New answers tagged

2

Ice can be denser than water for certain values of $P,T$. Look at these two pictures taken from here: The darker areas in the second picture denotes areas of greater density. So you can clearly see that when pressure is increased, ice becomes denser than water along the coexistence line. For example at $T=400$ K ice VII is clearly denser than water ...


0

Accordimg to me NO. **Explaination* As water freezes to ice its density becomes almost constant as it becomes a solid. So whatever be the temperature of ice it will always remain 0.98g/ml. Now it is found that even at 100℃ ( this is the largest temperature at which we can have water in liquid form and at this temperature water has minimum density in its ...


1

Equilibrium thermodynamics should in principle only deal with equilibrium states. In an equilibrium state, the thermodynamic variables (for example $P,V,T$) are by definition constants: this is why the time variable $t$ never appears in thermodynamics. However, when we use an equation of state $f(P,V,T)=0$ such as the ideal gas equation of state, $PV=nRT$, ...


1

This question is quite profound, but I didn't think it would go that far. I'll post my views for feedback, and since I'll write some lines it is best to leave it as an answer. As I mentioned briefly in my first comment, time derivatives won't make sense in thermodynamics. I'll try to develop this comment a little bit more. The entities we are talking ...


2

Let's calculate engine efficiency by using the ratio of work and gasoline energy (assuming it can be completely converted to CO2 and H2O). And let's calculate cell (muscular) efficiency by the ratio of work (from gym equipment) and fat energy (assuming it can be completed converted to CO2 and H2O). These two efficiency are comparable. A gasoline engine ...


0

If a fluid system is experiencing a transient irreversible change, the temperature and pressure within the system will typically be non-uniform spatially. This is the result of two factors not present in (slow) reversible changes: the inertia (mass) of the fluid (which allows non-uniformities in velocity and density) and the viscous nature of the fluid (...


4

Nothing in the laws of thermodynamics forbids multiple liquid phases for a single substance. The only limit is the simultaneous coexistence of at most three phases (at triple points). Water has a solid-liquid-gas triple point and several soid-solid-liquid and solid-solid-solid triple points; see the phase diagram of water and ice. In addition, although not ...


4

For quasi-stationary processes you can take the time derivative, no problem. Apart from the "state equation", there is a "process equation", for example, for an adiabatic expansion $PV^{\gamma}=\rm{const}$ or so. It is also quasi-stationary. What you can loose in taking the time derivatives are quick processes like sound or shock wave propagation and their ...


1

There is actually only one disordered phase - from a physicist's perspective, the liquid and the gas are actually the same phase because one can continuously vary the external parameters (temperature and pressure, in this case) to get from the liquid to the gas without passing through any phase transition, because the phase transition line terminates within ...


12

The most immediate answer would seem to be that a great variety of different crystal phases can exist because their long-range order makes it possible to classify them based on the different symmetries of their lattice structure. Since the liquid (or amorphous solid) phase only has short-range order and the gaseous phase doesn't even have that, it seems ...


0

a diesel engine produces 2.68 kg CO2 per liter diesel. The stoichiometric mixture is 14.5 by mass. Air is 0.78 N2 (28 gr/mol) and 0.209 O2 (32 gr/mol) -> 11.2 kg N2 and 3.3 kg O2 per kg diesel. energy needed per kg liquid N2: currently about 0.5 kWh (according to this) -> total 5.6 kWh liquefaction of CO2: 0.09 to 0.14 kWh per kg. -> total energy needed 0....


3

P-V work is not the only kind of work that can be done on the contents of your system. In the case of your fan example, the fan is doing work on the gas within the container by exerting force on it through a displacement (of the fan blade). The kinetic energy imparted to the gas by the fan is then converted to internal energy by viscous dissipation (a ...


3

In statistical mechanics, the most important behavior of a system (and the starting point for any more detailed analysis) is the bulk behavior in the limit where its size is macroscopic (i.e. $N\sim 10^{23}$). Statistical mechanical predictions of macroscopic thermodynamic properties (like total energy and entropy) can be decomposed into distinct sectors ...


1

Thermodynamics was developed largely with gases in mind. In this case work can be done on the gas, the $p\Delta V$ term. But there is also the internal energy U to consider, so when one wants to compare experiments done on the same substance under different conditions it is useful to define a new quantity, which is the enthalpy, as the internal energy plus ...


11

Why is the efficiency of human cells less than efficiency of an Otto engine? It's not. You are comparing two very different things. The low value of 18 to 26% efficiency you found for the human body is the energy produced by an exercising human compared to the energy consumed by that person. The high value of 56 to 61% efficiency is for an ideal Otto cycle ...


44

Can we compare alive cells with heat engines at all? No, not really, because the living being isn't only a heat engine. There are three main points I want to make here. 1. Homeostasis Requires Constant Energy Input This statement is especially true and obvious of homeotherms Mammals (Mammaliaformes, descended from the Therapsid Synapsid Amniotes), and ...


-1

First. Carnot engine is the ideal heat engine. It is already highest efficiency possible for its kind Second. Efficiency need high maintenance cost. You need specific material, specific fuel, specific condition to work and most important it could do only specific task. To make something just work and durable and reuse what it already there is better to ...


0

As the liquid and vapor temperature increases, the liquid density goes down, and the liquid expands. Also, as the temperature goes up, the vapor density goes up, due to the higher vapor pressure of water at the higher temperature. Both liquid and vapor density can be obtained from steam tables, or from a quadratic or cubic curve fit on steam table data. ...


1

In physics one of the most fundamental concepts is the conservation of energy and in thermodynamics we systematize, in an ideal manner how to account for the energy and changes in energy in systems. The units of enthalpy are energy units such as Joules. And for a homogeneous system, the enthalpy is the sum of the system internal energy and the pressure ...


0

If a body (substance) has zero kinetic energy then we should suppose that it is a perfect crystal at absolute zero and all motion is in the vibrational ground state if it is a molecule. If an atomic solid then only zero point motion in the lattice. So yes it has a temperature. The potential energy is that within and between the molecules or atoms. In your ...


1

Ok, let's take the 1st law (in any reference frame) as $$\frac{D}{Dt}E_t(V) = P + Q $$ where $V$ is a part of the body, $P$ is the work rate by mechanical processes and $Q$ is the change of heat content. They will hopefully become more clear when I say how they can be calculated: $$P= \int_V dV \rho \vec{v} \vec{b} + \int_{\partial V} da \vec{v} \mathbf{T}\...


2

The starting point for the development of this equation is an energy balance on a fixed control volume of fluid: $$\int{\frac{\partial E}{\partial t}dV}=-\int{E(\vec{v}\cdot \vec{n})dA}+\int{(\vec{v}\cdot \vec{\sigma} \cdot \vec{n})dA}-\int{\vec{q}\cdot\vec{n}}dA$$ where dV is differential volume, dA is differential surface of the control volume, and $\vec{n}...


0

It is for the same reason that you need to use a dashed line for irreversible process which is higher than reversible process. This is because we know it is higher but don't know how high. If the dashed line is lower, we will conclude this is not feasible. This is the first thing what we want to know from the diagram.


2

For a general closed systems containing only a single chemical species, you have, if you express $U$ as a function of its natural variables $\{S,V,N\}$ $$U=U(S,V,N)=TS-PV+\mu N$$ from which we obtain $$d U = \frac{\partial U}{\partial S} dS + \frac{\partial U}{\partial V} dV + \frac{\partial U}{\partial N}dN=T dS - P dV + \mu dN$$ $U$ is a function of $\{...


2

Otto cycle consists of two isochoric and two isentropic processes. If $T_4$ approaches to $T_1$, then $T_3$ will approach to $T_2$, because for Otto cycle, line $3\to 4$ in $T-s$ diagram must always be vertically. Thus, although the numerator of the Otto efficiency approaches to zero, but the fraction doesn't approach to zero, because the denominator ...


4

Before mixing the average kinetic energy of the molecules which make up the tea is greater than the average kinetic energy of the molecules which make up the milk. This is restatement of the fact that the tea is hotter than the milk as the temperature of a substance depends on the average kinetic energy of the molecules which make up the substance. When you ...


1

You are talking about two different cases. At first you say: We notice that our liquid is no longer as hot as before adding milk. And your meaning about "our liquid" is the tea. Then you say: The total thermal energy of the system (the cup) is conserved. And your meaning about "system" is tea + milk. The total thermal energy of tea and milk is ...


-1

well , the thermal efficiency of otto cycle can be reduced to e (otto) = 1- T1 / T2 when T1= TL e (carnot)= 1- T1 / T3 T3 is greater than T2 , then carnot is still higher efficiency than a reversible otto cycle


1

When you say that the system (ie the contents of the cup) is not as 'hot' as before, you are talking about temperature. However, the system was initially at two different temperatures (hot tea, cold milk). There is no obvious way of comparing the two initial temperatures with one final temperature. To decide if there has been a change in temperature, you ...


1

Clausius' statement about heat not being able to flow spontaneously from a cold body to a warm body is sufficient to prove that no engine can have an efficiency greater than that of a perfectly reversible engine. But it's not enough to prove that the Carnot engine is the only reversible engine. For example, there could be a perfectly reversible engine where ...


1

But for an ideal gas, internal energy is only a function of temperature and so internal energy remains constant here,no change in average kinetic energy of gas particles takes place, so where does the chaos come from to increase entropy of the system. 'Chaos' is not a very well defined term in context of statistical physics. It is not necessary to use it ...


0

The chaos comes from by changing of volume or pressure of the system. The average kinetic energy doesn't change, but number of collisions increases (if pressure increase) or length of paths increases (if volume increases).


3

The heat death of the universe is the idea you are describing (this idea is also known as the Big Freeze). The problem with this idea is for it to work the cosmological constant has to be zero...and it isn't zero. It's very tiny, but it isn't zero. The other problem with your idea is the belief that because it all "freezes", so to speak, it'll all collapse ...


3

The Maxwell-Boltzmann distribution and the Boltzmann distributions are probability distributions, i.e. functions $\rho(\vec x,\vec v)$ of velocity and position of a particle, that say what is the probability density that the velocity and position belong to the small cube around the given value of them. The Boltzmann distribution is the more general one, $\...


0

Greiner's Thermodynamics and Statistical Mechanics is pretty good from a few short readings I did. Also, it has better reviews from almost all of the other popular textbooks on the subject in goodreads.com


4

The question doesn't say exactly what is meant by darkest, but it seems reasonable to interpret the brightness as the intensity of radiation from the object. In that case the relevant equation is the Stefan-Boltzmann law: $$ J = \varepsilon\sigma T^4 $$ where $\sigma$ is the Stefan-Boltzmann constant and $\varepsilon$ is the emissivity. The emissivity of a ...


1

You are correct that quantum mechanics is the basic framework of nature, but not everything in this basic level is quantized, in the sense of coming in a discrete spectrum. Even the spectrum of the hydrogen atom at very high n has such close spacing where it can be called a continuum. The first quantum level is bound atoms. The second level is bound ...


0

The continuum radiation from a material is the conversion of thermal energy into electromagnetic energy. As per Wikipedia All matter with a temperature by definition is composed of particles which have kinetic energy, and which interact with each other. These atoms and molecules are composed of charged particles, i.e., protons and electrons, and kinetic ...


0

After some searches I found this little paper. I will give only a short answer to the question, for details you can read the paper. So, according to the results obtained for the model in the paper: the flame length: increases as the gravity level increases from $0g_e$ to $3g_e$ decreases from $3g_e$ to $60g_e$ and blows off at higher gravity levels ...


2

I think there may be some confusion as to terminology: compressibility is defined as $$ -\frac 1V \frac{\partial V}{\partial p}$$ where something like temperature or entropy is held constant, whereas the compressibility factor is defined as a certain ratio. (ref: https://en.wikipedia.org/wiki/Compressibility) In the ideal gas, particles do not interact ...


1

The compressibility factor was originally derived from empirical testing of gases to correct for the observed non-ideal behavior at more extreme pressures and temperatures. Although it cannot be derived from first principles in the kinetic theory of gases, the experimentally derived factors can be reconciled using Van Der Waal's equation that deals with the ...


0

first of all, human body temperature is maintained by the body. If it is lower, body will burn some fat to keep it up. If it is high, the body will do something like sweating. Wearing a wet cloth in wind can remove the heat from the body. The amount of heat may be calculated. 1. convection heat transfer by wind between water (a thin layer) to air 2. mass ...


3

Yes. An equivalent way of saying this is that if an ice cube (or iceberg...) melts, the water level remains unchanged. (I.e.: the melted iceberg exactly fits in the 'hole' it creates underwater.) To see this, think of what is holding the ice up: it's buoyancy, which is the upward force due to the pressure of the surrounding water. This force is directly ...


1

Imagine a portion of the(liquid) water. It will be in equilibrium with the whole fluid. The buoyancy force $E_1$ over this portion matches its weight. Now freezes this same portion. The buoyancy force $E_2$ on the ice equals its weight. Since the weights are the same, the buoyancy forces are equal. This implies the volume of fluid dislocated in both ...


1

If you isolate the balloon from everything else then after filling it with gas the balloon will soon attain the temperature of the gas. Then the balloon and gas will be in thermal equilibrium and no more heating/cooling will take place. Upon colliding with balloon skin the atom is not always lose energy but half the time it will gain energy (if the two ...


1

Your balloon is not an isolated system, it can exchange heat with the environment as the "skin" material is not a perfect insulator. The expanding gas inside a balloon will cool as it expands, but the outside air will warm it up again after some time, the speed will depend on the balloon material and volume, but eventually the heat exchange between ...


0

Your premise "but higher the velocity, greater is the temperature and pressure must be high" is wrong. To see why, you must recognize that the velocity field $\mathbf v $ that you calculate in fluid dynamics refers to a "fluid particle", which is composed of maybe some $10^{10\pm 5}$ molecules. On the other hand, the (local) temperature, pressure, density ...


2

Water evaporates if it has a higher temperature and or if the humidity of the air isn't at 100%. This is a nonlinear process as the vapour pressure of water is nonlinear with temperature and the rate of evaporation is linear with partial pressure difference (the partial pressure in the water is equal to the vapour pressure). The evaporation reduces the ...


0

I have went through a Thermodynamics Course, one of the best book which had used to clear my problem is: Thermal Physics by " Michael Sprackling" it describe the entropy in terms of both "Microstate" and "Macrostate". I have lot of doubts and which was very helpful for me


0

Assuming that your preliminary calculations are correct, you now have everything that you need: n, R and T. And unless you have been given a modulus of elasticity for the tyre, you must assume that it is rigid, so you have V as well (as a commentator hinted above). Just plug them into the ideal gas equation. If you know the modulus of elasticity of the tyre,...



Top 50 recent answers are included