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 $$\left(\frac{t_2}{t_1} \right)$$ is ___.

Let the object have surface area $$A$$ and a temperature-independent heat capacity $$C$$ (in J K-1).
Because the hollow sphere is at $$0\;\text{K}$$ and both bodies are ideal black bodies, the only mode of heat exchange is the radiation emitted by the object.

Radiative power lost
Stefan-Boltzmann law: $$P = \sigma A T^{4}$$, where $$\sigma$$ is the Stefan-Boltzmann constant.

Differential energy balance
Energy lost in time $$dt$$ = decrease in internal energy:
$$\sigma A T^{4}\,dt = -C\,dT$$

Rearranging, $$dt = -\dfrac{C}{\sigma A}\,T^{-4}\,dT$$

Time required to cool from $$T_0$$ to $$T$$
$$t = \dfrac{C}{\sigma A}\int_{T_0}^{T}\!\!-T^{-4}\,dT = \dfrac{C}{\sigma A}\int_{T}^{T_0}T^{-4}\,dT$$

Since $$\displaystyle\int T^{-4}\,dT = -\dfrac{1}{3}T^{-3}$$, we get
$$t = \dfrac{C}{3\sigma A}\left(T^{-3} - T_0^{-3}\right) \quad -(1)$$

Time $$t_1$$ : cooling from $$T_0 = 200\;\text{K}$$ to $$T = 100\;\text{K}$$
$$t_1 = \dfrac{C}{3\sigma A}\bigl(100^{-3} - 200^{-3}\bigr)$$

Time $$t_2$$ : cooling from $$T_0 = 200\;\text{K}$$ to $$T = 50\;\text{K}$$
$$t_2 = \dfrac{C}{3\sigma A}\bigl(50^{-3} - 200^{-3}\bigr)$$

Required ratio
$$\dfrac{t_2}{t_1} = \dfrac{50^{-3} - 200^{-3}}{100^{-3} - 200^{-3}} = \dfrac{\left(\dfrac{1}{50^{3}}\right) - \left(\dfrac{1}{200^{3}}\right)} {\left(\dfrac{1}{100^{3}}\right) - \left(\dfrac{1}{200^{3}}\right)}$$

Multiply numerator and denominator by $$200^{3}$$ to clear fractions:
$$\dfrac{ \left(\dfrac{200}{50}\right)^{3} - 1 } { \left(\dfrac{200}{100}\right)^{3} - 1 } = \dfrac{4^{3} - 1}{2^{3} - 1} = \dfrac{64 - 1}{8 - 1} = \dfrac{63}{7} = 9$$

Therefore, $$\displaystyle\left(\frac{t_2}{t_1}\right)=9.$$