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Question

Physics Question on Heat Transfer

The SISI unit and dimensions of Stefan?s constant σ\sigma in case of Stefan?s law of radiation is

A

Jm3s4,[M1L0T3K4]\frac{J}{m^{3}\,s^{4}}, \left[M^{1}L^{0}T^{-3}K^{-4}\right]

B

Jm2s4K,[M1L0T3K3]\frac{J}{m^{2}\,s^{4}\,K}, \left[M^{1}L^{0}T^{-3}K^{3}\right]

C

Jm3sK4,[M1L0T3K4]\frac{J}{m^{3}s\,K^{4}}, \left[M^{1}L^{0}T^{-3}K^{4}\right]

D

Jm2sK4,[M1L0T3K4]\frac{J}{m^{2}s\,K^{4}}, \left[M^{1}L^{0}T^{-3}K^{-4}\right]

Answer

Jm2sK4,[M1L0T3K4]\frac{J}{m^{2}s\,K^{4}}, \left[M^{1}L^{0}T^{-3}K^{-4}\right]

Explanation

Solution

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=σT4E=\sigma T^{4}
where, E=E= energy emitted/area/second, T=T= absolute temperature in kelvin
and σ=\sigma= Stefan's constant. σ=ET4\Rightarrow \sigma=\frac{E}{T^{4}}
Unit of Stefan's constant
J/m2sK4=J/m2sK4\Rightarrow \frac{ J / m ^{2} s }{ K ^{4}}= J / m ^{2} s K ^{4}
Dimensions of Stefan's constant,
σ=[ Energy ] [Area] ×[ Time ]×[ Temperature ]4\sigma=\frac{[\text { Energy }]}{\text { [Area] } \times[\text { Time }] \times[\text { Temperature }]^{4}}
=[ML2T2][L2T×K4]=[M1L0T3K4]=\frac{\left[ ML ^{2} T ^{-2}\right]}{\left[ L ^{2} T \times K ^{4}\right]}=\left[ M ^{1} L ^{0} T ^{-3} K ^{-4}\right]