New answers tagged

0

Heat energy and thermal energy are pretty much the same thing. My science teacher taught me, that because potential energy is related to height, it would be the kinetic energy that is converted into th


0

Do the answers need to be long? It can get away from equilibrium, it just has a very low probability of doing so.


1

Background Generally, all phase transitions require some input energy in order for the transition to occur. For instance, the transition from solid-to-liquid or vice versa requires what is called the enthalpy of fusion or latent heat of fusion. This is the amount of energy needed to change the total interal energy (i.e., enthalpy) of a substance in order ...


1

You need to collect some data to answer this question. Heat pumps can provide more heat to a building than the electricity they use. Whether yours does is something that needs to be measured. Real world efficiencies are often different from lab efficiencies. The radiator coils can get dusty and be installed in places with poor air flow, for example. You ...


4

You have to distinguish between "different states" and "number of states" - or, in the words of @Numrok, between "macrostates" and "microstates". The fundamental theorem refers to "accessible micro states". If I have three white balls and two buckets to put them in, I could put two balls in one and one in the other (that is a macro state); there are in fact ...


-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

The ratio of emissive power to the absorbitity is constant when the substance is at thermal equilibrium with surrounding. Or The emissive power of a substance is equal to its absorbtivity under the same conditions.


0

Exactly the experiment you describe is shown in this video around 8 minutes in. This is called evaporative cooling and indeed has been used for cooling things for all of recorded history. indeed, assuming you possess sweat glands your own body uses evaporative cooling. Actually your refrigerator uses evaporative cooling as well, but it keeps the evaporating ...


1

I'll show you how the book derives that solution. First, we start with the definition for energy fluctuation: $$ \langle(\Delta E)^2\rangle =\frac{\partial^2}{\partial \beta^2}\ln(Z) $$ where $Z$ is the Canonical Partition Function. The heat capacity, $C$, can be defined as: $$ C=\frac{1}{k_BT^2}\langle (\Delta E)^2\rangle $$ Next, let's calculate $Z$ ...


0

How much does g vary between ground level and 10 km altitude. The heat capacity of dry air is also pretty constant over the temperature range of the troposphere. At ground level, the globally averaged temperature is about 25 C, and at about 10 km, the temperature of the air is on the order of about -60 C. Look up the heat capacity of air at these ...


0

I'll start by saying that correctly stating Kirchhoff's law is quite tricky. "Emissivity equals absorptivity" in a certain sense, but they may depend on wavelength, and angle of incidence (or emission), and polarization. In magneto-optic materials, you can have high absorptivity from one direction balancing high emissivity into a different direction!! (This ...


1

If you carry out an irreversible adiabatic process starting out at state A and ending at state B, you will not be able to identify a reversible path between the same two states that does not involve an exchange of heat with the surroundings. In other words, there is no adiabatic reversible path between states A and B. Here is the simplest example I can ...


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 ...


0

Roughly, in the integrands you have something like $$J^{(3)}+\alpha^2J^{(1)}\sim (x^2-\alpha^2)^{3/2}+\alpha^2(x^2-\alpha^2)^{1/2}=(x^2-\alpha^2)^{1/2}(x^2-\alpha^2+\alpha^2)=(x^2-\alpha^2)^{1/2}x^2$$


0

The liquid will not boil and when temperature attains critical temperature, two phases becomes indistinguishable.


1

You say: We shouldn't care about how we reached that equilibrium In fact the entire point of the example is that we do. I shall try to explain why. Jaynes and Gull both work in the framework of Bayesian inference (I can recommend the introductory text: http://www.amazon.co.uk/Data-Analysis-A-Bayesian-Tutorial/dp/0198568320. The title may seem ...


0

It depends on what one defines as an ideal gas. In the physics literature, an ideal gas is defined as one which has constant heat capacity from absolute zero all the way up to the temperature T. In the engineering literature, we consider an ideal gas to have temperature-dependent heat capacity, just as real gases do at low pressures. In both cases, the ...


0

I found an answer in the book Monte Carlo simulations in statistical physics - an introduction (by K.Binder, D.W.Hermann), page 35. To determine equilibration we need to run the simulation a few times, let's say $n_{run}$ times. we define the average $<>_T$ as an average after $t$ steps of the simulation: $$<A(t)>_T = \frac{1}{n_{run}} ...


1

The answer to your query is Equipartition of Energy. Equipartition Theorem: At temperature $T\,_,$ the average energy of any quadratic degree of freedom is $\frac{1}{2} kT\;.$ For each degree of freedom, the ideal gas molecule can store $\frac12 kT$ of energy on average. For monatomic ideal gas molecule, there are only three degrees of freedom: ...


1

It depends on gas whether its monoatomic diatomic nonrigid ,rigid as $\alpha$ is degree off freedom which depends on gas for mono it s $3/2$


0

As an experimentalist, I know that black body is dependent on materials, which is taken into account by the emissivity . I always took the cavity model of the black body formula as an idealized tool to fit observed experimental distributions, the only contact with reality being the oscillators envisaged. The formula fits , and the experimentally best fit ...


1

This equation describes 3D transient heat conduction in a material, where $\alpha$ is the thermal diffusivity. The thermal diffusivity is related to the thermal conductivity, the heat capacity, and the density by $\alpha=\frac{k}{\rho C}$. The equation is derived by performing a differential heat balance on an infinitesimal volume of the material. It can ...


0

According to the Stefan-Boltzmann law, radiated power is proportional to the fourth power of the absolute temperature of the body. With T=0 the radiated power would be zero.


0

The vapor pressure of your water column at 50 degrees C is approximately 100 Torr, which is 1/7.6 of one atmosphere. The pressure in the water column is given by the equation $P = \rho g h $ , where $\rho$ is the density of the water, g is the local acceleration due to gravity, and h is the height of the water column above a stated point. The water will ...


0

consider ten kg substance .Take few kg substance and measuring mass density,the density is same as before substance. so we can say that from above explanation,density is an intensive property.


2

If z is a function of x and y, then:$$dz=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy$$So, $M=\frac{\partial z}{\partial x}$ and $N=\frac{\partial z}{\partial y}$. Since the order of taking partial derivatives is immaterial,$$\frac{\partial^2 z}{\partial x \partial y}=\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$$


0

Calculate the derivative of $dz$. $$\begin{align} d(dz) &= \partial_yM\;dy\wedge dx + \partial_xN\;dx\wedge dy \\ &= \left(\partial_xN - \partial_yM\right)\;dx\wedge dy \end{align}$$ It is zero iff. the condition $(\cdots)=0$ holds, meaning $dz$ is a closed 1-form. On a simply connected domain - that is one where every loop is contractible to a ...


0

Does the specific heat of a metal should be then related to the electrons and not the atom of the metal? Specific heat (just like conductivity) depends on the molecular structure of the solid, and that depends on the atomic number, i.e. the shell of electrons. Whether the electrons are the ones conducting the heat or exchanges of vibrational ...


0

As the metal heats up at one location, eg, a laser pulse, the motions of the atoms in the crystal lattice increase. If these motions are coherent you have sound, but if they are random it is heat. Most motion starts as coherent, resulting in quantized sound (phonons) but they soon exchange momentum with the free electrons, and become randomized in a few ...


0

I don't quite understand what answer you exactly want, since there are too many perspectives to deal with the problem. In classical physics, you might use the degree of freedom to deal with this question, and you will find out the result fail to explain the experiment. And this problem comes from the "incorrect" Maxwell-Boltzman distribution when dealing ...


0

The derivation is not correct. The mass within the control volume is $\rho A\Delta x$. The rate of energy accumulation within the control volume is $\rho A\Delta xC\frac{\partial T}{\partial t}$. So the heat balance should be:$$\rho A\Delta xC\frac{\partial T}{\partial t}=Q_x-Q_{x+\Delta x}$$Dividing by $\Delta x$ and taking the limit as $\Delta x$ ...


0

First of all: you and the people in your course are most certainly not the only students encountering this problem. In my 2nd and 3rd university year I had it myself. The reason is that physicists use abusive notation. And they do it a lot. Mathematicians have less of a problem with these things. In this answer I will try to use intuitive terms and be ...


1

DanielSank's answer is 100% rigth about the temperature issue. The question deserves much more argumentation because the book you use, and many others, are plain wrong. I will use numbers and the laws of physics to show my point: it is not true that If the mass of an electron, the Planck constant, the speed of light, or the mass of a proton were even ...


2

In the method 1, you expressed increase of energy during adiabatic process as $$ dU=C_A dT $$ where $C_A$ is presumably capacity when the area $A$ is constant: $$ C_A = \left(\frac{dU}{dT}\right)_A. $$ But $C_A$ is not the right factor to use for adiabatic process, because in this process $A$ is not constant, but $S$ is. The energy increase formula is ...


2

Your error in method 1 was assuming the U = U(T) and not U = U(T,A). But, you would have been much better off starting out directly with entropy S = S(T,A), so that $$dS=\frac{C_A}{T}dT+\left(\frac{\partial S}{\partial A}\right)_TdA$$Then from the Maxwell relationship, $$\left(\frac{\partial S}{\partial A}\right)_T=-2\left(\frac{\partial \sigma}{\partial ...


1

Not sure about your question. Here are some suggestions: In thermodynamics we define internal energy as $U = U(S, V)$ (number of particles is constant). Then we write it's differential:$$dU = \frac{\partial U}{\partial S}dS + \frac{\partial U}{\partial V} dV$$ Now we recall 1st law of thermodynamics $dU = \delta Q + dW $ together with 2nd (or it's ...


0

After a long discussion with "curiousone" I would like to like to share the relevant points of our discussion (hopefully I will do them justice) and some extra bits I added after thinking it over First Law of thermodynamics While the equation $$TdS = k_b T \ln N dN + dU -PdV $$ is quite general to any system where particle number is not conserved. We ...


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} ...


-3

These answers are way too complicated. It's not rocket science. When a charged particle moves it creates electromagnetic radiation. Matter is made of charged particles. Since everything not at absolute zero is moving around and everything is made of charged particles you're gonna get a spectrum of EM radiation coming off matter. If the thing is ...


1

Phase changes depend not only on temperature but also pressure. Like any substance, if you get the right combo of pressure and temperature you can keep it at a solid, liquid or gas or if you're really good keep it at the triple point and get all three phases in equilibrium with each other. For C02, the liquid phase is very likely outside of most temperature ...


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 = ...


3

The simple answer to this question is that the specific heat capacity of a hot solid body you touch really doesn't matter too much for whether you burn yourself; what is much more important is its heat conductivity. The reason for this is that the surface of a hot body with low heat conductivity will rapidly cool down when you touch it (the blood in ...


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

The maximum potential energy is limited by the energy content released by the one liter of boiling water. you are able to boil away all of the water - this gives you the total energy released. The obvious answer is to follow Watt and build a steam engine - the height of the piston is the measure of the potential energy extracted, U=mgh. You can use a ...


9

TL; DR The material with the greater effusivity will be more likely to burn you upon contact. Analysis of a simplified case First consider the case where your palm comes into contact constant temperature wall. Often, we can consider your palm as a semi-infinite solid. The requirement for the semi-infinite approximation is that $T(x \rightarrow \infty, t) = ...


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.



Top 50 recent answers are included