Question

Two spheres of the same material have radii \(1\,m\) and \(2\,m\), and temperatures \(2000\,K\) and \(1000\,K\) respectively. What is the ratio of energy radiated per second by the first sphere to that by the second?

• A. \(1:2\)
• B. \(2:1\)
• C. \(1:4\)
• D. \(4:1\)

Hint:

Thermal radiation power is \[ P\propto r^2T^4. \] A small increase in temperature greatly increases radiation because of the fourth-power dependence on \(T\).

Answer 1

The Correct Option is D

Step 1: Radiation law.
By Stefan's law a body radiates power \[ P=e\sigma A T^4, \] and for a sphere $A=4\pi r^2$. Same material means the emissivity $e$ is the same for both.

Step 2: Keep only what changes.
Dropping the common constants, \[ P\propto r^2 T^4. \]

Step 3: Write the ratio.
\[ \frac{P_1}{P_2}=\frac{r_1^2 T_1^4}{r_2^2 T_2^4}. \]

Step 4: Substitute the numbers.
With $r_1=1,\ r_2=2,\ T_1=2000,\ T_2=1000$, \[ \frac{P_1}{P_2}=\frac{(1)^2(2000)^4}{(2)^2(1000)^4}. \]

Step 5: Handle each factor.
The radius part gives $\dfrac{1}{4}$. The temperature part gives $\left(\dfrac{2000}{1000}\right)^4=2^4=16$.

Step 6: Multiply them.
\[ \frac{P_1}{P_2}=\frac{1}{4}\times16=4. \] So the ratio is $4:1$.
\[ \boxed{P_1:P_2=4:1} \]