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12

CMB spectrum has near-perfect black-body Planck spectra. Planck distribution has temperature as a parameter. By fitting observed spectrum by Planck function, we can "measure" temperature of CMB. This way we observe 2.7 K temperature today. In the context of photons temperature describes degree of disorder of photons of radiation in thermal equilibrium ...


12

The universe was a thermal plasma prior to the recombination epoch, consisting mostly of photons, protons, electrons, and alpha particles (helium-4 nuclei). There were also a small portion of deuterium, helium-3, and lithium-7 nuclei. All of this primordial stuff was in thermal equilibrium. The temperature of a gas is a result of the random velocities of ...


7

Next to the very detailed and good qualified answers, here is a simplified alternative: If you had a very dark body very far from any radiating thing, the cosmical background heated it until 2.7K.


7

The sign of a good fit is that the residuals have the same distribution as your model for the errors. Usually the assumption that goes into fitting methods is that the errors are normally distributed. That is, given perfect inputs $x_i$, and an ideal relation $y_i = f(x_i)$, you will measure $y_i + \epsilon_i$, where $\epsilon_i$ are distributed normally. ...


5

What is the temperature of the clear night sky from the surface of Earth? It's much closer to 273 K than 2.73 K. The answer depends on the surface temperature, the humidity, the temperature gradient through the atmosphere, and what exactly you mean by "the temperature of the clear night sky". The Swinbank formula provides an ad hoc expression for the ...


4

Here is a link to a study comparing heating water in a microwave to heating water in a conventional oven. Depending on the power of the microwave, the volume of the water, and time it's placed inside, the temperature will vary approximately linearly with time until either the system reaches equilibrium (for low power microwaves and large volumes of water) or ...


4

255 values sounds like the value that can be contained in a single byte. The person who created this "code" wanted to be able to represent "reasonable" temperatures with a single byte - they decided they wanted resolution better than 1°C, and they wanted to go down to "about as cold as you can get". This means that the conversion is as follows: From "C" ...


3

The units are probably in degrees Celsius. Whenever you add physical quantities together, they must have the same units. You can't add meters to kilograms (although you can multiply them or divide them). The result of such a thing would be nonsensical. However, you can add meters to meters. In your case, you are multiplying degrees by 2. If '2' is ...


3

Using the linearization procedure has become so common in experimental chemical kinetics that practitioners have taken to using it to define the activation energy for a reaction. That is the activation energy is defined to be (-R) times the slope of a plot of ln(k) vs. (1/T ) Thus, you should go with the linearized version because it is the common ...


2

Your idea is correct but the calculation is not. For $N$ moles of an ideal gas you have $U = cNRT$, look that if $T$ doesn't change, $U$ also doesn't change. So you start at the state $(T_1,V_1)$ and want to go to the state $(T_2,V_2)$. You then use two process corresponding to the following sequence of states: $$(T_1,V_1)\to (T_2,V_1)\to (T_2,V_2)$$ On ...


2

Simple answer: When you blow harder, more surrounding air gets mixed in with the stream of air from your mouth. The faster air moves, the lower pressure it has (Bernoulli's principle). So when you blow faster, your stream of air is lower pressure than the surrounding air. Thus the surrounding air fills in the stream. The surrounding air is obviously cooler ...


2

Indeed spherical harmonics are inappropriate, since they are not orthogonal on the restricted domain. This is particularly noticeable in small-scale surveys like ACT and BOOMERanG, but even "full-sky" surveys mask bad data. COBE for instance masked out the entire galactic plane, so the problem has been known since then. The solution presented by Górsky ...


2

If a cold metal object is standing still, and then if we move it, all of its particles gain some energy, kinetic to be precise. All of the molecules gain this energy. But, there is no increase in temperature. Why? Because we connect temperature with the chaotic motion on a molecular level, on the atomic level. This motion of a solid object is on a ...


2

You are probably most interested in one of the following: Flash point: lowest temperature at which a material will ignite in a normal atmosphere with an external source of ignition. Autoignition temperature: lowest temperature at which a material will spontaneously ignite in a normal atmosphere without an external source of ignition. For example: if a ...


2

How can I calculate the gas pressure given particles per cubic centimeter, and its temperature in Kelvin? as pointed out in comment by KyleKanos $PV=Nk_BT$ where $P$ is pressure, $V$ is volume (in $m^3$), $N$ is the number of particles, $k_B$ is Botzmann's constant and $T$ is temperature in Kelvin. If you rearrange it $P= {N \over V}~k_BT$ so ...


2

Several things are happening here that may make the sensations of touching metal and touching water similar when they are at room temperature (~ 25 C), although the thermal conductivities are a couple of orders of magnitude different. The sensation of coldness comes from the loss of heat from the part of your body contacting the material. The rate of heat ...


2

All internal energy such as thermal, rotational, and internal potential energy contributes to the rest mass of an object. In fact the vast majority of the mass of an atom is due the internal energy between quarks that make up the nucleus rather than the rest mass of the quarks themselves. So yes, a hot objected has greater rest mass and would weigh more ...


2

The formula is actually better written $$ \Delta S = \frac{Q}{T}. $$ That is, the change in entropy associated with the flow of heat is inversely proportional to the temperature at which the heat flow occurs. Note that $Q$ is already a change itself: it is not a state variable, but rather something more like $\Delta W$. Physically, this is because adding ...


1

The formula $$ L = \epsilon A \sigma T^4,$$ where $L$ is the luminosity in Watts, can be used for a "grey body" i.e. one that has a constant emissivity with frequency. Here you were told to "assume the Sun to be a perfect blackbody". This means that its emissivity is 1 because a blackbody absorbs everthing incident upon it and because it is in thermal ...


1

The parts of your body that generate heat and that can sense temperature and the loss of heat are insulated from the environment by a layer of dead skin cells. The total thermal conductivity to the environment is the thermal conductivity of the materials that you touch in series with the thermal conductivity of this layer of skin. Since this layer has a ...


1

The definition of the thermal coefficient of resistance (TCR) is the change in resistance per change in temperature divided by the resistance at a specified, fixed reference temperature: $$ \mathrm{TCR} = \frac{1}{R(T_\mathrm{ref})} \left.{\frac{\mathrm{d}R}{\mathrm{d}T}}\right|_{T=T_{ref}}. $$ Note that $R(T_\mathrm{ref})$ is a fixed value. It is the ...


1

In the standard homogeneous cosmological models the total energy in an expanding volume is zero. This is true for positive, negative or zero curvature and it must take into account the gravitational energy (which is negative), dark energy, matter and heat. Since the gravitational energy is negative the heat can be positive and increasing as you go back ...


1

For the sake of doing a strict units analysis in data conversions, to confirm that the code converts correctly; for that, each variable must be documented with the correct units. Then you're straight outa luck (or some cruder version of SOL). This unsigned eight bit integer contains a value that represents a temperature in a non-standard unit. The value ...


1

Absolute temperature relates only to translational degrees of freedom (connection to pressure via momentum exchange with a supposed exterior membrane). Since energy is constantly being randomly reshuffled between translational and non-translational degrees of freedom, the molar heat capacity is greater.


1

Just to add an explanation to sh37211's good empirical answer: the microwave transfers energy to the water at a constant rate. For a microwave rated 1,300 Watts and a typical 50% efficiency$^1$ that means a constant 650 Watts of heat energy delivered to the water. To calculate the change in temperature $(\Delta T)$ of a mass of material $(m)$ based on how ...


1

The first law of thermodynamics, is as the name suggest a law. It states that if you consider some process a thermodynamic system undergoes then $$\Delta U = W + Q$$ The point is that $W$ and $U$ are things that depend on what you are studying. Pick the infinitesimal version just for simplicity $$dU = \delta W + \delta Q$$ Then for a gas it trully makes ...


1

Suppose you have 5 cells with 1 ghost cell (related to your particular boundary conditions, whatever they may be) on either side. In this case, you can write your steady-state heat equation as \begin{align} T_0-2T_1+T_2 &= dx^2\,f(T_1)\\ T_1-2T_2+T_3 &= dx^2\,f(T_2)\\ T_2-2T_3+T_4 &= dx^2\,f(T_3)\tag{1}\\ T_3-2T_4+T_5 &= dx^2\,f(T_4)\\ ...



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