From a6ba656affa3fba31612fb20df196061e18a6faf Mon Sep 17 00:00:00 2001 From: Krishna Date: Sun, 17 Nov 2024 17:18:58 -0600 Subject: [PATCH] thermo stuff --- content/physics/thermodynamics.md | 26 +++++++++++++++++++++++++- 1 file changed, 25 insertions(+), 1 deletion(-) diff --git a/content/physics/thermodynamics.md b/content/physics/thermodynamics.md index 119ecd5..1bd3a53 100644 --- a/content/physics/thermodynamics.md +++ b/content/physics/thermodynamics.md @@ -65,5 +65,29 @@ $$ \lambda_{\text{peak}}=\frac{b}{T} $$ -Where $b$ is Wien's displacement constant, equal to $2.897771955\cdot 10^{-3} \mathrm{m}\cdot\mathrm{K}$. +Where $b$ is Wien's displacement constant, equal to $2.897771955\cdot 10^{-3} \mathrm{m}\cdot\mathrm{K}$. + +> [!NOTE] +> The material does not matter. All black bodies emit radiation over all wavelengths. + +## Stefan Boltzmann Law of Radiation +This law describes the emissive power of a Black Body per unit area. + +$$ +E_b=\varepsilon \sigma\cdot T^4 +$$ + +In the above equation,: + +- $E_b$ is the **Emissive Power** of a black body, per unit time, per unit area. +- $\varepsilon$ is the emissivity of an object +- $\sigma$ is the Boltzmann constant, about $\approx 5.670374419\cdot 10^{-8} \mathrm{W}\cdot\mathrm{m}^{-2}\cdot \mathrm{K}^{-4} +- A perfect black body has an emissivity of $1$, and an albedo of $0$ +- always work in default SI units + +If you desire the total power across the entirety of the surface: +$$ +P_b=A\varepsilon \sigma\cdot T^4 +$$ +Where $A$ is the area of the exposed surface.