MCQ
The unit of Stefan-Boltzmann s constant $(\sigma )$ is
- A$\frac{{wat{t^4}}}{{m \times {K^4}}}$
- B$\frac{{calorie}}{{{m^2} \times {K^4}}}$
- ✓$\frac{{watt}}{{{m^2} \times {K^4}}}$
- D$\frac{{joule}}{{{m^2} \times {K^4}}}$
✓
Answer
Correct option: C.
$\frac{{watt}}{{{m^2} \times {K^4}}}$
c
According to $Stefan-Boltzmann's$ Law, energy radiated per sec per unit area is:
According to $Stefan-Boltzmann's$ Law, energy radiated per sec per unit area is:
$E=e \sigma\left(T^{4}-T_{0}^{4}\right)$
Hence, unit of $\sigma=\frac{\text {joule} / \mathrm{sec}-m^{2}}{K^{4}}=\frac{\text {watt}}{m^{2} \times K^{4}}$
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