Hot answers tagged

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 ...


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 ...


4

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 ...


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 ...


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 ...


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 ...


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 ...


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 ...


2

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 ...


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 ...


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 ...


2

Inside the bathroom after a shower, the humidity rises and thus, the feeling of more warmth as it pertains to higher temperatures(warm air holds more moisture than cold air). In other words, it may be a similar temperature just outside the bathroom but the rise in humidity inside will make it feel much warmer for a short time. The size of the room and other ...


2

Since the melting point and especially the boiling point of a substance depend on the pressure you had to specify a pressure. Also not all substances do have a melting or boiling point. Some, e.g. carbon at standard pressure sublimate. A good canditate for the substance with the highest difference at standard pressure is neptunium with a value above 3200 K. ...


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 ...


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

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 ...


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 ...


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 ...


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 ...


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 ...


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 ...


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

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

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

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 ...


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 ...


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

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 ...



Only top voted, non community-wiki answers of a minimum length are eligible