Ratio of Cooling Times for Ideal Black Body

According to the Stefan-Boltzmann law, the total energy radiated per unit time by a black body of surface area $A$ and temperature $T$ into an evacuated environment at temperature $T_0 = 0\text{ K}$ is given by: $-\frac{dQ}{dt} = \sigma A T^4$

where $\sigma$ is the Stefan-Boltzmann constant.

Since the heat capacity $C$ of the object is independent of temperature, the heat loss can also be expressed as $dQ = C \, dT$. Substituting this into the rate equation yields: $-C \frac{dT}{dt} = \sigma A T^4$

Rearranging the terms to separate variables: $\frac{dT}{T^4} = -\frac{\sigma A}{C} dt$

Let $k = \frac{\sigma A}{C}$. Integrating both sides from the initial time $t = 0$ (at initial temperature $T_i = 200\text{ K}$) to a time $t$ (at temperature $T$): $\int_{200}^{T} T^{-4} \, dT = -k \int_{0}^{t} dt$

$\left[ -\frac{1}{3T^3} \right]_{200}^{T} = -kt$

$\frac{1}{3} \left( \frac{1}{T^3} - \frac{1}{200^3} \right) = kt$

$t = \frac{1}{3k} \left( \frac{1}{T^3} - \frac{1}{200^3} \right)$

1. For time $t = t_1$, the temperature drops to $T = 100\text{ K}$: $t_1 = \frac{1}{3k} \left( \frac{1}{100^3} - \frac{1}{200^3} \right) = \frac{1}{3k \cdot 100^3} \left( 1 - \frac{1}{2^3} \right) = \frac{1}{3k \cdot 100^3} \left( 1 - \frac{1}{8} \right) = \frac{1}{3k \cdot 100^3} \left( \frac{7}{8} \right)$

2. For time $t = t_2$, the temperature drops to $T = 50\text{ K}$: $t_2 = \frac{1}{3k} \left( \frac{1}{50^3} - \frac{1}{200^3} \right) = \frac{1}{3k \cdot 100^3} \left( 2^3 - \frac{1}{2^3} \right) = \frac{1}{3k \cdot 100^3} \left( 8 - \frac{1}{8} \right) = \frac{1}{3k \cdot 100^3} \left( \frac{63}{8} \right)$

Taking the ratio of $t_2$ to $t_1$: $\frac{t_2}{t_1} = \frac{\frac{63}{8}}{\frac{7}{8}} = \frac{63}{7} = 9$