# Tag Info

0

When the balloon is heated, it increases its volume so to keep the pressure constant. The balloon does work to its surrounding. The heat increases its internal energy as well as is used to do work. You might think that the molecular velocity increases, but at the same time, the volume increases as well, so it will take longer time for molecular to travel to ...

0

static pressure at B can be larger than at A and can be lower. This is not used for estimating stagnation pressure change. If there is no frictional loss, the fact, that the two total pressures are the same, means the energy is conserved. When this is loss, of course energy at A is larger than that at B in order to conserve the energy.

1

The way it usually works is by showing that the relative fluctuations of the variable that becomes a random variable when changing ensemble ($N$ in this case) vanish for large system sizes. This then shows that to a single value of the chemical potential, one can assign in very good approximation a single value to the random variable $N$ that corresponds to ...

0

No. The frequency of EM radiation such as visible light is many orders of magnitude greater than the frequency of mechanical vibrations. You would not be able to detect any difference in the spectrum even if there was a difference.

4

What you are hearing is mains hum: mains electricity is alternating current (ie the voltage is approximately sinusoidal and symmetric about zero), with a frequency of 50Hz or 60Hz. things like kettles and heaters use a lot of power and parts of them will mechanically change shape at this frequency, which is audible. This kind of physical noise from things ...

0

Maybe I haven't read correctly the pdf you put as a link but it does not write exactly what you write. That being said, let me try to explain the Maxwell's construction without appealing to the Gibbs free energy function. To this, I just consider a van der Waals fluid with the various isotherm PV curves that you have in the first pages of the pdf you put ...

-1

Maxwell construction is about a system transitioning from one phase to another phase but not about two phases in equilibrium. When changing from one phase to another phase, there is a path that pressure and volume can change. However, because the Gibbs energy of the change is higher than the simple phase change, the pressure and volume make abrupt change ...

1

One purpose of the glass beads is to disrupt the flow of vapour by convection to improve mixing and attainment of thermal equilibrium at each level of the fractionating column. This creates a steeper, smoother temperature gradient. Without the beads (or similar 'baffles') convection currents would bring vapour at different temperatures and compositions to ...

0

What is exactly the canonical ensemble? Thermodynamic ensembles are ensembles in the mathematical sense, so your option no. 2 is the correct one. Consider a system of non-identical particles, this will appear much more clearly. What do "thermal average" and "thermal fluctuation" mean? "Average" is not something per se, one should speak about the thermal ...

0

I think, the comment of lucas is the best answer; I'll elaborate a bit If I mix these two containers eventually both the Oxygen and Nitrogen will be at the same temperature. Why is that? By temperature you mean the mean energy per particle. We observe experimentally, that if you mix two heaps of particles with different mean energies, after some time ...

0

If you mix two gases, their atoms/molecules exchange with energy, momentum, etc. It is not surprising that a hot gaz heats a cold one. It (heating) can only stop when the mutual energy exchange becomes equal to both gases. So the energy distributions of "subsystems" are equal in the end, but the momentum distributions are still not. P.S. The subsystem ...

1

Yes, it applies, and it's not really related to the Stefan-Boltzmann law. The energy radiated from a blackbody at temperature $T$ does indeed scale like $T^4$. Any object (blackbody or not) can absorb radiated energy, and that is the part which increases the temperature. The inverse square law is a statement about the density of radiation (or intensity, ...

1

In a system of many particles, we essentially observe the most probable configuration, and relative fluctuations around it are negligible. Here I will prove that the most probable state of a 2-particle system is this with equal energies. The probability of a state is proportional to the volume of the corresponding part of phase space. If a particle has ...

1

There is something they forgot to mention in your notes (either from ignorance, or out of omission). The temperature within the system is spatially non-uniform during an irreversible process. So what value of the temperature are you supposed to use in the integral of dq/T? The Clausius inequality calls for the use of the temperature at the boundary ...

0

Saying that a body possesses heat energy is as absurd as saying it possesses work! that is work and heat are merely names given for energy transfers. when you heat an object and say heat energy is transferred, we mean that the molecules in the heated object are becoming more jiggly(vibrations) that is their kinetic energy is increasing. that is what is meant ...

0

I think you have a typo in your question. A reversible process will have a smaller entropy change than an irreversible process. Your interpretation that the equality refers to a reversible process, while the inequality refers to in irreversible process is correct. Looking at the specific equations in that notes document, the integral in 8.31 applies to an ...

0

This is Clausius inequality. The term ds>dQ/T. Holds an impossible reaction which does not obey Clausius statement in 2nd law of thermodynamics. The things in your class thought is about this ds<0. Where you can add c that is entropy generation term. Hope it will help you

3

I will be blunt. As fas I know, nobody knows a priori for which systems equilibrium statistical mechanics will work or not. Part of the current effort to determine which systems are fine being described by equilibrium statistical mechanics focuses on various proofs of ergodicity for such systems. For now, they are somewhat limited to either a restrictive ...

0

There are several ways to characterize a system of particles. You can see them as whether they are isolated or in thermal equilibrium or in mechanical or chemical equilibrium. Similarly, you can ask whether a system is made up of identical bosons, identical fermions or distinguishable particle. In each of this cases, different statistics prevails. Simply ...

0

The work can completely go to internal energy. For example, if you stir fluid in a fixed volume. The work can also completely go to PV. For example, if you very slowly blow a balloon. So with limited information, one cannot determine how much goes where. For ideal gas, it may be simpler, but you still need fluid dynamic computation to get change in P, V and ...

-1

The boundary lines are, 1. isolated system 2. equilibrium 3. no composition change such as chemical reaction

0

Here's another, simpler answer. The BB analyses (like the one above) are typically based on some very modest definition of life, like an unpressurized brain floating in vacuum. If you include all the other equipment necessary for a brain to actually function then the odds are much different. Furthermore, the universe actually doesn't contain much ...

1

For all three bullets, you need to use Neumann boundary conditions. open door: $Q = k(T_{out door}-T_{room})$ insulated wall $Q = 0$ heater boundary (i.e. hole) $Q = P_{heater}$ where $P_{heater}$ is the power of heater. Summation of Q can then be used to calculate the room temperature change,i.e. $$C_p m\frac {dT}{dt} = \sum Q$$

0

Because hydrostatic pressure depends on depth, the problem requires integration. The pressure as a function of $y$ according to Pascal's Law is: $$p(y)=p_0+\rho g y\sin \theta$$ Where $p_0$ is the atmospheric pressure and $\theta=45\:\mathrm{degrees}$. $y\sin \theta$ is the depth. On an infinitesimal piece of door of length $dy$, at position $y$ and ...

1

The question has asked for Heat capacity(which is independent of mass) and NOT specific heat capacity(which depends on mass) thus the answer is $\ C = \frac {Pt}{ \Delta T}$ Remember two things: Heat capacity or thermal capacity is a measurable physical quantity equal to the ratio of the heat added to (or removed from) an object to the resulting ...

0

U is a property that can be calculated for 'ideal gases' only. For every other fluid, change in U can be calculated. But the physical properties, P and T can always be measured using pressure guages and thermometers etc. Enthalpy change could be calculated using the heat supplied and the heat rejected. Hence the change in U could also be calculated using ...

0

if we assume $E_1$ (zero state) and $E_2$ (state 1), from this question, the system has n units of energy and N particles. The multiplicity is then $$\Omega (N,n) = \frac {(n+N-1)!}{n!(N-1)!}$$ and entropy can be calculated, $$S=k \ln \Omega(N,n)$$

0

Work done by the coil can be calculated by $$\ W = Pt$$ This amount of work is contributed to heat the body (in this case the efficiency is 100%). $$\ W = Q_{heat}$$ Also, ($\ c$ is the thermal capacity) $$\ Q = cm\Delta T$$ Therefore, combine the equations above, we get: $$\ c = \frac {Pt}{m \Delta T}$$

0

$$F\delta r=\delta U = U(a_0+\delta r)-U(a_0)$$ Use Taylor expansion and remember that $U^\prime (a_0)=0$ at the stable position (minimum potential). $$F\delta r=\frac {U^{''}(a_0)}2{{\delta r}^2} + o({\delta r}^2)$$ Thus, $$\delta r = \frac {2F}{U^{''}(a_0)}$$ There is a factor of 2 missing in your question.

0

Let me answer your second question first. In a steam power plant, you have a pump, turbine, boiler and condenser. The pump consumes power while the turbine generates power. Now it doesn't make sense to have a separate power source for the compressor when you have a power-producing turbine a few metres away from it. Hence, the part of the power produced by ...

2

The earth's atmosphere can be considered as a thin sheet of air extending from the earth's surface to about an altitude of 60 miles. It is the earth's gravity that holds the atmosphere. The interconnection between temperature, pressure and density with altitude is as follows. About temperature variation with altitude:- The sun heats our earth's surface. ...

1

I would like to add to what Procyon said in his(her) answer, which is right on target. The first equation in the OP should be an equality, not an inequality. For an irreversible process, the temperature of the system is typically non-uniform, and when we write $\int{\frac{dQ}{T}}$ for the Clausius inequality, what we really mean is ...

1

Entropy is a property of the system. The change in entropy of a system as it traverses from an initial state(1) to a final state(2) is independent of the path by which the system is taken from state 1 to state 2. The path can be a reversible one, or even irreversible, the change in entropy is always the same as long as the initial and final states are the ...

0

Further to the correct answer of @Zeeshan Ali: A difficulty with this type of problem is that the equation you need to solve to find the final state depends on the final state! Is it slightly cooled liquid water, a smaller block of ice floating in water at $0$ C, or a slightly warmer block of ice with all the liquid water frozen to it. If the block of ice ...

1

Your instructor is correct. This is because you have not included latent heat of fusion in your equations which is the heat required to melt 50g ice at 0ºC to 50g water at 0ºC. Your true equation is: (400)(4.2)(40-T)=(50)(2.1)(10) +(50)(334) + (50)(4.2)(T)

4

Are all heaters (same wattage, electric to thermal, no geothermal or other extra energy source) exactly as efficient as each other? No. Let's focus just on electrically powered heaters. If you have a heater that basically consists of a resistor with a current passing through it, you have 100% efficiency of electrical energy to heat energy conversion. ...

0

It is just understanding the system: closed or open. A closed system doesn't interact with the environment, hence there is no change in pressure and volume. In other words, no external work is done on the system neither by the system. In the second case, there is pressure change due to interaction with environmental, hence work is done by the system or on ...

0

I found the answer. the d(PV) is the work done to push new mass in the system

2

$kT$ is related to the kinetic translation energy by the equipartition theorem. You are saying that the mean kinetic energy, is much greater than the rest energy. The particle has a large or relativistic velocity. The limit $kT>> mc^2$ is called ultrarelativistic limit. It means you can approximate the energy momentum relation $E^2=(pc)^2+(mc^2)^2$ by ...

4

Three processes are involved: Conduction: Heat flows from the object to its environment. Removal rate of heat from the interface further away from the object is proportional to the coefficient of conductivity (0.024 for air, 205 for aluminum -see http://www.engineeringtoolbox.com/thermal-conductivity-d_429.html). Convection: The interface between the object ...

1

This is in general a very complicated problem. There is three processes by which your object will cool down : conduction and convection which you already mentionned, as well as emission of blackbody radiation. You can treat blackbody radiation with the help of Stefan's law that gives you the power emission with respect to temperature. Conduction and ...

0

The (partial) answer is the Electrocaloric Effect

0

Sign convention is important, particularly, for the work W. We can define it as positive work if the work done by the surrounding is positive. Or we can define it as positive work if the work done by the system is positive. Check your text book about this sign convention. With the first definition, the first law can be written as $$Q=E_{int}-W$$ thus, if ...

1

For example, rotational kinetic energy of gas molecules stores heat energy in a way that increases heat capacity, since this energy does not contribute to temperature. This description is misguiding in two ways. First, the statement that rotational energy does not contribute to temperature makes an impression that temperature is a quantity that is ...

-1

This question is surprisingly difficult to answer and requires clear thinking. I am not sure that I have got it entirely right myself. I don't think the usual explanations - which state that the energy in rotational and vibrational motions are also proportional to temperature - solve the problem of why only the translational motion determines temperature. ...

-1

The rotation does contribute to the temperature. The heat capacity (at constant volume) is the derivative of the internal energy of the gas with respect to the temperature. If the internal energy increases (by adding more degrees of freedom), the heat capacity increases accordingly. The heat you provide to the system is now equally partitioned over all ...

1

If it was a clear sky then your motorbike most probably lost heat by radiative cooling and the ground usually cools more rapidly than the air. If you have placed your motorbike under a tree you would have probably found that there was no frost on your motorbike. You would have noticed the opposite of radiative cooling, radiative heating, when the Sun was ...

3

The (long-term) temperature of an object depends on the heat transfer between it and all of the environment. Air isn't a great conductor of heat. So if there is little air movement, the radiation environment may dominate the heat transfer. A cold calm day may feel quite balmy under full sunlight. On a cold evening, the sky may have a radiation ...

0

It ultimately depends on the context. The energy is the energy; the way the different contributions to the energy are named depends on how these contributions affect the dynamics of the system. In a thermodynamical context, the denomination "internal energy" usually refers to the energy due to chemical bonds and due to molecular motion (i.e. the vibration ...

3

Energy stored as heat, by itself, is neither low- nor high-quality. What matters is the temperature at which the heat is stored, and the relationship of that temperature compared to the heat sink that will absorb the excess energy in the process. To be more specific, say you have a heat sink at $T_S=20°\:\mathrm C$, such as the atmosphere for a car engine. ...

Top 50 recent answers are included