Tag Info

5

When two quantities of water ($m_1$ and $m_2$) at different temperatures (resp. $T_1$ and $T_2$) are mixed in adiabatic conditions (no heat loss and no external heating during mixing) the temperature $T$ of the resulting mixture can be calculated from the heat balance (no heat is lost or added so the heat contained in both masses is found again in the ...

3

At the classical framework , i.e. no General relativity and astrophysical observations of the 18th century , this is a valid question. When talking of a "Universe" one must have a model , and the model depends on the state of physics knowledge at the time of the model. The second law states that entropy always increases or stays the same. One can make a ...

2

The minimal counterexample seems to me to be the following: Take two materials, placed next to each other: ____________________ | | | | Material|Material | | 1 | 2 | ____________________ E1 _ _ _ E0 _ _ They have energy levels as indicated above- both have states at E0 and E1, but one has two excited states. ...

2

As rockets have finite fuel capacity and important parameter is how much rocket velocity can be achieved in a given situation per unit of fuel mass consumed. The higher the exhaust velocity the more effectively the fuel mass is utilised. Issues such as energy required are also important but generally exhaust velocity or "specific impulse" is amongst the most ...

2

Basically a "molecule" of water cannot heat up ice. I think what you are trying to say is, how does heat transfer take place on a molecular level? If that's the case, then its something like this. In the interface between water and ice, water molecules are moving, while ice molecules are static. on contact, some molecules of ice acquire velocity (due to no ...

1

Here's a heuristic answer that may help. Imagine that water is in contact with an extremely cold ice cube (so the molecules in the ice are barely moving). When a liquid molecule collides with an ice interface, it excites an ice molecule in the crystal structure, causing a small wave to propagate in the material, kind of like how a pulse propagates in ...

1

Actually the human body emits more than thermal radiation. The Czech military did a study on measuring the extreme low frequency radio band emitted by the nervous system. It can be found on www.measurement.sk by searching Human electromagnetic emission in the ELF band - Measurement ... www.measurement.sk › Lipkova This makes perfect sense when you consider ...

1

I'm fairly certain that the energy difference between the two is negligible. Any heat as a result of light absorbed is localized on the roof, with layers of insulation between the inside of your house and the additional heat generated. Most of the additional energy absorbed would heat up the air rather than the inside of your house. Additionally, the optical ...

1

The color of an object says something about its emissivity/absorbance for visible radiation. But it has virtually no correlation to the behavior at IR wavelengths. A white object may reflect significantly more visible light than a black one, but may reflect identical amounts in IR. Since your roof will not be cooling via visible radiation, the color ...

1

Nuclear rocket motors work by heating a gas and allowing it to expand out of the exhaust. To get the most thrust from your gas you want the momentum of the gas molecules to be as high as possible, because the force is equal to the rate of change of momentum of the gas molecules. Suppose the nuclear reactor heats the gas to a temperature $T$, then the ...

1

Ultimately, Newton's law of cooling is a simplification that can be obtained from the full heat equation, i.e. $$\rho c\frac{\partial T}{\partial t} = - \kappa \nabla \cdot T.$$ The heat equation itself can be derived from first principles, assuming Fourier's law for heat flow, namely that it is proportional microscopically to the difference in temperature ...

1

Is the Sun absorbing energy from it's surroundings? No, of course not in a net sense. The Sun loses far more energy than it absorbs from its surroundings. It is not in thermal equilibrium. The Sun is also not a blackbody at a single temperature, even though it most definitely absorbs nearly all radiation that is incident upon it. That is because the Sun is ...

1

From a practical standpoint, some processes can be considered adiabatic because they happen so quickly that there isn't time for any substantial heat transfer. The best common example of this is the process inside the cylinders of your auto engine when the spark plug ignites the air-fuel mixture. The resulting combustion, compression, and expansion work ...

1

Heat, a measure of thermal energy, can be transferred from one point to another. Heat flows from the point of higher temperature to one of lower temperature. The heat content, Q, of an object depends upon its specific heat, c, and its mass, m. The Heat Transfer is the measurement of the thermal energy transferred when an object having a defined specific heat ...

1

Newton's law of cooling is a corollary of Fourier's law of heat conduction: $$q=-\kappa \nabla T,$$ where $q$ is the heat flux, $\kappa$ the heat conductivity and $\nabla T$ the temperature gradient (in a single dimension $\nabla T=\frac{dT}{dx}$). In essence this law tells us that heat flows from hot to cold and that the heat flow is proportional to the ...

1

You can circulate chilled water through the pipe which should distribute the temperature. Rate of circulation, along with the thermal properties of the pipe and embed media will control the precise temperature gradient - if thats even important for your application. Lastly, the problem of steady state temperature distribution in a metal rod is well ...

1

The summation indices on $Z$ must match the summation indices on $\bar{E}$, because $Z$ is the normalization constant for the total probabilities of being in every state. The $n=0$ state has zero wavenumber and doesn't exist. You're right to discard the analogy with para hydrogen. In that case, the peak is caused by interaction of the nuclear spins of the ...

1

When normalized, $A$ is just equal to $1,$ so that $f(E)$ varies between $0<f(E)<1.$ Addendum for the edited question: The prefactor $\frac{2}{(2\pi\hbar)^3}$ crops up in the volume integration of density of states performed in k-space for the computation of number of states $N$ (i.e. all available energy states up to a certain maximum (fermi level) ...

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