Question

If the absolute temperature of a black body is tripled, the total radiant energy emitted per second per unit area increases by a factor of:

• A. 3
• B. 9
• C. 27
• D. 81

Hint:

Radiation depends strongly on temperature: \[ E \propto T^4 \Rightarrow small T change causes large energy change \]

Answer 1

The Correct Option is D

Step 1: Understand black body radiation.
A hot black body gives off energy as radiation. The hotter it is, the much more energy it sends out each second from each unit of area.

Step 2: Recall the Stefan-Boltzmann law.
The energy radiated per second per unit area is proportional to the fourth power of the absolute temperature: $E \propto T^4$.

Step 3: Write the two states.
Let the first energy be $E_1$ at temperature $T$. The temperature is tripled, so the new temperature is $3T$ and the new energy is $E_2$.

Step 4: Take the ratio.
Dividing the new by the old removes the constant: \[ \frac{E_2}{E_1} = \left(\frac{3T}{T}\right)^4. \]

Step 5: Simplify the power.
\[ \frac{E_2}{E_1} = 3^4 = 81. \]

Step 6: State the answer.
Tripling the temperature raises the radiated energy by a factor of $81$. \[ \boxed{81} \]