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Step-by-Step Solution
Step 1: Identify the relevant formula
The power (energy radiated per unit time) by a sphere at temperature $T$ and surface area $A$ is given by the Stefan-Boltzmann law:
$P = e \sigma T^4 A$
where:
$e$ is the emissivity of the material (assumed the same for both spheres),
$\sigma$ is the Stefan-Boltzmann constant,
$T$ is the absolute temperature of the sphere,
$A$ is the surface area of the sphere.
Step 2: Express the ratio of the powers
For two spheres (labeled 1 and 2) having the same emissivity and material, the ratio of their powers is:
$\displaystyle \frac{P_1}{P_2} \;=\; \frac{e \sigma T_1^4 A_1}{e \sigma T_2^4 A_2}
\;=\; \frac{T_1^4 A_1}{T_2^4 A_2}.$
Step 3: Substitute the surface areas in terms of radii
Each sphere’s surface area is given by $A = 4 \pi r^2$. Therefore,
$A_1 = 4 \pi r_1^2 \quad \text{and} \quad A_2 = 4 \pi r_2^2.$
Step 4: Plug in the given values
Given:
Sphere 1: radius $r_1 = 1\,\text{m}$, temperature $T_1 = 4000\,\text{K}$
Sphere 2: radius $r_2 = 4\,\text{m}$, temperature $T_2 = 2000\,\text{K}$
Hence,
$\displaystyle \frac{P_1}{P_2}
= \frac{T_1^4 \cdot 4\pi (r_1)^2}{T_2^4 \cdot 4\pi (r_2)^2}
= \frac{T_1^4 (1^2)}{T_2^4 (4^2)}
= \frac{(4000)^4}{(2000)^4} \times \frac{1}{16}.$
Step 5: Simplify the expression
Note that $\left(\frac{4000}{2000}\right)^4 = (2)^4 = 16.$ So,
$\displaystyle \frac{P_1}{P_2}
= 16 \times \frac{1}{16}
= 1.
This simplifies to $1:1$.
Answer: The ratio of energy radiated per second by the first sphere to that by the second is $1:1$.
