Hot answers tagged

14

The main reason is due to the fact that the prism refracts light in such a way that the "blue" part is more spread than "red" part. So that overall the energy hitting the thermometer is greater in the infrared and red part than on the blue part of the spectrum. Edit: I have just seen your edit. You're right. There you can see the details. Here's a quote ...


5

Static discharge has a strong correlation with humidity. In winter, due to low temperature, the dew point is lowered and as a result, the water content in the atmosphere is lowered. In summer, correspondingly the dew point is higher and water doesn't condense as easily and so there is a lot more content of water in the air. Now the moist air being weakly ...


4

Neutrons interact with each other only via exchange interaction Neutrons interact via the strong and weak forces. At low energies the interaction is principally via the nuclear force, or residual strong force, which derives ultimately from the strong force interactions between quarks. This can be described as an effective force due to the exchange of ...


4

There are some serious issues with your question, as others have pointed out. If you're in a perfectly protective space suit, you could be light years away from the sun and be perfectly comfortable since it is perfect and doesn't allow heat to escape. Oooorrrrrrr, maybe you would be hot at unreasonably large distances for precisely the same reason; there ...


4

There is a limited sense in which this correct. When inflation ended there was a temperature rise known as reheating, and we believe it was at this stage that the standard model particles were first created, or at least the majority of them. If you are measuring temperature by the energies of the standard model particles then this would have been the hottest ...


3

it has to do with the temperature lapse with altitude. since the speeed of sound is related to temperature by: $a = \sqrt{\gamma RT}$, where $\gamma$ and R are gas properties and T is temperature and the temperature profile follows (generally) like the left of these three plots: The area of interest for airliners is in the lowermost region where the ...


3

Water has a very narrow range of temperatures over which it expands when cooling rather than contracting (IIRC +2 to 0 Celsius). This occurs due to the way the highly polar molecules "line up" with each other near the freezing point. For a more interesting example, consider a number of rubber compounds which shrink as they warm up. In this case it's due ...


3

The Planck temperature isn't the hottest possible temperature in the same sense that zero Kelvin is a theoretical minimum. It is simply the temperature at which it's black body radiation is of the order of the Planck length. The Planck length is the length scale at which it is theorised that quantum-gravitational effects become significant. Quantum-gravity ...


3

The zero-range nuclear interactions felt by neutrons makes free neutrons, outside a nucleus or a neutron star, an excellent implementation of an ideal gas. However a neutron gas is a little unusual since neutron gases mostly are at such low density and pressure that the neutron-neutron interaction is very unlikely to occur before the neutron either decays ...


3

The term absolute isn't strictly defined, but most of us would agree that an absolute temperature scale has to have its zero at absolute zero. You are free to define any unit of temperature you want. There is nothing special about the size of the degree Kelvin, it was chosen to be the same as the degree Celcius i.e. one one hundredth of the difference ...


3

Yes, agitation will generally promote heat transfer and reduce heating times (although quantifying the effect is not easy). But the effect is not related to the bulk speed of the kettle. When the water is heated a diffuse (poorly defined) boundary layer is formed on the bottom of the vessel. This layer is at a temperature that is slightly higher than the ...


2

Assuming no heat is lost to the environment the heat balance on adding some boiling water ($212\:\mathrm{F}$) is given by: $$m_{bath}cT_{bath}+m_{added}cT_b=(m_{bath}+m_{added})cT_f$$ where: $m_{bath}=300\:\mathrm{Gall}$ is the initial amount of water, $T_{bath}=32.2\:\mathrm{Celsius}$, $m_{added}$ the amount of boiling water added, ...


2

It's not the question asked, but looking at the power requirements might give some insight. Raising the water temperature requires a specific amount of energy, and the time constraint gives a required power. $$P = \frac{m C T} { t}$$ $$P = \frac{300\text{gallon }(1000\text{kg/m^3})(4.186\text{J/g K}) 14\text{degF}}{1 \text{hour}}$$ $$P = ...


2

If you measure with sufficient accuracy, you will see a difference. Of course, any object you use to measure the air temperature might be exposed to sunlight and therefore experience significant difference in "apparent" temperature; but if we assume that you have a well shielded device that admits air but is insensitive to the effects of radiated heat (from ...


2

While a funny-looking coincidence, this is not a valid alternative expression for entropy in general, since the entropy of a probability distribution (which are what rigorously hides behind the strange word "macrostate") is more generally given by $$ S = - k_B \sum_i p_i\ln(p_i) \tag{1}$$ and becomes only $$S = k_B \ln(\Omega) \tag{2}$$ in the case of a ...


2

Well essentially there is no much difference, at least if we are talking of temperature and we mean "average kinetic energy" as we generally do. Why? Because neutrons as "every-day particles" interact via the nucleon-nucleon potential which similarly to the one between molecules, it is repulsive at short distance and atractive at longer distances. Of course ...


2

The book is wrong in the sense that it has made a lot of assumptions without actually accounting for all of them. So you are right that the question has incomplete information. The inaccuracies in the question are as follows: In such questions, you must refer to the principle of calorimetry. $$m_1s_1t_1=H=m_2s_2t_2$$ where $H=$heat absorbed by the body, ...


2

An egg has to reach an inner temperature of 100C to cook and in water the egg shell is kept at 100C for five minutes in a heat bath. Thus your question is answered by "can the egg shell be heated to 100C by the much hotter thin gas in the thermosphere" In vacuum the egg will radiate away and go close to 0 kelvin, so in the thermosphere it will be a fight ...


2

To continue Matas' comment: the definition of temperature is a quantity dependent on mean energy. It's sort of ugly when the particles all go near-relativistic, but there is no upper bound. Interestingly, you can extend the definition to things like magnetic particle orientation states. In this case, beyond a certain magnetic field strength, all the ...


2

It should be obvious that the ideal gas law $$ pV=nRT$$ is not invariant under the transformation $T\mapsto T+T_0$. But that is exactly what the relation between Kelvin and Celsius is - they are not, like most other choices of units, a different scaling, but they choose a different zero point of temperature, so you may not use them interchangably in formulae ...


2

Well, you could do it in any arbitrary units if you choose. Let's look at the perfect gas law: $P = \rho R T$. I can pick any consistent set of units that I want there, but it may involve offsets. For example, let's keep $P$ in Pascals, $\rho$ in kilograms per unit volume, $R$ in joules per kilogram per kelvin and let's use $T$ in Celsius. We know that $T_k ...


2

The main point to grasp is that the tiny inhomogeneities seen in the CMB are too small (by a factor of 100) to grow into the structures we see today without something like "cold dark matter". The CMB was formed at the epoch when normal (baryonic) matter and the radiation filling the universe decoupled. Only at this point was normal matter free to start ...


2

The ambient temperature comes out of many different factors, and daytime normally is warmer due to sunlight, but the largest single factor is simply how warm the air is... And that air can blow in from other places. The easiest time to notice this is when a front blows through your area in the middle of the day: the sunlight stays the same, but the air ...


1

Aside from the trivial possibility that a warm front passes through during the night, a frequent occurrence in winter (where I live) would be that you have a cold, sunny day, but that cloud cover begins to increase some time after the Sun has set. The cloud cover acts as a blanket, preventing infrared radiation from the ground escaping into space. This has ...


1

The rate of evaporation depends critically on the rate at which water vapor is removed from the water/air interface. If your cup was closed, water would evaporate until you reach the saturated vapor pressure of water at the prevailing temperature, and then things would reach equilibrium - water vapor would go into the liquid as fast as it evaporated. The ...


1

From a non-technical viewpoint, I would say that the simplest way I understand this is the following one. Yes, it has to do with pressure. Actually, it might be easier to think that it has to do with density. Consider the dominoes effect. If the dominoes are more apart from each other, that is, if the density of dominoes is lower, it will take longer for a ...


1

A sound wave moving through a gas requires a small scale bulk movement of gas molecules back and forth as pressure at any locations builds or falls. Therefore, the sound wave can not possibly move through the gas at a speed greater than that of the individual molecules themselves, and in fact must move at a lower speed than that due to the random nature of ...


1

Each individual blue photon is higher energy than a red photon, but the distribution of intensity in sunlight (or any source) is not equal for all wavelengths. "This light source emits in blue wavelengths and also in red wavelengths" is not enough to conclude that the total energy delivered by the blue will be more than that by the red.


1

In principle yes, though the situation isn't as clear cut as you describe. If you could confine a volume of matter within some volume then gradually heat it by adding energy to it then at some point the total energy density would exceed the density required to form a black hole and at that point the matter would start to collapse into a black hole. However ...


1

We need to think about what it means for one system to be "hotter" than another, or to have a temperature $T$. Thermodynamics defines the temperature of systems in thermal equilibrium. If an energy level $i$ of energy $U_i$ has degeneracy $g_i$, its occupation level $\propto g_i \exp \left( -\beta U_i \right) $ for some constant $\beta$ at thermal ...



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