A blackbody with initial temperature of 300^ C is allowed to cool… - Vidyadip

Question

A blackbody with initial temperature of $300^{\circ} \mathrm{C}$ is allowed to cool inside an evacuated enclosure surrounded by melting ice at the rate of $0.35^{\circ} \mathrm{C} /$ second. If the mass, specific heat and surface area of the body are $32 \mathrm{grams}, 0.10 \mathrm{cal} / \mathrm{g}{ }^{\circ} \mathrm{C}$ and $8 \mathrm{~cm}^2$ respectively, calculate Stefan's constant. (Take J $=4200 \mathrm{j} / \mathrm{kcal}$.)

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Answer

$
\begin{aligned}
& \text { Data }: \mathrm{T}=273+300=573 \mathrm{~K}, \mathrm{~T}_0=273 \mathrm{~K} \\
& \frac{d Q}{d t}=0.35^{\circ} \mathrm{C} / \mathrm{s}=0.35 \mathrm{~K} / \mathrm{s} \text {, at } \\
& \mathrm{M}=32 \mathrm{~g}=32 \times 10^{-3} \mathrm{~kg}, \mathrm{~A}=8 \mathrm{~cm}^2=8 \times 10^{-4} \mathrm{~m}^2 \mathrm{C}=0.10 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}=0.10 \mathrm{kcal} / \mathrm{kg} \cdot \mathrm{K}=420
\end{aligned}
$
$\mathrm{j} / \mathrm{kg} \cdot \mathrm{K}$
since $\mathrm{J}=4200 \mathrm{~J} / \mathrm{kcal}$
$
\begin{aligned}
& \frac{d Q}{d t}=\sigma A\left(T^4-T_0{ }^4\right) \quad \therefore M C \frac{d Q}{d t}=\sigma A\left(T^4-T_0{ }^4\right) \\
& \therefore \sigma=\frac{M C(d Q / d t)}{A\left(T^4-T_0{ }^4\right)}=\frac{32 \times 10^{-3} \times 420 \times 0.35}{8 \times 10^{-4} \times\left[(573)^4-(273)^4\right]} \\
& =\frac{1680 \times 3.5}{\left[(5.73)^4-(2.73)^4\right] \times 10^8}=5.754 \times 10^{-8}
\end{aligned}
$
Stefan's constant, $\sigma=5.754 \times 10^{-8} \mathrm{~W} / \mathrm{m}^2 \cdot \mathrm{K}^4$

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