# Tag Info

2

Polymers such as wax definitely undergo phase transitions. You inquired about paraffin wax, which is oil-based. Paraffin wax is made of long rod-like molecules called linear straight-chain alkanes. It's solid at room temperature, but when refined as liquid paraffin and combined with water, it can act as liquid crystal, which complicates its phase diagram. ...

0

I've dealt with this same issue in the past. To some extent, the issue you are describing is a nomenclature issue. In other words, some thermo text books say work leaving a system is positive, and some thermo text books say that work leaving a system is negative (I'm writing about high school physics and college undergrad text books - I don't know about ...

0

Ice has a sharp melting point temperature because it is a pure substance. Wax is a mixture of higher molecular weight hydrocarbons, so it doesn't have the same sharp melting point. While I have petrochemical experience in my background, I didn't work much with waxes, so I can't give you a firm estimate of the composition, but I suspect that web searches ...

1

This comes down to the fact that ice is a crystal, and wax (and chocolate) is a (glassy) polymer-like material. Here is the chart of stiffness of a polymer as it is heated. When it is cold, the polymer is in a glassy phase. As it is heated, it enters the "leathery phase", and it begins to soften. As it warms up more, the material becomes progressively ...

0

Your question is somewhat more "general" than what is implied. You are essentially talking about cavitation (or getting very close to talking about it). Cavitation does not require slugs of liquid traveling downward. In fact, there are internet pictures showing ship propeller designs where the engineers didn't account for the pressure drop on the ...

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Sometimes it's just faster to measure. Bought 2 exactly same type of thermometer, left both in wrapping. They both said same temp. Walked home 12 minutes, both ended up at above 88. Put one inside, put other in shaded area near porch, under the "tree" that gives us all the cool shade, and the stairs, and otherwise surrounded by walls, in the shade. (also on ...

0

Your problem has to do with your example system rather then sign convention. To understand all of the energy dynamics of the human body is rather complex and increase in body temperature doesn't imply increase in internal energy. The increase in temperature is due to reactions taking place in your muscle but if were to add up to energy due to Kinetic ...

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With no heat input (Q=0) and KE and PE of the system =0 it can't do work on the surrounding.. note: rise in temperature, rise in internal energy is in fact rise in KE of gas molecules in a P-V-T system. Don't complicate things, the 1st law is simply conservation of energy.. if you do work then you lose energy; as simple as that. The example you gave is a ...

0

This has nothing do to with thermodynamics. Your momentum equation is wrong. Where did you get your $(m₁−m₂)×v$ from? You have presupposed that your final velocity is the same. It should be: $$(m_1×v_{1,i}))+(m_2×v_{2,i}) = (m_1×v_{1,f})+(m_2×v_{2,f})$$ Where i is initial and f is final. The directions are included in the velocities. There's also the ...

0

Zeroth law of thermodynamics? Zeroth law of thermodynamics is across systems. If they are colliding they are in the same system. Temperate is not a common velocity. It is a measure of the average kinetic energy. Common velocity? (m₁−m₂) × v is neither a proper application of zeroth law or temperature. You asserted a constraint of a common speed ...

0

Temperature is a statistical property of a large collection of particles moving and interacting at random. Talking about temperature for a pair particles colliding doesn't really make any sense. Your particles simply don't have a temperature and your result doesn't have anything to do with the zeroth law. What you seem to have done is calculated the ...

1

The Fermi level in the Richardson-Dushman equation is already the Fermi level at the temperature your system has. You do not have to account for thermal excitations seperately. Usually the equation is applied with a high temperature and is intended for just this.

2

I am sorry to disappoint you, but there is no such formula that you can just apply. This is because it strongly depends on how and under what exact conditions and with wwhich tools you did the experiment. Think of this: If every methanol molecule reacts (burns) at once all at the same time, then the exact same amount of energy is spent, but it went really ...

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A better question in this case is to ask if we drop a feather from a height, we see that the feather settles on the ground quietly. I you can under this which is a lot easier to explain and much mor intuitive. So as any object falls in a fluid (in this case air) it firstly accelerates and finally reaches constant velocity termed as terminal velocity, at ...

1

Just before the ball reaches the ground, all of its molecules are coming down with almost an equal speed that is the speed of the ball.(Although, due to the non-zero temperature of the ball, the molecules are also vibrating about their mean position wrt COM frame of the ball).And thus the ball possesses a systematic macroscopic kinetic energy. Now when ...

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If you're dropping a projectile into sand, the potential energy that you began with ends up being converted into kinetic energy (from the sand thrown out from the collision), sound energy, and thermal energy. Ultimately, the thermal energy is the only surviving energy after any appreciable time though.

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A somewhat longer answer, since I'm afraid my comment may have seemed a bit abrupt... Lets look at a fairly simple thermodynamic system, the Ag-Ge binary phase diagram. This consists of 3 phases only, fcc Ag, diamond cubic Ge, and the liquid. Taking the published thermodynamic model from J. Wang et al. in Thermochimica Acta 512 240-246 (2011), one can ...

1

You can try ice cubes made of soda for having only soda. You can try having spherical cubes instead of small cubical ones, which take more time to melt. Or else try adding some salt to the water and freezing it. It reduces the freezing temperature of ice which makes the ice freeze faster and melt slower.(But 10 gm of salt in 10 litres make a change of 0.1 ...

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If you look at this table you will see that water, even without the phase transition, actually has the highest volumetric heat capacity of any of the substances listed (absent any phase transition materials). This means that an ice cube is indeed the best you can do for cooling down your drink. Even when you take account of the heat of fusion, water is ...

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Regarding the first question, I believe that you missed a couple of sentences in the Efficiency section of the Wikipedia article: It also requires energy to overcome the change in entropy of the reaction. Therefore, the process cannot proceed below 286 kJ per mol if no external heat/energy is added. If you use 286 kJ/mol, you "don't get something for ...

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yes it would, because of the moons atmosphere, even the craters have not faded away so the footsteps would not fade away. Best type of legacy one can leave

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So lets start with the first part of your question: Black holes radiate away by the famous Hawking process. Hawking radiation has been interpreted in many ways i.e. as pair creation near the black hole, tunnelling from the black hole and almost every other physicist will have a nice way of explaining this. What is the temperature of a black hole? It ...

0

I hope that you don't seriously intend to try this project in the real world. There are several issues that WILL lead to failure and/or serious bodily harm: 1) The water will expand as it is heated. If you start with a container that is too full of water, it will burst the sides of the container before you get to 300 C. 2) Pressure vessels lose some ...

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Yes the core will warm gradually. Heat transfer in a solid is conduction. Ice has a known thermal conductivity and will have a linear temperature profile from all paths from surface to center. There will be concentric rings of constant temperature at all times. It would be impossible to warm just the surface and not warm up the molecules next to the ...

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Short quick answer: From Wikipedia Entropy In thermodynamics, entropy (usual symbol S) is a measure of the number of specific ways in which athermodynamic system may be arranged, commonly understood as a measure of disorder. According to the second law of thermodynamics the entropy of an isolated system never decreases; such a system will spontaneously ...

1

I guess the Alka Seltzer is an effervescent kind of tablet. When it is immersed in water, gas bubbles form on the tablet surface and surface tension effects prevent the bubbles from separating them from the tablet surface. By this effect the global density of the system tablet plus bubbles is going down overtime, until the global density approaches the ...

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Heat and sound, which then also is converted into heat. The total energy of the system is thus conserved.

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The energy gets converted into the form of heat and sound. In this way the energy is conserved.

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When your house heats up, it is receiving the contribution from the hot air outside, plus the sun's radiated heat. When cooling in the Winter, the factors are the cooler air and the radiation loss, which is not comparable to the sun's.

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If all of physics is reversible like some say, then the the increasing entropy of the Universe is causing us to burn fossil fuels as much as we are increasing the entropy.

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In fact, the water would act as a neutron moderator, speeding up the reaction. However, reactor pressure vessels are quite sturdy, and it would be very unlikely for the salt water to enter the pressure vessel.

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In principle, no, you cannot make a Dyson sphere which is indistinguishable from the CMB. The reason is fairly simple. Let's start with a blackbody DS which encloses nothing at all, and is so far from any nearby stars that no noticeable radiation reaches it. Since it is surrounded by CMB with an effective temperature of 2.75 K, it will reach an equilibrium ...

0

I don't think so Dirk. GRBs don't happen for nothing. See this 2001 paper by Friedwardt Winterberg. I think it's essentially correct. Why? Take a look at Einstein saying light curves because the speed of light is spatially variable. Also see Shapiro, and this Baez article: Einstein talked about the speed of light changing in his new theory. In the ...

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If you correct plank units by using Einstein's Appendix 2 of "Relativity" relation meters = i c seconds, and carry the i through all equations, converting any seconds units to meters, you get better insight. For example, this gives E= -mc^2 instead of E=mc^2, which cosmology agrees with. So you also convert all mass units to negative energy units. You then ...

1

A gas torch can reach thousands of degrees (on your favorite temperature scale) without any electricity. A large mirror or lens positioned appropriately with respect to the sun can project light onto a focal point, reaching a temperature up to that of the surface of the sun, which is ~ 5800 K. This does not require any other energy source or electrical ...

0

The answers seem too scattered, abstract, and complex. It is important to keep in mind temperature is a direct measure of the average (or RMS) kinetic energy of the particles. First let's be clear and say each container contains the same number of particles in the same size volume. Let their temperature be different by a factor of two. If you add enough heat ...

1

If you built the sphere, at the optimum radial distance, then insulated the exterior as much as possible, would the gravitational redshift provided by the black hole not act to sort out your problem for you, if you want to dissipate it, as far as co-ordinate observers at any reasonable distance were concerned? Would accretion discs and the massive gravity ...

2

Yes, bonds have mass, like every other kind of energy. This can be significant; if you had a glueball (a hypothetical particle made of massless gluons), it would have mass, and all of the mass would be from the bond energy! Same would go if you somehow managed to bind photons together.

20

As far as the theory goes, you are absolutely correct, the (negative) binding energy between atoms in a molecule contributes to the total mass of that molecule, so a stable molecule is less massive than the sum of the masses of its constituent atoms. However (as you yourself calculated), the mass difference is absolutely tiny, and as far as I know, it has ...

1

$f(v)$ is a probability density, so it is convenient to set $$\int_0^{2V{_0}} \mathrm{d}v \, f(v) = 1$$ In your case this will give you an expression for $A$ in terms of $V_0$. Now, e.g. $$\int_{V_0}^{2V{_0}} \mathrm{d}v \, f(v)$$ is the probability that a particle has velocity between $V_0$ and $2V_0$. The average value of any function of $v$ can then ...

0

$N$ is equal to the sum of the two integrals you have: $$\begin{eqnarray} N = \int_0^{2 V_0} dv \ f\left(v\right) &=& \int_0^{V_0} dv \ A + \int_{V_0}^{2 V_0} dv \left(-\frac{A}{V_0} v + 2 A\right) \\ &=& A V_0 + \left[ -\frac{A}{2 V_0} v^2 + 2 A V\right]_{V_0}^{2 V_0} \\ &=& \frac{3}{2} A V_0 \\ \end{eqnarray}$$

0

Consider the forces acting on the piston. Since there is no friction, the only forces are $F_{\text{gas on piston}}$ and $F_{\text{atmosphere on piston}}$. So the net force on the piston is $F_{\text{piston, net}} = F_{\text{atmosphere on piston}} + F_{\text{gas on piston}}$. Since the piston is massless, Newton's 2nd law says $F_\text{piston, net} = ... 0 Yes, the temperature of the gas would decrease quite fast, given that the molecules in the container are still, which implies zero temperature for container. However, if the container's temperature is non-zero, it sometimes happens that gas molecules will instead gain energy because the molecule it collides with is moving fast enough in the opposite ... 0 If you assume that the gas is ideal then each collision of a molecule of gas with the wall conserves the kinetic energy. Hence the temperature will stay the same. 0 The temperature of the gas will eventually reach equilibrium with the walls of the container, and since a perfect insulator is not possible, the gas, walls and outside environment will, given enough time, be at the same temperature. 3 There is no particles in space so how can it have a temperature since it does not really pass the heat around and if so how can a object cool down in space? When space is said to be cold, it does not always mean its temperature is low; it may not have temperature if the radiation present is not equilibrium radiation (which never is exactly; the ... 2 The variables involved here are classical and you resolve them classically. They enter into the operator because they are parameters of the wavefunction. So let's do this a little more broadly. For continuous systems, we want a family of solutions based on some parameters which I'll collectively identify as$\alpha \in A\$; the solutions are then labeled ...

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I don't see anything wrong except the extra minus sign. The idea that heat gained equals heat loss is spot on but both quantities need to be positive in this case. One way I like to think about this would be that the change in heat of the metal plus the change in heat of the water must add to zero. In this equation, "H_metal + H_water = 0", the sign of ...

0

You are correct that space is not filled with jiggly stuff with energy ½ KT per degree of freedom. However, space is filled with cosmic microwave background photons. Their distribution in energy is that of a black body distribution with temperature 2.7 degK. If you put some material out in space (away from other heat sources), the material will absorb and ...

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Intergalactic space is filled with a photon gas at temperature 2.7K. For heat transfer you don't necessarily need atoms, other particles, such as photons suffice.

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