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If temperature of a back body increases from 7 0 C to 287 0 C, then the rate of energy radiation becomes: 2 times 16 times 4 times (287/7) 4

If temperature of a back body increases from 70C to 2870C, then the rate of energy radiation becomes:
  1. 2 times
  2. 16 times
  3. 4 times
  4. (287/7)4

Grade:12

3 Answers

Eshan
askIITians Faculty 2095 Points
4 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
184 Points
4 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
3007 Points
4 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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