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If temperature of a back body increases from 70C to 2870C, then the rate of energy radiation becomes:\t2 times\t16 times\t4 times\t(287/7)4

Indrani , 7 Years ago
Grade 12
anser 3 Answers
Eshan

Last Activity: 6 Years ago

T_1=7^{\circ}C=280K
T_2=287^{\circ}C=560K

Rate of energy radiation\propto T^4

Hence the rate of energy radiation becomes(\dfrac{560}{280})^4=2^4=16times.

Gitanjali Rout

Last Activity: 6 Years ago

 
T_1=7^{\circ}C=280K
T_2=287^{\circ}C=560K

Rate of energy radiation\propto T^4

Hence the rate of energy radiation becomes(\dfrac{560}{280})^4=2^4=16times
so hope that you find the answer useful.

Khimraj

Last Activity: 6 Years ago

Rate of energy radiation is directly proportional to T4
where T is temperature
THen the rate of energy radiation become (560//280)4 = 16 times.
Hope it clears.

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