The thermal conductivity of a material is a measure of its ability to conduct heat. It is commonly denoted by
k
{\displaystyle k}
,
λ
{\displaystyle \lambda }
, or
κ
{\displaystyle \kappa }
.
Heat transfer occurs at a lower rate in materials of low thermal conductivity than in materials of high thermal conductivity. For instance, metals typically have high thermal conductivity and are very efficient at conducting heat, while the opposite is true for insulating materials like Styrofoam. Correspondingly, materials of high thermal conductivity are widely used in heat sink applications, and materials of low thermal conductivity are used as thermal insulation. The reciprocal of thermal conductivity is called thermal resistivity.
The defining equation for thermal conductivity is
q
=
−
k
∇
T
{\displaystyle \mathbf {q} =-k\nabla T}
, where
q
{\displaystyle \mathbf {q} }
is the heat flux,
k
{\displaystyle k}
is the thermal conductivity, and
∇
T
{\displaystyle \nabla T}
is the temperature gradient. This is known as Fourier's Law for heat conduction. Although commonly expressed as a scalar, the most general form of thermal conductivity is a second-rank tensor. However, the tensorial description only becomes necessary in materials which are anisotropic.
Hi,
I am reading "An Introduction of Solid State Physics" from Ibach Lüth and don't understand the integration process.
They write $$\sigma=\frac{e^2}{8\pi^3 \hbar}
\int df_{E}dE \frac{v^2_x(\bf{k})}{v(\bf{k})} \tau(\bf{k}) \delta(E-E_F)
$$
$$
= \int_{E=E_F}^{}df_{E}...
Hello guys!
I have to solve a problem about crystal symmetry, but I am very lost, so I wonder if anyone could guide me.
The problem is the following:
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If some voltage is traveling from a low conductivity metal to a high conductivity metal, does it impose limitations on the high conductivity metal such that it will only do as much as the low conductivity metal?
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Homework EquationsThe Attempt at a Solution
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Homework EquationsThe Attempt at a Solution
Part 1.1) I managed to find the energy needed to melt the ice, before using ΔU = Nf½KΔT to solve for the new temperature, which was correct
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Greetings Denizens!,
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Hi Dear,
First of all I thank the administrators and the whole team that runs this forum for having accepted me to be part of this great family and also for the quality of this forum.
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Hi,
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why there's no horizontal conductivity in this question ?
Homework EquationsThe Attempt at a Solution
IMO , there's also horizontal conducitivity in this question because of the water can also flow horizontally across layer A , B and also C . [/B]
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Hi all,
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Homework Statement
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2. Homework Equations [/B]
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The problem is given.
Homework EquationsThe Attempt at a Solution
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Thank you
Hi folks,
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Hi everbody
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Homework Statement
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Homework Equations
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Hi I am new to the forum,
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