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Question: If temperature of a black body increases from \(7 ^ { \circ } \mathrm { C }\) to \(287 ^ { \circ }...

If temperature of a black body increases from 7∘C7 ^ { \circ } \mathrm { C } to 287∘C287 ^ { \circ } \mathrm { C }, then the rate of energy radiation increases by

A

(2877)4\left( \frac { 287 } { 7 } \right) ^ { 4 }

B

16

C

4

D

2

Answer

16

Explanation

Solution

For a block body rate of energy Qt=P=AσT4\frac { Q } { t } = P = A \sigma T ^ { 4 }

⇒ P∝T4P \propto T ^ { 4 } ⇒ P1P2=(T1T2)4={(273+7)(273+287)}4=116\frac { P _ { 1 } } { P _ { 2 } } = \left( \frac { T _ { 1 } } { T _ { 2 } } \right) ^ { 4 } = \left\{ \frac { ( 273 + 7 ) } { ( 273 + 287 ) } \right\} ^ { 4 } = \frac { 1 } { 16 }