A) 0.52
B) 0.31
C) 0.81
D) 0.42
Correct Answer: A
Solution :
If temperature of surrounding is considered, then net loss of energy of a body by radiation \[Q=Ae\sigma ({{T}^{4}}-T_{0}^{4})\Rightarrow \,\,Q\propto ({{T}^{4}}-T_{0}^{4})\] \[\therefore \] \[\frac{{{Q}_{1}}}{{{Q}_{2}}}=\frac{T_{1}^{4}-T_{0}^{4}}{T_{2}^{4}-T_{0}^{4}}\] \[=\frac{{{(273+327)}^{4}}-{{(273+27)}^{4}}}{{{(273+427)}^{4}}-{{(273+27)}^{4}}}\] \[=\frac{{{(600)}^{4}}-{{(300)}^{4}}}{{{(700)}^{4}}-{{(300)}^{4}}}=0.52\]You need to login to perform this action.
You will be redirected in
3 sec