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Three processes are involved: Conduction: Heat flows from the object to its environment. Removal rate of heat from the interface further away from the object is proportional to the coefficient of conductivity (0.024 for air, 205 for aluminum -see http://www.engineeringtoolbox.com/thermal-conductivity-d_429.html). Convection: The interface between the object ...


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Are all heaters (same wattage, electric to thermal, no geothermal or other extra energy source) exactly as efficient as each other? No. Let's focus just on electrically powered heaters. If you have a heater that basically consists of a resistor with a current passing through it, you have 100% efficiency of electrical energy to heat energy conversion. ...


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What you are hearing is mains hum: mains electricity is alternating current (ie the voltage is approximately sinusoidal and symmetric about zero), with a frequency of 50Hz or 60Hz. things like kettles and heaters use a lot of power and parts of them will mechanically change shape at this frequency, which is audible. This kind of physical noise from things ...


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I will be blunt. As fas I know, nobody knows a priori for which systems equilibrium statistical mechanics will work or not. Part of the current effort to determine which systems are fine being described by equilibrium statistical mechanics focuses on various proofs of ergodicity for such systems. For now, they are somewhat limited to either a restrictive ...


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One benefit of scaling the heat capacity with another extensive variable is that you end up with an intensive property -- heat capacity per # of particles. Similarly specific heat refers to the heat capacity per unit mass so that the value of the intensive property can be compared between samples of the same material but with different sizes or geometries ...


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Energy stored as heat, by itself, is neither low- nor high-quality. What matters is the temperature at which the heat is stored, and the relationship of that temperature compared to the heat sink that will absorb the excess energy in the process. To be more specific, say you have a heat sink at $T_S=20°\:\mathrm C$, such as the atmosphere for a car engine. ...


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The (long-term) temperature of an object depends on the heat transfer between it and all of the environment. Air isn't a great conductor of heat. So if there is little air movement, the radiation environment may dominate the heat transfer. A cold calm day may feel quite balmy under full sunlight. On a cold evening, the sky may have a radiation ...


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Your second thesis is right. At least the result. The explanation not, as the example in the answer of Diracology might illustrate. Assume, that the piston does not move (we fix it). So $V$ didn't change on either side. Obviuosly, $n$ didn't change on either side. Therefore, $p$ is proportional to $T$ (in each side separately!). So if $T$ rises by the same ...


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Let $T_{20}$ be the initial temperature of tank 2 and $T_{10}=T_{20}+\Delta T$ be the initial temperature in tank 1. Let $\delta T$ be the equal rise in the temperature of both thanks. Assuming that the piston does not move, we would have $$p_{2f}=p_2\frac{T_{20}+\delta T}{T_{20}}$$and$$p_{1f}=p_1\frac{T_{20}+\Delta T+\delta T}{T_{20}+\Delta T}$$ Since ...


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$kT$ is related to the kinetic translation energy by the equipartition theorem. You are saying that the mean kinetic energy, is much greater than the rest energy. The particle has a large or relativistic velocity. The limit $kT>> mc^2$ is called ultrarelativistic limit. It means you can approximate the energy momentum relation $E^2=(pc)^2+(mc^2)^2$ by ...


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The Clausius' statement of the second law of thermodynamics states that the only effect of a cyclic process cannot be the transfer of heat from a cold body to a hot body. That is, for heat to be transferred from a cold to a hot body, work has to be expended, such as in a refrigerator, where the energy to remove heat from inside to the outside is derived from ...


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The earth's atmosphere can be considered as a thin sheet of air extending from the earth's surface to about an altitude of 60 miles. It is the earth's gravity that holds the atmosphere. The interconnection between temperature, pressure and density with altitude is as follows. About temperature variation with altitude:- The sun heats our earth's surface. ...


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The question has asked for Heat capacity(which is independent of mass) and NOT specific heat capacity(which depends on mass) thus the answer is $\ C = \frac {Pt}{ \Delta T}$ Remember two things: Heat capacity or thermal capacity is a measurable physical quantity equal to the ratio of the heat added to (or removed from) an object to the resulting ...


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For all three bullets, you need to use Neumann boundary conditions. open door: $Q = k(T_{out door}-T_{room})$ insulated wall $Q = 0$ heater boundary (i.e. hole) $Q = P_{heater}$ where $P_{heater}$ is the power of heater. Summation of Q can then be used to calculate the room temperature change,i.e. $$C_p m\frac {dT}{dt} = \sum Q$$


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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 ...


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There is something they forgot to mention in your notes (either from ignorance, or out of omission). The temperature within the system is spatially non-uniform during an irreversible process. So what value of the temperature are you supposed to use in the integral of dq/T? The Clausius inequality calls for the use of the temperature at the boundary ...


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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 ...


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Yes, it applies, and it's not really related to the Stefan-Boltzmann law. The energy radiated from a blackbody at temperature $T$ does indeed scale like $T^4$. Any object (blackbody or not) can absorb radiated energy, and that is the part which increases the temperature. The inverse square law is a statement about the density of radiation (or intensity, ...


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One purpose of the glass beads is to disrupt the flow of vapour by convection to improve mixing and attainment of thermal equilibrium at each level of the fractionating column. This creates a steeper, smoother temperature gradient. Without the beads (or similar 'baffles') convection currents would bring vapour at different temperatures and compositions to ...


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Ice, at least at atmospheric pressure, cannot form above the melting point of water (0 Celsius). The phenomenon of water freezing on objects like the ground, parked cars, motorbikes etc, is due to thermal inertia. On a long, cold spell these objects will cool down to below 0 Celsius. But when the ambient air temperature rises above 0 Celsius, the actual ...


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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)


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Entropy is a property of the system. The change in entropy of a system as it traverses from an initial state(1) to a final state(2) is independent of the path by which the system is taken from state 1 to state 2. The path can be a reversible one, or even irreversible, the change in entropy is always the same as long as the initial and final states are the ...


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I would like to add to what Procyon said in his(her) answer, which is right on target. The first equation in the OP should be an equality, not an inequality. For an irreversible process, the temperature of the system is typically non-uniform, and when we write $\int{\frac{dQ}{T}}$ for the Clausius inequality, what we really mean is ...


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Rigorous calculation The first hypothesis is hand-wavy Aristotelian logic. Whatever does happen has to be explainable by the Ideal Gas Law. At all times, $p_1 = p_2$ must hold, otherwise, the piston would move to equalize the pressures. So, $$\begin{eqnarray} p_1 =& p_2 \\ \frac{N_1 k_b T_1}{V_1} =& \frac{N_2 k_b T_2}{V_2} \\ \frac{V_2}{V_1} ...


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$$P_1V_1=m_1RT_1$$ $$P_2V_2=m_2RT_2$$ $P_1=P_2$,$V_1=V_2$,$T_1=T_2+\Delta T$ Then, we have:$$\frac{m_2}{m_1}=\frac{T_1}{T_1-\Delta T}$$ After heating: $$P'_1V'_1=m_1RT'_1$$ $$P'_2V'_2=m_2RT'_2$$ $P'_1=P'_2$,$T'_1=T'_2+\Delta T$ Then, we have:$$V'_2=V'_1\left(\frac{m_2(T'_1-\Delta T)}{m_1T'_1}\right)=V'_1\left(\frac{T_1}{T_1-\Delta ...


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If it was a clear sky then your motorbike most probably lost heat by radiative cooling and the ground usually cools more rapidly than the air. If you have placed your motorbike under a tree you would have probably found that there was no frost on your motorbike. You would have noticed the opposite of radiative cooling, radiative heating, when the Sun was ...


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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 ...


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This is in general a very complicated problem. There is three processes by which your object will cool down : conduction and convection which you already mentionned, as well as emission of blackbody radiation. You can treat blackbody radiation with the help of Stefan's law that gives you the power emission with respect to temperature. Conduction and ...



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