# Tag Info

1

Assuming the water can't cool, then after about 30 minutes, all the neutrons have decayed to protons and electrons, the resultant antineutrinos escape, but an average of around 0.5 MeV per decay will get thermalised in the water (the range of the electrons will be of order 1cm). As Floris points out, it takes 15.8kJ to heat a gallon of water by 1K, so this ...

2

One really, really energetic one might do it. But it would need to be very close to the speed of light, and I'm not sure you could capture its energy to boil the water. Assuming you want to convert the mass of a non-relativistic neutron to energy, it's about 940 MeV per neutron. Boiling a gallon of water (starting at room temperature, 20°C) requires raising ...

0

Read the "Formula" section of that Wikipedia article again. It gives a second set of coefficients which are good for humidities from 0-80%.

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

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 ...

0

(Thanks to Drebin J.) Let's suppose there's a building with many floors, labelled G, 1-10, and after that continues with A1, A2... A10, B1,B2 etc. A1 is floor 1 but actually the 11th, while A2 is 2nd but actually 12th. (Similar to Celsius in Kelvin) Supposing I jump from A2 to A1 (from the staircase), it would obviously hurt. The damage incurred ...

1

In an exam, Alice scored $50$, Bob scored $40$, Eve scored $10$. Now to raise the class mean, the teacher decided to add $20$ points to every students. So Alice's score becomes $70$, Bob's score becomes $60$, and Eve's score becomes $30$. Now Alice complained to the teacher, "my score was higher than Bob's by $10$ marks, and you see, according to the ...

0

This seems to me like the general "Problem", that differences of Observables behave in a different way, than the observables itself do: One other example (that has nothing to do with additivity or multiplicativity, as you stated) are differences of frequencies: You know that you can convert a frequency to a wavelength by dividing c by the frequency: ...

1

Celsius and Kelvin are two scales that differs only for an additive factor, but the single increment corresponds to the same temperature difference. In other words, an object become "hotter" in the same way if you rise its temperature by 1K or 1°C. You can use conversion formula in differences, just make sure you use it for both terms and keep in mind that ...

0

A temperature change of 1$^\circ$C is the same as a temperature change of 1K. So if you start at 30$^\circ$C (= 303.15K ) and increase the temperature by 5$^\circ$C (5K) the new temperature is 30 + 5 = 35$^\circ$C (303.15 + 5 = 308.15K).

0

temperature difference is 5°C, which is convertible to 278.15K. This is where you go wrong. A difference of 5°C is a difference. Express it as you will in Celcius, Kelvin, Rankine or even in Fahrenheit. But it is a difference. Saying 5°C, which is convertible to 278.15K mean that you are not looking at a difference. Instead you are calculating how to ...

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)

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 ...

0

Yes, the amount of radiation (i.e. power) would decrease $N^4$ times, since the whole geometry of the system does not change, only the power emitted by the sun. But the temperature of the earth would decrease only $N$ times, since the ratio of the temperatures remains equal - again because of the unchanged geometry. The earth radiates a power proportional ...

1

Background The specific intensity or brightness, $I_{\nu}$, is defined as: $$I_{\nu} = \frac{ dE }{ dA \ dt \ d\Omega \ d\nu } \tag{1}$$ where $\nu$ is the frequency, $dE$ the differential energy, $dA$ the differential area, $dt$ the differential time, $d\Omega$ the differential solid angle, and $d\nu$ the differential frequency. We can define a net flux ...

-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 ...

2

The important point here is that there is no thermodynamic limit for gravitating systems, and thus there is no well-defined temperature. This is, perhaps, not a completely intuitive result, but it comes from work on the stability of matter. This is not as glamorous as it sounds, but revolves around the need to show that the energy of matter is an extensive ...

6

A typical velocity dispersion in a globular cluster is 10 km/s. For a typical 1 solar mass subgiant in an old globular, then equating the kinetic energy to $3kT/2$, we get $T = 5\times 10^{60}$ K. Doesn't seem that helpful really... The concept of temperature is only ever applied in a relative sense - i.e. some component is hotter than another. Can't say ...

2

Temperature is not useful concept for describing clusters of stars or other gravitational systems, because such systems are not in the realm described by thermodynamics. There is no way to set up thermodynamic equilibrium - globular clusters partly evaporate and core implodes. Also the velocity distribution can't be Maxwell-Boltzmannian, because very fast ...

1

Let's start with the physical interpretation. We are considering an ideal gas of particles in equilibrium at some temperature $T$. Let's ask the following question: if the system is in equilibrium, why don't all particles have the same speed? Answer: because the particles interact through collisions. Imagine that one could prepare a system in such a way that ...

4

In relativity we only use the rest mass, also known as the invariant mass, of an object. In days past the concept of a relativistic mass was used, but this is now strongly deprecated as it has caused endless confusion. For example an obvious question is whether the increase of relativistic mass with speed can cause an object to become a black hole (tl;dr it ...

0

Yes it will do. You can take the potential energy to be zero when the spring is neither compressed or stretched. In special relativity the total invariant mass of the system would then include a contribution from the potential energy / $c^2$. The concept of mass in general relativity is quite subtle, but for weak gravitational fields, the Newtonian limit for ...

0

I'm not familiar with the math, but am wondering if the 4 should be a 2 instead, (for hemisphere). The Moon's day is about 13.5 of our days, so at its highest the lit side and dark side of the Moon can vary up to 500 degrees F, from about 250F on the lit side to about -250F on the dark side. I would suggest that our atmosphere - which is literally mass ...

2

The Maxwell-Boltzmann Distribution is a probability distribution - eg the distribution of speeds of particles in a gas. The area under the curve represents probability, not energy, so the total area under the curve must be 1, regardless of temperature. The y axis represents a probability density function, eg probability per unit interval of speed, while the ...

1

The way mobility depends on average scattering time of the carriers is given here: A simple model gives the approximate relation between scattering time (average time between scattering events) and mobility. It is assumed that after each scattering event, the carrier's motion is randomized, so it has zero average velocity. After that, it accelerates ...

0

Assuming that the 'whole system' is the total of 400g of water, the temperature of the 'whole system' immediately after mixing is what you call the constant temperature Tfo. It is not 'instantaneously' any other value before the water is thoroughly mixed, and does not cool down to Tfo from some other value. The 'whole system' of water does not necessarily ...

5

Edited because I had misread the question If the goal is to keep the tea hot, you add the milk first. This will bring the temperature of the tea down by some amount $\Delta T$, and the cooler tea will now lose heat more slowly while you dissolve the sugar. I am assuming that since you cannot see the sugar in the milky tea, you will do what I do - you add ...

2

Getting cooler is the key phrase in you question. If you vigorously stir cold water you might get it to warm up if it was well insulated - duplicating Joule's clasic Mechanical equivalent of heat experiment. But if you stir a cup of hot coffee or a pot of hot water the far more significant effect effect it that the stirring accelerates cooling. Stirring ...

1

Moving water convects heat better than static water. Take the reference frame of the water being static and the air moving. In this case, new cooler air is always sweeping in to take heat away. If the air is static, hot air remains at the interface and will not accept heat as well as cooler air. The same argument can be made for the moving water. Edit: The ...

2

It needs energy to solve sugar in water because the enthalpy of solvent (water) and solute (sugar) is lower than the enthalpy of the final solution, solving is an endothermic reaction in this case[1]. Milk has already some sugar in it, the (in)famous lactose, so the enthalpy of tea+milk might be higher than the enthalpy of tea alone and the order "milk ...

4

If the processes are instantaneous, and you drink the tea at once after that, then it doesnt matter. A more interesting question would be, when to put the milk in the tea. Now it does matter if you wait first and then add the milk and drink - or if you add it at once and then wait and drink. Do you see, why? Also, in your formulation "you were asked to ...

1

The order doesn't matter. The reason is conservation of energy. The tea, milk, and sugar before the mixing have some initial energy, and the final tea will have some energy that depends only on its state (the tea doesn't have any kind of memory of how it got to that state). The energy difference between these two states is the additional energy associated ...

1

The metal bar cools by conduction to container A, and loses heat at a rate proportional to the temperature difference so $\frac{dT}{dt}=-k_1(T(t)-T_A(t))$ Meanwhile container A has heat incoming from the metal bar, and is losing heat to container B, so $\frac{dT_A}{dt}= - k_2(T_A(t)-T_B) + k_1(T(t)-T_A(t)) = - k_2(T_A(t)-T_B) -\frac{dT}{dt}$ Then solve ...

2

A possible explanation is simply that the air and water have not equilibrated due to the high heat capacity of water. It's the same reason you can jump in a lake in the early summer and it still be cold. So if you filled up a glass from the sink and measured it 15 minutes later, it's entirely possible that it would be cooler. A more detailed description of ...

0

This could be explained by evaporative cooling. That would match all of the (extremely scant) information you have provided about your experimental setup and results. If you would like a more thorough analysis of possible causes, you are going to need to write at least a couple sentences (preferably even more than that!) about your experimental setup. ...

0

What is important to understand here is that the system is always going to try to attain equilibrium. So if you're temperature surroundings are the same as before , your tank will still attain the same temperature as it did before. However it is necessary to mention the time taken to reach this temperature or at least whether the temperature varies very very ...

0

If you are using a good temperature controller it should hold that temperature at the set point almost regardless of external atmospheric temperature. As the external temperature rises it would use less power to maintain the fixed set point. As external temperature fall it will compensate by adding more power to match the heat loss from the tank.

0

According to what I know, if the temperature of the room is increasing that is, heat is being input into the system, the water will gain heat from the air regardless of the temperature difference between the water and the Room. according to the equation H=M x c(Specific Heat capacity) x change in temp. However due to the difference in the final temperatures ...

0

The entropy defined by Boltzmann can be applied in systems with fluctuating energy. The only requirement is the sharp definition of $\Delta\Gamma$, the number of microstate inside the macrostate. When the time is very long, the distribution of probability in phase space tend to be constant at every allowed point. This justifie the microcanonical ensemble. ...

0

What bothers me is that if (ferro)magnetism stems from the arrangement of electrons in various orbitals (the imbalance in total electron spin), why don't a lot more materials, including basic elements, have a Curie point? What makes the very few ferromagnetic materials so unique? Not all materials display magnetism or paramagnetism. Materials may ...

2

Yes, the increased air density in winter will increase drag and thus fuel consumption, especially on the highway. The density is increased by even more than the 10% you propose, since the air is usually drier in the winter than in the summer, and dry air is denser than wet air (because the molar mass of N2 is much greater than that of H20). Also, the ...

2

I think you are right about the 10% increase air density in the winter and this thorough answer on the Bicycle Stack Exchange supports your observation about the effect when cycling. Drag forces are directly proportional to air density so that would have an effect on fuel consumption. There is another factor this is also significant - in regions where ...

0

You said "at constant pressure." I think you mean without wind, correct? (Because you also mentioned wind.) Just so there won't be confusion let's remove wind from the equation and assume the air is completely motionless at colder temperatures. Air molecules condense or shrink in cooler temperatures - causing higher pressure. So yes, cooler ...

1

I think you misunderstand the definitions. The critical temperature is the temperature above which no amount of pressure will cause a gas to liquefy. The critical pressure is the pressure which will cause a gas to liquefy at its critical temperature. A supercritical fluid is another state of matter. A liquid and a gas phase have been subjected to ...

0

Remember the heat needed to melt ice is higher than that to boil water. At the point when she inserted the ice into the container,the heat absorbed was enough to stop the water from boiling(reduced the latent heat of vaporization of the boiling water)

1

First of all, note that one cannot associate a temperature to a single quantum state (cf "vacuum state of the theory is defined as having zero energy and zero temperature"), and having a zero energy vacuum state is just a convention (as it is cut-off dependent, and thus renormalized). Furthermore, the OP is confused. Standard (i.e. zero-temperature) QFTs ...

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