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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 ...


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 ...


0

You would expect it to radiate roughly a black body spectrum, but with lower intensity due to the thinness. You would also be able to see through it to things beyond. So if you had a container like that in front of you, you would see the background with a slight overlay of black body spectrum.


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Yes, when we heat an object its mass increases. The complete equation is $$E^2=(pc)^2 + (mc^2)^2$$ And from this equation if a system has zero momentum (p=0), then it has energy $E=mc^2$ When you heat an object the molecules or atoms begin to vibrate, rotate with more kinetic energy. But this doesnt increase the momentum of the system (the object is a ...


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

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 ...


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 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 ...


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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.


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 ...


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 ...


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.


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 ...


0

As long as the temperature of an ideal gas does not change its internal energy will remain the same. Change in volume without changing the temperature as in an isothermal process will not bring about any change in the internal energy of the gas. However a change in temperature of the gas by any means would result in change in internal energy of the gas as ...


0

No, your solution is not correct. The energy difference is by definition the first term in your formula, you should drop the second term. The work done during the process is path dependent, so you don't have enough information to calculate the work done, neither the transferred heat (which is the sum of the energy difference and the work). One possible way ...


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 ...


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)\\ ...


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 ...


0

As you wrote $W = \int_{\gamma} p(s) ds$ you are implying that the pressure, I assume you denoted pressure by $p$, is function of only the volume $V$ and the dummy integration variable is in fact the volume. That is not the case, the thermal equation of state even for the simplest system involves the (absolute or empirical) temperature, as well. That is ...


0

If there were a net force at the interface, the interface would move (i.e. one rod would expand and the other contract while they remain touching). If the interface remains in a fixed position then the net force across it must be zero. Actually it seems quite plausible to me that the interface could move slightly from the point at which the rods first ...


0

As far as I can tell, the question is "Why is assuming a zero net force at the interface less appropriate than assuming a net force that is balanced by forces from the walls". The reason is because if we assume a zero net force, then that means that we're assuming each rod is providing the same amount of force. With them being different sizes and materials, ...


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 ...


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. ...


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 ...


0

The voltage of a Galvanic cell can both increases with temperature or decreases with temperature. You can prove using thermodynamics or using your intuition that: for Galvanic cells that get hot when working, the voltage decreases as the temperature increases; for Galvanic cells that get cold when working, the voltage increases as the temperature increases. ...


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 ...


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 ...


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 ...


0

In addition to Mike's answer, there are a few things that need to be noted. First, I would say that the existence of TCR is a consequence of the assumption that the resistivity is linear. It is not because the TCR is constant that the resistivity is linear. A given TCR is valid only for a specific range of temperature. Basically that is the slope of the ...


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 ...


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 ...


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

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

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 ...


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 ...


0

When the temperature differences are "smooth" enough, i.e., locally there is a reasonable definition of temperature (local equilibrium), then the temperature gradient determines the heat flux. In the opposite case, it is molecular kinetics who determines the energy transfer. The latter happens much faster and local equilibrium gets established quickly.


1

In Zero-Energy Model, negative energy associated with Gravity counterbalances positive energy associated with matter, photons, etc. So, No, Big Bang wasn't cold. You are just looking at partial picture (you just ignored Gravity). This is what Zero-Energy Model says: With traditional Big Bang model (which doesn't contain Inflation), the universe started out ...



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