20$ points to every students. So Alice's score becomes $70$, Bob's score becomes $60$, and Eve's score becomes $30$. Now Alice complained to the teacher, "my score was higher than Bob's by $10$ marks, and you see, according to the ...


1

Celsius and Kelvin are two scales that differs only for an additive factor, but the single increment corresponds to the same temperature difference. In other words, an object become "hotter" in the same way if you rise its temperature by 1K or 1°C. You can use conversion formula in differences, just make sure you use it for both terms and keep in mind that ...



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