The rectangular surface area (12\ cm×6\ cm) of the black body, at a temperature 127^ ^ C , emits heat energy at the rate of Q units per second. If the length and breadth of the surface area are each reduced to half of

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Correct answer: D. (81Q)/(64)

According to Stefan-Boltzmann law, the rate of heat emission is given by

E = \sigma A T^4

.

Initial area

A_1 = 12 \times 6 = 72\ \text{cm}^2

Initial temperature

T_1 = 127^{}^{\circ}\text{C} = 400\ \text{K}

Initial rate of heat emission

Q_1 = Q

Final area

A_2 = \dfrac{12}{2} \times \dfrac{6}{2} = 18\ \text{cm}^2 = \dfrac{A_1}{4}

Final temperature

T_2 = 327^{}^{\circ}\text{C} = 600\ \text{K}

Using the relation

Q \propto A T^4

, we get:

\dfrac{Q_2}{Q_1} = \dfrac{A_2}{A_1} \left(\dfrac{T_2}{T_1}\right)^4

\dfrac{Q_2}{Q} = \dfrac{1}{4} \left(\dfrac{600}{400}\right)^4

\dfrac{Q_2}{Q} = \dfrac{1}{4} \left(\dfrac{3}{2}\right)^4 = \dfrac{1}{4} \times \dfrac{81}{16} = \dfrac{81}{64}

Q_2 = \dfrac{81Q}{64}

Answer:

\dfrac{81Q}{64}

The rectangular surface area (12\ cm×6\ cm) of the black body, at a...

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