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A copper ball $2 \mathrm{~cm}$ in radius is heated in a furnace to $327^{\circ} \mathrm{C}$. If its emissivity is 0.3 , at what rate does it radiate energy?
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Verified Answer
The correct answer is:
$11.0 \mathrm{~W}$
The surface area of the ball
$$
A=4 \pi r^2=(4 \pi)(0.02 \mathrm{~m})^2=0.005 \mathrm{~m}^2
$$
and its absolute temperature is
$$
\begin{aligned}
T & =327+273=600 \mathrm{~K} . \text { Hence } \\
P & =e \sigma A T^4 \\
& =\left(0.3 \times 5.67 \times 10^{-8} \times 0.005 \times(600)^4\right. \\
& =11.0 \mathrm{~W}
\end{aligned}
$$
$$
A=4 \pi r^2=(4 \pi)(0.02 \mathrm{~m})^2=0.005 \mathrm{~m}^2
$$
and its absolute temperature is
$$
\begin{aligned}
T & =327+273=600 \mathrm{~K} . \text { Hence } \\
P & =e \sigma A T^4 \\
& =\left(0.3 \times 5.67 \times 10^{-8} \times 0.005 \times(600)^4\right. \\
& =11.0 \mathrm{~W}
\end{aligned}
$$
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