ThermodynamicsHard
Question
If the temperature of a black body increases from 7oC to 287oC then rate of energy radiation increases by
Options
A.
B.16
C.4
D.2
Solution
By Stefan′s law energy radiated per sec a black body is given by E = AσT4
where A = area of black body. σ = Stefan′s constant. For a black body at temperature
T1, E1 = AσT14, at T2E2 = AσT4
(Since A, σ all same)
∴
⇒ E2 =
E1
T2 = 287oC = 287 + 273 = 560K
T1 = 7oC = 7 + 273 = 280K
∴ E2 =
E1 = 24 E1 = 16E1
∴ Rate of energy radiated increases by 16 times.
where A = area of black body. σ = Stefan′s constant. For a black body at temperature
T1, E1 = AσT14, at T2E2 = AσT4
(Since A, σ all same)
∴
⇒ E2 =
E1T2 = 287oC = 287 + 273 = 560K
T1 = 7oC = 7 + 273 = 280K
∴ E2 =
E1 = 24 E1 = 16E1∴ Rate of energy radiated increases by 16 times.
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