New answers tagged

0

Yes. According to the accepted theory of gases, a diatomic molecule (like N2) has both translational kinetic and rotational kinetic energy, five partitions (two rotation axes, and three directions of travel in space), which each can hold kT/2 energy, on average, for a total thermal average energy of 5kT/2. A neon atom, on the other hand, cannot hold ...


3

Your question is how the differential quotient $\frac {dU}{dS}$ can mean anything in equilibrium when the quantities $U$ and $S$ are supposed to be constant, it is equilibrium after all... Indeed, $dU$ or $dS$ do not mean changes over time in a physical sense, ie., over time during some process, Instead they mean the differentials of the respective ...


-2

Lets not complicate it with the use of thermodynamic terms.Temperature is a measure of the internal energy of a substance and internal energy is related with the change is the heat which is turns relate with entropy change.


0

How do we define temperature? You have to start with the zeroth law of thermodynamics which is all about bodies in thermal equilibrium. The strange name is because after the first and second laws of thermodynamics were formulated suddenly somebody realised there was another law of thermodynamics which in some ways was more fundamental than one and two, so ...


0

There is no easy way to calculate this for liquids because the heat exchange will depend on whether there is any convection in the liquid or not. You can calculate the solution for the heat (conduction) equation for your geometry, but this may or may not give the right answer. The problem is a lot better defined for solids which can not convect. The ...


1

$CO_2$: Without greenhouse gases, Earth would be about 30 degrees colder. Everyone sensible agrees that $CO_2$ contributes to this, but also that it only directly causes part of the warming. Water vapor is the other important greenhouse gas. There are feedback relationships between temperature, $CO_2$ levels, and water vapor levels, e.g. if the world gets ...


0

Typically, you need an integral over a scalar field each time you want to know the total amount of an extensive value fiven in the field: total mass, total load, etc from the density field... To compute how a light ray is absorbed as traversing a semi-opaque media, you how have to write the integral giving the optical depth along the ray. etc.


1

A CFT is still a QFT, and the way to put it at finite temperature is standard for any quantum system - you take your Hamiltonian $H$ and compute $Z=\mathrm{tr}\,e^{-\beta H}$, where the trace is over the Hilbert space of states living on $\mathbb{R}^{d-1}$ if your CFT is in $d$ dimensions. The thermal correlators are computed in a similar way, ...


1

Instead of just adding boiling water as @Gert did, lets drain cool water and replace it with boiling. This way we'll keep the total volume of water constant $V=300$ gal. We can determine what volume of water to add $v_\mathrm{add}$. $$(V-v_\mathrm{add})\cdot T_\mathrm{orig} + v_\mathrm{add}\cdot T_\mathrm{add} = V\cdot T_\mathrm{target}$$ $$v_\mathrm{add} ...


2

It's not the question asked, but looking at the power requirements might give some insight. Raising the water temperature requires a specific amount of energy, and the time constraint gives a required power. $$P = \frac{m C T} { t}$$ $$P = \frac{300\text{gallon }(1000\text{kg/m^3})(4.186\text{J/g K}) 14\text{degF}}{1 \text{hour}}$$ $$P = ...


1

A typical hot tub will be coming to an equilibrium based on some forcing term $F$ adding heat to the system plus some proportional response $\lambda$ which loses heat to the environment:$$\rho \frac{dT}{dt} = F - \lambda (T - T_0)$$This is a linear ODE whose equilibrium temperature is $T = T_0 + F/\lambda.$ To increase $T$ as fast as possible you should: ...


2

Assuming no heat is lost to the environment the heat balance on adding some boiling water ($212\:\mathrm{F}$) is given by: $$m_{bath}cT_{bath}+m_{added}cT_b=(m_{bath}+m_{added})cT_f$$ where: $m_{bath}=300\:\mathrm{Gall}$ is the initial amount of water, $T_{bath}=32.2\:\mathrm{Celsius}$, $m_{added}$ the amount of boiling water added, ...


1

Heat transfer can occur by conduction, by convection, and by radiation. If you consider conduction through the bulk of the cup, the rate of heat transfer is directly proportional to the temperature difference across the material of the cup. As your experiment holds all variables equal except the temperature difference, cup A will lose heat at a faster rate ...


3

Yes, agitation will generally promote heat transfer and reduce heating times (although quantifying the effect is not easy). But the effect is not related to the bulk speed of the kettle. When the water is heated a diffuse (poorly defined) boundary layer is formed on the bottom of the vessel. This layer is at a temperature that is slightly higher than the ...


2

Coherent motion does not add to the temperature; so you would have to shake it violently, with random motions. Consider sound in air - this is a coherent motion - when you can no longer make out the sounds in a closed room, the energy of the sound waves has been transformed into heat.


3

The term absolute isn't strictly defined, but most of us would agree that an absolute temperature scale has to have its zero at absolute zero. You are free to define any unit of temperature you want. There is nothing special about the size of the degree Kelvin, it was chosen to be the same as the degree Celcius i.e. one one hundredth of the difference ...


2

While a funny-looking coincidence, this is not a valid alternative expression for entropy in general, since the entropy of a probability distribution (which are what rigorously hides behind the strange word "macrostate") is more generally given by $$ S = - k_B \sum_i p_i\ln(p_i) \tag{1}$$ and becomes only $$S = k_B \ln(\Omega) \tag{2}$$ in the case of a ...


1

From a mathematical perspective this means that it is not differentiable. The problem is that you need the discreteness to be able to count states. If you replace the discreteness by something smooth you get something differentiable, but your definition of entropy no longer makes sense. This is just one of the points where mathematicians cringe, but it works ...


1

In ideal gas model, temperature is the measure of average kinetic energy of the gas molecules. In the kinetic theory of gases random motion is assumed before deriving anything. If by some means the gas particles are accelerated to a very high speed in one direction, KE certainly increased, can we say the gas becomes hotter? Do we need to ...


5

Static discharge has a strong correlation with humidity. In winter, due to low temperature, the dew point is lowered and as a result, the water content in the atmosphere is lowered. In summer, correspondingly the dew point is higher and water doesn't condense as easily and so there is a lot more content of water in the air. Now the moist air being weakly ...


1

In principle yes, though the situation isn't as clear cut as you describe. If you could confine a volume of matter within some volume then gradually heat it by adding energy to it then at some point the total energy density would exceed the density required to form a black hole and at that point the matter would start to collapse into a black hole. However ...


2

Well, you could do it in any arbitrary units if you choose. Let's look at the perfect gas law: $P = \rho R T$. I can pick any consistent set of units that I want there, but it may involve offsets. For example, let's keep $P$ in Pascals, $\rho$ in kilograms per unit volume, $R$ in joules per kilogram per kelvin and let's use $T$ in Celsius. We know that $T_k ...


2

It should be obvious that the ideal gas law $$ pV=nRT$$ is not invariant under the transformation $T\mapsto T+T_0$. But that is exactly what the relation between Kelvin and Celsius is - they are not, like most other choices of units, a different scaling, but they choose a different zero point of temperature, so you may not use them interchangably in formulae ...


2

To continue Matas' comment: the definition of temperature is a quantity dependent on mean energy. It's sort of ugly when the particles all go near-relativistic, but there is no upper bound. Interestingly, you can extend the definition to things like magnetic particle orientation states. In this case, beyond a certain magnetic field strength, all the ...


0

Background Generally speaking, in plasma physics one uses the terms cold and hot to refer to the ratio of the thermal, $V_{T,j}$, to bulk flow speeds, $V_{o,j}$. Meaning, a cold(hot) plasma has $V_{T,j} \ll V_{o,j}$($V_{T,j} \gg V_{o,j}$). A quick reference of the possible thermal speeds can be found in this answer. For more information on cold vs. hot, ...


2

An egg has to reach an inner temperature of 100C to cook and in water the egg shell is kept at 100C for five minutes in a heat bath. Thus your question is answered by "can the egg shell be heated to 100C by the much hotter thin gas in the thermosphere" In vacuum the egg will radiate away and go close to 0 kelvin, so in the thermosphere it will be a fight ...


3

The zero-range nuclear interactions felt by neutrons makes free neutrons, outside a nucleus or a neutron star, an excellent implementation of an ideal gas. However a neutron gas is a little unusual since neutron gases mostly are at such low density and pressure that the neutron-neutron interaction is very unlikely to occur before the neutron either decays ...


1

EDIT : The egg I am considering doesn't have a shell Temperature is a measure of the average kinetic energy of particles in a gas. In the thermosphere, the highly energetic solar radiation collide with the (having very less pressure) air, and are thus given very high velocities and so have a high temperature. However, the temperature shouldn't play a big ...


0

Units of temperature are K, degrees C, degrees F. Units of temperature change are K, C degrees, and F degrees.


1

We need to think about what it means for one system to be "hotter" than another, or to have a temperature $T$. Thermodynamics defines the temperature of systems in thermal equilibrium. If an energy level $i$ of energy $U_i$ has degeneracy $g_i$, its occupation level $\propto g_i \exp \left( -\beta U_i \right) $ for some constant $\beta$ at thermal ...


3

The Planck temperature isn't the hottest possible temperature in the same sense that zero Kelvin is a theoretical minimum. It is simply the temperature at which it's black body radiation is of the order of the Planck length. The Planck length is the length scale at which it is theorised that quantum-gravitational effects become significant. Quantum-gravity ...


3

Water has a very narrow range of temperatures over which it expands when cooling rather than contracting (IIRC +2 to 0 Celsius). This occurs due to the way the highly polar molecules "line up" with each other near the freezing point. For a more interesting example, consider a number of rubber compounds which shrink as they warm up. In this case it's due ...


0

I imagine alot of negative expansion coefficients occour at phase transitions. Probably the temperature increase causes the material to overcome an activation energy and a change in lattice structure occours. In these cases one would not expect these processes to be reversible.


1

Aside from the trivial possibility that a warm front passes through during the night, a frequent occurrence in winter (where I live) would be that you have a cold, sunny day, but that cloud cover begins to increase some time after the Sun has set. The cloud cover acts as a blanket, preventing infrared radiation from the ground escaping into space. This has ...


2

The ambient temperature comes out of many different factors, and daytime normally is warmer due to sunlight, but the largest single factor is simply how warm the air is... And that air can blow in from other places. The easiest time to notice this is when a front blows through your area in the middle of the day: the sunlight stays the same, but the air ...


0

They are 2 different concepts. According to the inflationary paradigm, the primordial fluctuations , visible with a contrast scale of $10^5$, are supposed to be the seeds of the galaxies and of the parent structures. Not all is clear but , with the informations which are available, this idea is enough credible to be intensively studied. On another hand, ...


2

The main point to grasp is that the tiny inhomogeneities seen in the CMB are too small (by a factor of 100) to grow into the structures we see today without something like "cold dark matter". The CMB was formed at the epoch when normal (baryonic) matter and the radiation filling the universe decoupled. Only at this point was normal matter free to start ...


4

There is a limited sense in which this correct. When inflation ended there was a temperature rise known as reheating, and we believe it was at this stage that the standard model particles were first created, or at least the majority of them. If you are measuring temperature by the energies of the standard model particles then this would have been the hottest ...


0

Along with all the other reasons posted, it's flat-out absurd to think that requiring double the thermal energy would somehow change the boiling point from 100 to 200 $^\circ C$ . Maybe from X to 2*x Kelvins, but even then the heat of vaporization appears to have been ignored. Edit: as thedude points out, the energy to reach boiling temperature is ...


2

The book is wrong in the sense that it has made a lot of assumptions without actually accounting for all of them. So you are right that the question has incomplete information. The inaccuracies in the question are as follows: In such questions, you must refer to the principle of calorimetry. $$m_1s_1t_1=H=m_2s_2t_2$$ where $H=$heat absorbed by the body, ...


0

I'd say the book is 50% wrong. For this problem, you have to assume that the specific heat is the same (or that the boiling point is the same). But either way, the problem is badly formulated. In order to solve it, you have to use the fact that this is a textbook problem and therefore must be solvable (and that's a really un-clean way of solving it). I ...


2

Well essentially there is no much difference, at least if we are talking of temperature and we mean "average kinetic energy" as we generally do. Why? Because neutrons as "every-day particles" interact via the nucleon-nucleon potential which similarly to the one between molecules, it is repulsive at short distance and atractive at longer distances. Of course ...


4

Neutrons interact with each other only via exchange interaction Neutrons interact via the strong and weak forces. At low energies the interaction is principally via the nuclear force, or residual strong force, which derives ultimately from the strong force interactions between quarks. This can be described as an effective force due to the exchange of ...


0

As heat is essentially a measure of the kinetic energy of the particles in a substance, technically a thermometer with excellent precision would measure a difference. The particles of air colliding with the thermometer (in Situation 2) will have a smaller relative speed to the thermometer (compared with situation 1), thus their collisions will have lower ...


2

If you measure with sufficient accuracy, you will see a difference. Of course, any object you use to measure the air temperature might be exposed to sunlight and therefore experience significant difference in "apparent" temperature; but if we assume that you have a well shielded device that admits air but is insensitive to the effects of radiated heat (from ...


0

I think I've actually got this after some work. According to (I.4) $$F_{AC}(A_1, A_2, ..., C_1, C_2, ...)-F_{BC}(B_1, B_2, ..., C_1, C_2, ...)=0.$$ For this to be equal to (I.5) the above needs to be true no matter what the C coordinates are. Therefore we ought to be able to decompose the above functions as $$F_{AC}(A_1, A_2, ..., C_1, C_2, ...



Top 50 recent answers are included