For a black body at temperature 727^ C , its radiating power is 60 ~W and the temperature of surrounding is 227^ C . If temperature of the black body is changed to 1227^ C , then its radiating power will be:

Correct answer: B. 320 ~W We can use the formula aligned & P (T_4-T_0^4) & (P_2)/(P_1)=((1500)^4-(500)^4)/((1000)^4-(500)^4) & =(500^4(3^4-1))/(500^4(2^4-1)) & (P_2)/(60)=(80)/(15) P_2=320 ~W aligned

NEET 2002

Exam: NEET
Year: 2002
Subject: Physics
Chapter: Thermal Properties of Matter

NEET 2002 · Physics · Thermal Properties of Matter

For a black body at temperature 727^ C , its radiating power is 60...

For a black body at temperature

727^{\circ} \mathrm{C}

, its radiating power is

60 \mathrm{~W}

and the temperature of surrounding is

227^{\circ} \mathrm{C}

. If temperature of the black body is changed to

1227^{\circ} \mathrm{C}

, then its radiating power will be:

Options

A $304 \mathrm{~W}$
B $320 \mathrm{~W}$
C $240 \mathrm{~W}$
D $120 \mathrm{~W}$

Correct answer

B. $320 \mathrm{~W}$

Step-by-step solution

We can use the formula

\begin{aligned} & P \propto\left(T_4-T_0^4\right) \\ & \therefore \frac{P_2}{P_1}=\frac{(1500)^4-(500)^4}{(1000)^4-(500)^4} \\ & =\frac{500^4\left(3^4-1\right)}{500^4\left(2^4-1\right)} \\ & \therefore \frac{P_2}{60}=\frac{80}{15} \Rightarrow P_2=320 \mathrm{~W} \end{aligned}

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