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(B) According to the Stefan-Boltzmann law,the rate of cooling is given by: $\frac{dQ}{dt} = \sigma A (T^4 - T_0^4)$.Since $dQ = mc \, dT$ and $m = \rh

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$\frac{7}{72} \frac{r \rho c}{\sigma}$

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(B) According to the Stefan-Boltzmann law,the rate of cooling is given by: $\frac{dQ}{dt} = \sigma A (T^4 - T_0^4)$.Since $dQ = mc \, dT$ and $m = ho V = ho (\frac{4}{3} \pi r^3)$,we have: $mc \frac{dT}{dt} = -\sigma (4 \pi r^2) T^4$ (as $T_0 = 0 \, K$).Substituting $m$: $(ho \frac{4}{3} \pi r^3) c \frac{dT}{dt} = -4 \pi r^2 \sigma T^4$.Simplifying: $\frac{dT}{T^4} = -\frac{3 \sigma}{ho c r} dt$.Integrating from $T_i = 200 \, K$ to $T_f = 100 \, K$: $\int_{200}^{100} T^{-4} dT = -\frac{3 \sigma}{ho c r} \int_{0}^{t} dt$.$[-\frac{1}{3} T^{-3}]_{200}^{100} = -\frac{3 \sigma}{ho c r} t$.$\frac{1}{3} [\frac{1}{100^3} - \frac{1}{200^3}] = \frac{3 \sigma}{ho c r} t$.$\frac{1}{3} [\frac{8 - 1}{8 	imes 10^6}] = \frac{3 \sigma}{ho c r} t$.$\frac{7}{24 	imes 10^6} = \frac{3 \sigma}{ho c r} t$.$t = \frac{7}{72} \frac{r ho c}{\sigma} 	imes 10^{-6} \, s$.Since the time is requested in $\mu s$ $(10^{-6} \, s)$,the value is $\frac{7}{72} \frac{r ho c}{\sigma} \, \mu s$.

$A$ solid copper sphere (density $\rho$ and specific heat capacity $c$) of radius $r$ at an initial temperature $200 \, K$ is suspended inside a chamber whose walls are at almost $0 \, K$. The time required (in $\mu s$) for the temperature of the sphere to drop to $100 \, K$ is:

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