Question 715: Thermodynamics & KTG

Question

Black sphere of radius R radiates power P at certain temperature $T$. If the temperature is doubled, the radius gets doubled. Now the power radiated would be

Accepted Answer

Let the initial radius be $R$ and initial temperature be $T$.
Step 1: Write the formula for power radiated by a black body
According to Stefan-Boltzmann law,
$ P = \sigma \cdot A \cdot T^4 $
where $\sigma$ is the Stefan-Boltzmann constant, $A$ is the surface area, and $T$ is the temperature.
For a sphere, $A = 4\pi R^2$.
So,
$ P = \sigma \cdot 4\pi R^2 T^4 $
Step 2: Write the expression for the new power
When temperature is doubled $(2T)$ and radius is also doubled $(2R)$:
Let the new power be $P'$,
$ P' = \sigma \cdot 4\pi (2R)^2 (2T)^4 $
Step 3: Expand $(2R)^2$ and $(2T)^4$
$ (2R)^2 = 4R^2 $
$ (2T)^4 = 16T^4 $
So,
$ P' = \sigma \cdot 4\pi \cdot 4R^2 \cdot 16T^4 $
Step 4: Multiply and simplify
$ P' = \sigma \cdot 4\pi \cdot 4 \cdot 16 \cdot R^2 T^4 $
$ P' = \sigma \cdot 4\pi \cdot 64 \cdot R^2 T^4 $
But the original power $P = \sigma \cdot 4\pi R^2 T^4$
So,
$ P' = 64 \cdot P $
Final Answer:
$ \boxed{64P} $
So,
Option D
is correct.

Answer

4 P

Answer

8 P

Answer

16 P

Answer

64 P

Thermodynamics & KTG

Black sphere of radius R radiates power P at certain temperature T. If the temperature is doubled, the radius gets doubled. Now the power radiated would be

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