Question 956: Thermodynamics & KTG

Question

The SI unit and dimensions of Stefan's constant $\sigma$ in case of Stefan's law of radiation is

Accepted Answer

According to Stefan's law, energy emitted by a body per unit area per second is proportional to fourth power of the absolute temperature.
$$E=\sigma T^4$$
where, $E=$ energy emitted/area/second, $T=$ absolute temperature in kelvin and $\sigma=$ Stefan's constant. $\Rightarrow \sigma=\frac{E}{T^4}$
Unit of Stefan's constant
$$\begin{aligned}
& 	ext { Unit of Stefan's } \
& \Rightarrow \quad \frac{\mathrm{J} / \mathrm{m}^2 \mathrm{~s}}{\mathrm{~K}^4}=\mathrm{J} / \mathrm{m}^2 \mathrm{~s} \mathrm{~K}^4 \
& 	ext { nimonsions of Stefan's constant, }
\end{aligned}$$
Dimensions of Stefan's constant,
$$\begin{aligned}
& \sigma=\frac{[	ext { Energy }]}{[	ext { Area }] 	imes[	ext { Time }] 	imes[	ext { Temperature }]^4} \
& =\frac{\left[\mathrm{ML}^2 \mathrm{~T}^{-2}ight]}{\left[\mathrm{L}^2 \mathrm{~T} 	imes \mathrm{K}^4ight]}=\left[\mathrm{M}^1 \mathrm{~L}^0 \mathrm{~T}^{-3} \mathrm{~K}^{-4}ight]
\end{aligned}$$

Answer

$\frac{\mathrm{J}}{\mathrm{m}^3 \mathrm{~s}^4}$,
$\left[M^1 L^0 T^{-3} K^{-4}\right]$

Answer

$\frac{\mathrm{J}}{\mathrm{m}^2 \mathrm{~s}^4 \mathrm{~K}}$, $\left[M^1 L^0 T^{-3} K^3\right]$

Answer

$\frac{\mathrm{J}}{\mathrm{m}^3 \mathrm{~s} \mathrm{~K}^4},\left[\mathrm{M}^{1} \mathrm{L}^0 \mathrm{~T}^{-3} \mathrm{~K}^4\right]$

Answer

$\frac{\mathrm{J}}{\mathrm{m}^2 \mathrm{~s} \mathrm{~K}^4},\left[\mathrm{M}^1 \mathrm{~L}^0 \mathrm{~T}^{-3} \mathrm{~K}^{-4}\right]$

Thermodynamics & KTG

The SI unit and dimensions of Stefan's constant \sigma in case of Stefan's law of radiation is

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