Hot answers tagged thermodynamics
9
It's because the Kelvin scale was and still is defined so that as a measure of temperature difference, one kelvin exactly coincides with one Celsius degree. So the temperature in kelvins was defined as the temperature in Celsius degrees minus $A$ where $A=273.15$ °C is the temperature of the absolute zero, without any additional multiplicative factor.
When ...
7
There is no such thing as $S(T,p)$. $T$ and $p$ are both intensive variables, while $S$ is extensive. Just knowing $T$ and $p$ tells you nothing about how large the system is, and therefore cannot tell you the entropy.
However, if you fix some extensive quantity, like the particle number $N$, you can have $S(T,p,N)$. This is implicitly what the book is ...
6
1 - From the equation of state you can express p as a function of the volume and the temperature
$$
p=p(T,V)
$$
then
$$
S(T,p)=S(T,p(T,V))=S'(T,V)
$$
defining a new function $S'$ that depends only on temperature and volume, now as is common in physics by abuse of notation we will note this new function as $S(T,V)$, but you are right that mathematically is ...
5
Wikipedia says:
The kelvin is defined as the fraction 1⁄273.16 of the thermodynamic temperature of the triple point of water (exactly 0.01 °C or 32.018 °F).
And:
1848
Lord Kelvin (William Thomson), wrote in his paper, On an Absolute Thermometric Scale, of the need for a scale whereby "infinite cold" (absolute zero) was the scale's null point, and which ...
5
A good question, here my attempt at an answer.
To describe a thermodynamic system, you can ask for the values of certain thermodynamic quantities:
Pressure $p$, Volume $V$, particle Number $N$, chemical potential $\mu$, temperature $T$, entropy $S$, internal energy $E$.
As it turns out, however, these quantities aren't entirely independent. For an ideal ...
5
Double the ammount of water does not need doulbe the ammount of time to heat, since while the energy needed is doubled indeed, losses due to vaporization and radiation from the kettle should be approximately constant.
You can plot the time needed for a given ammount of water to boil and try to fit a function into that. With two data points you can manage to ...
4
Evaporative cooling works by removing the high-velocity tail of the kinetic energy distribution. That is, only the fastest molecules escape the liquid, leaving the rest to thermalize at a lower temperature. If there is capillary action taking water to the outside of the pot and that is evaporating, then the pot cools down as it is losing heat to the leaving ...
3
The entropy of a black hole is given by the area $A$ of its event horizon according to the formula $S=\frac{kA}{4l_P}$ where $k$ is Boltzmann's constant and $l_P$ is the Planck length.
For a rotating black hole with mass $M$ and a Kerr parameter $a$ the area is $A=\frac{8\pi G^2}{c^4}M(M+\sqrt{M^2-a^2})$. This is largest when $a=0$ corresponding to the ...
3
When multiplying or dividing units, all you need to do is put the units in the numerator or denominator (wherever they appeared) of the answer. So:
$$[e/M]={J\over kg}$$
$$[M/shc]={kg\over{J\over kg^oC}}={kg^2\,^{\circ}\rm C\over J}$$
But this is not the correct way of analyzing your units. You have
$t = e / M / shc = e / (M * shc)$
The units of this are:
...
3
Good question. The rate of temperature increase scales as the power absorbed by the food divided by mass of the food. So to understand your question, you need to understand how power is absorbed.
There is a finite amount of power in the microwaves being produced. These microwaves bounce around in the metal cage where you put your food, until they come ...
2
The integral carries units of $[momentum]^{3N}[space]^{3N}$ which is exactley the same as $1/h^{3N}$, so this is the factor you need for making the phase space volume dimensionless. I don't understand why you say that it has to be $[momentum]^{3N-1}$, just look directly at the integral.
Furthermore, and maybe this is the main problem, the phase space VOLUME ...
2
This is a very difficult question to answer in detail because:
melting produces a layer of cold water around the berg
the iceberg changes its geometry while melting
the iceberg will roll in response to changes in geometry
the iceberg is transported, changing boundary conditions
This is a nonlinear nightmare. In any case, the back-of-the-envelope ...
2
I think it is a mistake, as often happens in popularizations of science.
A water or any molecule may lose kinetic energy and acquire potential energy, but it is the kinetic energy distribution that gives the temperature of an ensemble of molecules. The shape of the distribution shows that there will always be individual molecules at very high energy , in ...
1
As the other answers have said, temperature is a collective property and can only be defined when you have an assemblage of particles. However by definition in a molecule you have an assemblage of atoms, and they have relative motions described by the vibrational excitations of the molecule.
So if you have a large enough molecule you can look at the ...
1
Thermodynamics makes sense when you have large numbers of particles. For example, the second law of thermodynamics has an extremely low probability of being violated when you have Avogadro's number's worth of particles. However, if you have a very small number of particles, the second law will frequently be violated.
This comes up in nuclear physics, where ...
1
It makes sense if all you know about the molecule is its expected energy. Then you can show that it's energy distribution is the Boltzmann distribution $p(E) = e^{-E/kT}$ for some constant $T$, which is related to the expected energy.
So the question reduces to a philosophical view of probabilities. Does it make sense to assign probabilities to a ...
1
Your method seems correct. Here are the details:
$Q_{Heat Water}=m\times c\times (T_2-T_1)=100 \times 1 \times 66 = 6600$ calories
From 1 gram of steam, $$Q=L_v+c\times(100-T_2)=540+1 \times 10=550$$
Therefore, grams of steam needed $=\frac{6600}{550}=12$ grams
1
It really depends on how more gas is being "pumped in". For example, a typical pump will take air in at (roughly) one atmosphere, and pressurize it by doing work on it, which will increase the temperature. On the other hand, if the gas is coming from a cylinder of compressed air, it will cool on entering your volume, so the temperature would actually drop. ...
1
Integrability of the inexact differential $\delta Q$ is a law of nature. Although in general Pfaffian differential forms like $\delta Q$ are not integrable, second law of thermodynamics guarantees that an integrating factor always exists and it is $1/T$ in all cases, $T$ being the absolute temperature.
1
The best way to see this is to realize that the zero heat capacity is a quantum effect. Classically, the heat capacity does not go to zero.
Quantally what happens is that at low enough temperatures all the particles are in their lowest possible energy states. To get even one particle into a higher energy state requires a small but finite energy ...
1
1 - yes the zero temperature limit is not reachable, so you can't measure the heat capacity at zero temperature, what this calculation tell you is that if you measure at smaller and smaller temperatures you will see that C converges towards zero
2- No the reversibility of the path is not important as the entropy is an exact differential
1
Your process will be reversible only if it is a) quasi-static and b) non-dissipative. It will be quasi-static if it is carried out infinitely slowly in such a manner that the pressure on either sides of the piston varies only infinitesimally. It will be non-dissipative if the piston is frictionless and there is no viscous heating of the gas as it expands.
1
Check the definition of "reversible". The process is only reversible if the external pressure and the internal pressure are the "same", where "same" means that for an expansion the internal pressure is infinitesimally greater than the external pressure.
Put another way, work $W$ is given by $W = \int_{V_1}^{V_2} p_{ext} dV$, where $p_{ext}$ is the external ...
1
Imho, this process is driven not by energy considerations but by kinetic considerations. That should be why it naively seems weird that water absorbs heat from a cooler object and evaporates. Note: This is an explanation I came up with on-the-fly and have no references to back up with.
Since the earthen pot has small pores, water "flows" through those ...
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