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19

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 ...


5

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 along all lines. 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 ...


3

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 ...


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

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.


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.


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 ...


1

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.


1

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


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 ...


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 ...


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 ...


1

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 ...


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 ...



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