Hot answers tagged thermodynamics
48
Ice cubes have three distinct cooling effects:
The cube, initially at sub-zero temperature, absorbs some heat to reach fusion point (0⁰C).
The cube absorbs more heat to switch phase: it takes some energy to turn 1 kg of ice at 0⁰C into 1 kg of liquid water at 0⁰C.
The water absorbs some heat to become warmer than 0⁰C.
The three effects occur more or less ...
11
A fundamental principle of thermodynamics is that heat flows from warm places to cold ones, through either convection, conduction or radiation, and it will continue to do so until the temperature equalizes across the system.
The stones are colder than the whiskey when you put them in the glass, so as the system heads towards equilibrium, the whiskey gets ...
9
TL;DR: Whiskey stones work by absorbing heat from the whiskey in an attempt to reach thermal equilibrium1.
As Thomas mentioned, ice has three cooling effects:
Ice itself takes 2.11 kilojoules of heat per g to have its temperature increased by 1 degree (Celsius). This number is known as "specific heat capacity"
Ice takes 334 kilojoules of heat per kg at 0 ...
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
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 ...
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 ...
5
Whiskey stones aren't necessarily designed to keep the drink cold, instead they are designed to allow flavor profiles to come out in the drink that might not be present at room temperature. Some whiskeys open up at a slightly cooler temperature and using stones allows you to experiment without diluting the flavor of the beverage. There are better math ...
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 ...
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 ...
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 $$pV=\dfrac13Nm(v^2)_{av}=\dfrac
{\color{red}{\huge{2}}}3N\bigg[\dfrac12 m (v^2)_{av}\bigg]$$
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 ...
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
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 ...
3
The first law of thermodynamics says "the increase in internal energy of a body is equal to the heat supplied to the body minus work done by the body". Assuming there is no heat flow (for simplicity), this says "the increase in internal energy of a body is equal to the work done on the body". Since you are doing work on the gas, the internal energy ...
3
Short answer: No.
As you pump more and more air into the container, the pressure rises and rises. At some point, the molecules are so close to each other that instead of a gas, you get a liquid. If you continue even more, eventually you'll get a solid.
In this solid, atoms/molecules are arranged in a regular pattern with well defined distances between ...
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 ...
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 ...
2
Great question; I myself got confused for a moment there. I'm gonna try to be somewhat thorough, so bear with me. First consider the differential form of the first law of thermodynamics which holds for any quasi-static process.
$$
dE = \delta Q - \delta W
$$
For an adiabatic process, $\delta Q = 0$ by definition, so one obtains
$$
dE = -\delta W
$$
On ...
2
The work done by the gas on the environment is $W = \int_{V_a}^{V_b} P(V)\text{d}V$, where $V_a$ and $V_b$ are the initial and final volumes of the gas and $P(V)$ is the pressure exerted by the gas on the environment when the gas has volume $V$.
There is no simpler expression unless you specify something else about $P$. Work is a path-dependent quantity; ...
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 ...
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 ...
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) ...
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
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
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
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 ...
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 ...
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