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Scale Conversions

A small objec...

Question

A small object is placed at the center of a large evacuated hollow spherical container. Assume that the container is maintained at $0 K$ . At time $t = 0$ , the temperature of the object is $200 K$ . The temperature of the object becomes $100 K$ at $t = t_{1}$ and $50 K$ at $t = t_{2}$ . Assume the object and the container to be ideal black bodies. The heat capacity of the object does not depend on temperature. The ratio $(\frac{t_{2}}{t_{1}})$ is

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Solution

$T_{s} = 0 K$
$T_{i} = 200 K, e = 1$
We know that,
$- m s \frac{d T}{d t} = \frac{d Q}{d t} = σ e A T^{4}$
$- \frac{d T}{d t} = \frac{σ A T^{4}}{m s}$
$\frac{σ A}{m s} \int_{t_{i}}^{t_{f}} d t = - \int_{T_{i}}^{T_{f}} \frac{d T}{T^{4}}$
$\frac{σ A}{m s} (t_{f} - t_{i}) = + \frac{1}{3} ⎛ ⎝ \frac{1}{T_{f}^{3}} - \frac{1}{T_{i}^{3}} ⎞ ⎠$
$\frac{\frac{σ A}{m s} (t_{1} - 0) = \frac{1}{3} (\frac{1}{100^{3}} - \frac{1}{200^{3}})}{\frac{σ A}{m s} (t_{2} - 0) = \frac{1}{3} (\frac{1}{50^{3}} - \frac{1}{200^{3}})}$
$\frac{t_{1}}{t_{2}} = \frac{\frac{(200)^{3} - (100)^{3}}{(100)^{3} (200)^{3}}}{\frac{(200)^{3} - (50)^{3}}{(50)^{3} (200)^{3}}} = \frac{\frac{4^{3} - 2^{3}}{2^{3}}}{4^{3} - 1^{3}} = \frac{56}{5 \times 63} = \frac{1}{9}$
$\Rightarrow \frac{t_{2}}{t_{1}} = 9$

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