# Tag Info

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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 ...

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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 ...

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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.

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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 ...

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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 ...

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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 ...

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In simple terms the internal energy can be thought of as the sum of the kinetic energy and the potential energy of the molecules. The kinetic energy of the molecules depends on the temperature - a higher temperature means that the molecules have more kinetic energy. The potential energy of the molecules depends on the bonds (interactions) between them - ...

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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 ...

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This kind of question is more easily, quickly and reliably answered by doing an experiment, rather than a calculation. The calculation requires you to decide what are the most important mechanisms for heat loss, what formulas to use, how to take account of the shape of the container, then measure all dimensions, find the appropriate data (thermal ...

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First get data lined up. It seems the container is not small :) volume of water: $4.80\times 10^{-4}~m^3$ heat transfer surface area (ignore wood supporting face): $0.038m^2$ mass of water: $0.48~kg$ heat capacity of water is 4.2kJ/kg-K calculate how much heat required for the water to reach room temperature. $$\bigtriangleup Q=m c (27-10)= 34.3kJ$$ ...

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We have reached temperature as low as 0.00000001K in the lab and as high as millions of degrees in a nuclear bomb. We achieve such low temperatures by using different techniques. Laser cooling is one of them. In laser cooling, where we fire photons in a certain direction which are accepted and reemitted by atoms in such a way that they retain a component of ...

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No, there is not because, contrary to your last sentence: The amount of water and the container itself I am not concerned with, strictly some sort of equation I could use to get a measurement of time. the boiling time is not a well defined quantity unless you take these things into account carefully. To calculate the rate of temperature rise you need ...

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A few clarifications on this thread in case anyone is reading in the future and is getting confused. I know however, that the temperature did in fact change, hence it's not an adiabatic process. That isn't really how an adiabatic process is defined. The temperature can change within a system (and often does) and it still be adiabatic. An adiabatic ...

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If you can take the environment to be one with constant specific heats $c_p$ and $c_V$, then you can write $$dS_{\textrm{env}}=\frac{\delta Q_{\textrm{env}}}{T_{\textrm{env}}} = \begin{cases} \frac{nc_VdT}{T} & \textrm{if isochoric} \\ \frac{nc_pdT}{T} & \textrm{if isobaric} \end{cases},$$ where the temperatures all refer to the environment. Then, ...

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Your analysis is totally correct (except for the sign of the change), provided cv refer and cp refer to the heat capacities of the environment, and Tf and Ti are the final and initial temperatures of the environment. In such an analysis, the "environment" becomes your system. The signs are, of course, incorrect. They should be +'s.

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The reason the hottest temperatures of the year are later than the solstice is because the land and oceans need time to warm up. Interestingly enough, there's a name for this phenomenon - "the lag of the seasons".

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A reversible process is characterized by a continuous sequence of thermodynamic equilibrium states for whatever system you are considering. So, for your system to experience a reversible process, its pressure and temperature must differ only slightly from that of its surroundings throughout the entire process. And there can be no spatial temperature or ...

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It depends on what you consider to be the system. If the system is the entire container, then there are no thermodynamic operations, quasistatic or not, on the system by the external environment. And as you said, the system is not in thermal equilibrium. If you talk about a thermodynamic operations you need to define a system and an environment, in this ...

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The question may be about the covariance or otherwise of temperature, in which case have a look here. As well, have a look at the paper "Temperature in special relativity" by J. Lindhard, Physica Volume 38, Issue 4, 5 June 1968, Pages 635-640.

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In the kinetic theory of gases, you only really define the temperature for molecules that are in constant, random, and rapid motion. So if you have a container with a gas at temperature $T$ you don't change the internal energy of the gas by uniformly moving the container. Uniformly moving the container gives all the molecules a non-zero average motion, but ...

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I think you can avoid all these troubles if you define the temperature as proportional to the variance of velocity, i.e. $$E[(v-\overline{v})\cdot(v-\overline{v})]=E(v\cdot v)-\overline{v}\cdot\overline{v}$$ Here $E$ means expected value, $v$ ranges over the velocities of the individual particles, and $\overline{v}=E(v)$. Clearly this is frame-...

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All kinds of weird things happen if you try to define temperature in a moving object. The paradox to me (not a generalized accepted answer) resolves by realizing that temperature should only be defined as measured when the object is stationary. Not only is not a scalar but it is not even well defined for areference frame in relative motion. Is temperature ...

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Temperature is related to kinetic energy in the rest frame of the fluid/gas. In non-relatvistic kinetic theory the distribution function is $$f(p) \sim \exp\left(-\frac{(\vec{p}-m\vec{u})^2}{2mT}\right)$$ where $\vec{u}$ is the local fluid velocity. The velocity can be found by demanding that the mean momentum in the local rest frame is zero. Then $\vec{u}... 7 [June 19,2016: thoroughly revised, giving a more detailed, comparative presentation and better references] General case. In relativistic thermodynamics, inverse temperature$\beta^\mu$is a vector field, namely the multipliers of the 4-momentum density in the exponent of the density operator specifying the system in terms of statistical mechanics, using the ... 2 I think it is wrong to define the temperature by the average energy of the molecule in all frames of reference. The reason for that is clear: take all of your particles and send them at$100 m/s$to the north. This won't make the gas hotter, just like the fan does not cool/heat the air (another great mystery!). The organized movement does not participate in ... 0 1 eV is not equivalent to 11,600K, rather it is equivalent to 1.602e-19J. Both eV and J are unit of energy. K is unit of temperature. Sometimes, you can see people using 11,600K to describe eV. This is because people are using Boltzmann constant k = 1.381e-23 J/K, which links the temperature to molecular kinetic energy. The conversion is numerical and does ... 1 Binding energy is usually not associated with temperature. Only a kinetic energy of random motion is related to temperature. You can imagine a bunch of particles at rest. Then go to some moving frame of reference. Now all particles are moving in the same direction and they have some kinetic energy. However, we will not say that the temperature of the system ... 0 The temperature is the same either way. It is crucial to remember that your skin feels heat transfer, not temperature. There are two things to consider: Size of opening and amount of hot air: at the same (wind) speed, the current from a large opening will be larger than from a small opening. This means that when you blow breath from a widely open mouth, ... 1 Probably wrong, but could it be that your breath is pulling a current of ambient air with it, which could explain why when you open your mouth more to exhale it creates a larger warmer current, and can more effectively insulate from the outside current around it. this could be tested with a thermal camera most effectively. but again probably completely wrong.... 1 Water is the people's most popular liquid but it is also one that differs from almost all other liquids in some almost qualitative ways. The most important reason why water is considered an "improper liquid" is that its density is higher than the density of its solid phase, the ice. For "proper liquids", it's the other way around – the solid phase is ... 2 I only find the expression "proper liquid" (in this context) in works from the 19th century. An obvious reason would be the highly non-linear expansion curve (note that the picture below doesn't show that curve itself, but rather its derivative). For mercury, the secant volumetric expansion coefficient varies very little, from$0.18165*10^{-3}$at 0°C to$0....

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Calibration is pretty easy: take an isothermal environment and measure two events with known physical effects: an ice-water bath and boiling point of water. Define your thermometer to read those well defined values at these well defined temperatures and, assuming your thermometer has a roughly linear scaling, you have just calibrated your thermometer. Its ...

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