New answers tagged thermodynamics
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
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
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 ...
0
There is a systematic approach to verify such identities. Usually, the equations of state of a thermodynamical substance are given by expressing $T$ and $S$ as functions of $p$ and $V$, symbolically, $T=f(p,V)$, $S=g(p,V)$ for suitable functions $f$ and $g$ of two variables---this is presumably because these variables can be measured directly without ...
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 ...
0
It's certainly possible, as this video shows. The problem is that the vapour pressure of water at 0°C is only 0.6kPa (abut 0.6 atmospheres) so even in a very good vacuum the evaporation rate is slow and it takes a long time to get rid of the latent heat of fusion. Cooling with conventional refridgeration is a lot quicker if you want large quantities of ice.
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 ...
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 ...
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 ...
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 ...
0
Just write it out like this:
$t = \frac{e}{\frac{m}{shc}}= \frac{J}{\frac{kg}{\frac{J}{kg\cdot{}^\circ{}C}}} = \frac{J}{J \cdot kg \cdot \frac{1}{kg\cdot^{\circ}C}} = \frac{1}{\frac{1}{^{\circ}C}} = ^{\circ}{\rm{}C}$
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:
...
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
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
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
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
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
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
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 ...
0
Simply because the rate of flow of heat slightly increased when you added more water:
Heat is transferred very quickly to the kettle (which I assume is made out of metal), as metal is a good conductor of heat. Air is a bad conductor, so no heat enters the water through the air. Water is a worse conductor of heat than metal (it also has a pesky habit of ...
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 ...
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 ...
0
I think the confusion arises due to difference in equations from Physics and Chemistry text books.
The physics text writes $U=q-w$ while chemistry says $U=q+w$.
From the physics text its mentioned that work done by the system is positive means $U=q-(+w)$
and work done on the system is negative means $U=q-(-w)$.
THEREFORE, both are correct in their own ...
0
My intuition was correct. For more information visit APS Website.
2
Because of the following. Disorder is usually equated with one's ignorance of the system - the less you know about the outcome of the random variable the more disordered it is. If the system turns out to be in a very unlikely state with low $p(x)$ you will naturally consider yourself to have been more ignorant than when it is in a state you consider very ...
1
Yes, colouring the water could make the pool heat faster, though whether the colour you noticed has an significant effect is debatable.
Swimming pools exchange heat with their environment by conduction through their walls, evaporation at the surface and absorption of sunlight. I have little direct experience of swimming pool thermodynamics, but some ...
1
One can see that a black hole in thermodynamic equilibrium with its environment is always unstable by looking at the Hawking temperature of a black hole, given by
$$
T = \frac{\hbar c^3}{8\pi k_B M G}
$$
The temperature being inversely proportional to the mass means that big black holes are cold, a black hole with the mass of the sun has for example a ...
1
Latent heat of evaporation cools the pot in the same way it cools your skin. If the pot is then at a lower temperature than the water in it, heat energy will be transfered to the pot from the water in it by conduction.
Using plant xylem water for evaporative cooling, the desert cicada Diceroprocta
apache can maintain a body temperature as much as 5°C ...
0
It is not possible, assuming that all tools we could use are "normal" in some particular sense (that is, they agree with some energy conditions). That means, they are made of particles and fields that cannot travel faster that light, for one example. Everything physicists discovered so far is "normal" in that sense, and it is highly unbelievable that we ...
5
Firstly, the logarithm needn't necessarily be to base 2. Changing the base just introduces a (scale) factor, so log10, log2 and ln are all equally useful. Log2 is convenient for people working with binary systems.
Let's deconstruct the formula. I will define entropy to be $H = E[-\log(p)]$. You can see that this will reduce to a weighted average which ...
0
Anyway, consider the following two examples.
(1) A random variable always has a single, definite value, i.e., it isn't really random. Then the Shannon entropy is zero. This means that you don't gain any information by being told that my puppy is cute -- puppies are always cute, with probability 1.
(2) Suppose a variable is equally likely to take on $n$ ...
5
The work in the first law is exactly the usual work $W=\int Fdx\rightarrow\int PdV$. For point particles, this is enough to completely specify the behavior of the system using Newton's first law, or energy methods. However, for macroscopic objects, the motion of the internal components (in thermodynamics these would be particles) have some additional degrees ...
4
There are several ways we can approach this, but I'll argue that the integral of the PV curve is a more general form of the force times distance concept of work:
$$ W = F \Delta x $$
This applies for pretty much any action over a distance. If you compress a spring, lift a box, drive a car, the above equation applies to formalize the work done. To ...
2
The $W$ term in the first law expression exclusively refers to the mechanical work done by a system and all other things , all other possible exchanges of energy are clubbed together in $Q$.
Suppose I am the system under consideration , and I apply a force on a block and that does some mechanical work (that is the point of application moves a distance) ...
0
see we use simple integration ..
$$dT/dt = -k(T - 100)$$
this is in accordance with Newtons Law Of Cooling which says that the negative rate of change of temperature is directly proportional to the difference of the initial temperature of the substance and the surrounding temperature of the environment.
$$dT/dt = -k(T- 100)$$
$$dT/(T- 100) = -k \,dt$$
...
1
The other answers tackle the statistical/thermodynamic aspect. I will tackle the "falling apple " aspect.
Why does the apple fall?
From this observation onwards nature was modeled mathematically as interactions between masses, in this case, charges in the electromagnetic case etc.
The observations of gravitational interactions led to a mathematical model ...
3
Large systems with many degrees of freedom (e.g. a ball consisting of many molecules) tend to settle into low energy states. This is a direct consequence of two fundamental laws, the first and second laws of thermodynamics: energy conservation and entropy increase.
A system with many degrees of freedom can be in many different microscopic states (think ...
1
I will address such sample systems as a point (or a small metal ball) rolling or bouncing on some hard surface with hills and pits, and an atom which can be either in excited or in basic state.
I. If we consider an ideal closed system, then the enegly is conserved. But real systems do not (exactly) behave this way. For a macroscopic mechanic motion we can ...
1
The probability of finding a system in a state with energy $E$ is $P(E) = \exp(-\beta E)/Z$, where $\beta = (kT)^{-1}$, $k$ being the Boltzmann constant and $T$ being the absolute temperature. $Z$ in the formula for $P(E)$ is the canonical partition function. For our purpose, we can consider it as a factor introduced to ensure that $0 \le P(E) \le 1$. The ...
0
In this case, $P_2 > P_1$. Therefore, of you look at the side tube, through which gas is supposed to flow, there is a pressure gradient that will oppose the flow of the gas.
But there is one more flaw in the experiment. The liquid itself will rise through the side tube to a height that matches the level in the main vessel. In that case, there will be ...
0
The most important thing about feeling is that its relative and depends on the temperature of sensor point on skin. That's the reason we need thermometer to measure temperature with which everyone could agree.
Energy transfer is involved here, but it has very little to do with Amanda's feeling. Amanda's tongue and mouth was at high temperature due to ...
0
One way is to find out the internal change energy of the system and infer the heat transfer to the system from that and the work done: $$\delta Q_\text{to}=dU-\delta W_\text{on}.$$
If you have a handle on the system's entropy, on the other hand, then you can use the Gibbs relation, $$\delta Q_\text{to}=TdS,$$to find the heat delivered.
In general, though, ...
2
Ordinarily, no. In fact, the liquid level would usually rise in the tube until it is at the same height as in the main tank, so your gas would be very far from the end of the tube. To see why, imagine the surface of the liquid directly above the the end of the tube. The pressure is P1. That pressure gets transmitted through the liquid down to the end of ...
1
This paper,
http://www.nature.com/ncomms/journal/v4/n4/full/ncomms2665.html
Shows how a thermodynamic cycle of a gas can generate work for free if generalized uncertainty relations are violated. It ties uncertainty principle to notions of information theory.
0
If you have a path on $p-V$ diagram that is parametrized by some parameter $x$, so that $p=p(x),\,V=V(x)$, then:
$$
dU=\delta Q+p(x)dV(x)
$$
here $Q$ is the total heat received by the system (it is negative if system releases heat). I write $\delta Q$ to indicate that $Q$ is not a function of state, and $\delta Q$ is not a full differential. Assume now that ...
0
AB-isothermal.
$\Delta W=Q$ ; area under the curve depicts the work done , ie. heat intake.
$\Delta W=n\Bbb RT\ln\dfrac{V_2}{V_1}$
BC-isobaric.
Can't be calculate directly from curve.
otherwise use ,$Q=\Delta W + \Delta U$
where $\Delta W=P(\Delta V)$ and $\Delta U=\dfrac f2 p\Delta V$
CD-isochoric.
$Q=\Delta U=\dfrac f2 V\Delta p$ .
DA-adiabatic .
...
4
Nice catch!
For reference here is the book page.
:
See , though it may error in printing or anything else.The final equation they get $$pV=\dfrac23K_{tr}$$ is very correct.
The correct form of $eq.(18.12)$ must be $$pV=\dfrac13Nm(v^2)_{av}=\dfrac
{\color{red}{\huge{2}}}3N\bigg[\dfrac12 m (v^2)_{av}\bigg]$$
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