If temperature of a black body increases from 300K to 900K , then the rate of energy radiation increases by

Correct answer: A. 81 According to Stefan's law ρ = (E)/(t) = σ AeT^ 4 So, we can write as (E2)/(E1) = ((T2)/(T1))^ 4 (E2)/(E1) = ((900)/(300))^ 4 ⇒ (E2)/(E1) = (3)^ 4 E _ 2 = 81 E _ 1 ⇒ (E2)/(E1) = 81

BITSAT 2017

Exam: BITSAT
Year: 2017
Subject: Physics
Chapter: Thermal Properties of Matter

BITSAT 2017 · Physics · Thermal Properties of Matter

If temperature of a black body increases from 300K to 900K , then...

If temperature of a black body increases from

300K

900K

, then the rate of energy radiation increases by

Options

A $81$
B $3$
C $9$
D $2$

Correct answer

A. $81$

Step-by-step solution

According to Stefan's law $ρ = \dfrac{E}{t} = \sigma AeT^{4}$

So, we can write as

$\dfrac{E2}{E1} = \left(\dfrac{T2}{T1}\right)^{4}$

$\dfrac{E2}{E1} = \left(\dfrac{900}{300}\right)^{4} ⇒ \dfrac{E2}{E1} = \left(3\right)^{4}$

${E}_{2} = 81{E}_{1} ⇒ \dfrac{E2}{E1} = 81$

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