Rate of Cooling Explained

This problem requires understanding how the size of an object affects its rate of cooling when heat is lost primarily through radiation. We need to compare the cooling speed of two spheres with different diameters but identical material properties and initial temperatures.

Radiation Cooling Principles

Objects lose heat via radiation according to the Stefan-Boltzmann Law. The rate of heat energy radiated ($P$) is given by the formula:

$$ P = \frac{dQ}{dt} = \sigma \epsilon A (T^4 - T_0^4) $$

Key variables are:

• $\sigma$: Stefan-Boltzmann constant (a physical constant).
• $\epsilon$: Emissivity of the surface (depends on the material and finish, same for both balls).
• $A$: Surface area of the radiating body.
• $T$: Absolute temperature of the body (same for both balls).
• $T_0$: Absolute temperature of the surroundings (same for both balls).

The question asks for the "rate of cooling". Based on the context and typical physics problem interpretations relating to radiation, this usually refers to the rate of heat energy loss ($dQ/dt$).

Sphere Properties Analysis

We have two spheres:

• Same Material & Surface Finish: This means their emissivity ($\epsilon$) is identical.
• Diameter Ratio: The ratio of the diameters ($d_1$ for the bigger sphere, $d_2$ for the smaller sphere) is $d_1 : d_2 = 2 : 1$.
• Same Initial Temperature: $T_1 = T_2 = T$.
• Same Surroundings: $T_{0,1} = T_{0,2} = T_0$.

Surface Area Ratio Calculation

The surface area ($A$) of a sphere is calculated using $A = \pi d^2$. Let $A_1$ be the surface area of the bigger sphere and $A_2$ be the surface area of the smaller sphere.

The ratio of their surface areas is:

$$ \frac{A_1}{A_2} = \frac{\pi d_1^2}{\pi d_2^2} = \left(\frac{d_1}{d_2}\right)^2 $$

Given that $d_1 / d_2 = 2 / 1$, the ratio becomes:

$$ \frac{A_1}{A_2} = \left(\frac{2}{1}\right)^2 = \frac{4}{1} $$

Thus, the larger sphere has 4 times the surface area of the smaller sphere.

Cooling Rate Ratio Determination

According to the Stefan-Boltzmann Law ($P = \sigma \epsilon A (T^4 - T_0^4)$), when $\sigma$, $\epsilon$, $T$, and $T_0$ are constant for both spheres, the rate of heat loss ($P$) is directly proportional to the surface area ($A$).

$$ P \propto A $$

Therefore, the ratio of the rates of heat loss ($P_1 / P_2$) is equal to the ratio of their surface areas ($A_1 / A_2$):

$$ \frac{P_1}{P_2} = \frac{A_1}{A_2} $$

Using the calculated surface area ratio:

$$ \frac{P_1}{P_2} = \frac{4}{1} $$

Final Ratio Conclusion

The rate of cooling (meaning the rate of heat loss by radiation) of the bigger ball compared to the smaller ball is in the ratio 4 : 1.