Question

The temperature of a black body is increased by 50%, then the percentage increase in the rate of radiation by the body is approximated as

• A. 50%
• B. 100%
• C. 400%
• D. 150%

Hint:

For scaling problems with a power of 4, you can quickly estimate the answer using simple fractions: a 50% increase scales the variable by a factor of $\frac{3}{2}$. Raising this to the fourth power gives $\frac{81}{16} \approx 5$. A final value that is 5 times the original implies a net increase of exactly 4 times the baseline, which immediately converts to 400

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The problem states that the absolute temperature of an ideal black body increases by 50%. We need to calculate the resulting percentage increase in its total emissive power or rate of thermal radiation.

Step 2: Key Formula or Approach:
According to Stefan-Boltzmann Law, the total radiant energy emitted per unit surface area per second ($E$) from a perfectly black body is directly proportional to the fourth power of its absolute temperature ($T$):
$$E = \sigma T^4 \implies E \propto T^4$$ The percentage increase in any physical parameter is calculated using the standard fractional ratio:
$$\text{\% Increase} = \left( \frac{E_2 - E_1}{E_1} \right) \times 100\% = \left( \frac{E_2}{E_1} - 1 \right) \times 100\%$$

Step 3: Detailed Explanation:
Let the initial absolute temperature of the black body be $T_1$. The temperature is increased by 50%, so the new final absolute temperature $T_2$ is:
$$T_2 = T_1 + 50\% \text{ of } T_1 = T_1 + 0.5T_1 = 1.5T_1 = \frac{3}{2}T_1$$ Using the fourth-power dependency from Stefan's law, let's find the ratio of the final radiation rate to the initial radiation rate:
$$\frac{E_2}{E_1} = \left(\frac{T_2}{T_1}\right)^4 = \left(\frac{1.5T_1}{T_1}\right)^4 = (1.5)^4$$ Let's compute $(1.5)^4$ by squaring $1.5$ twice:
$$1.5^2 = 2.25 \implies (2.25)^2 = 5.0625 \approx 5$$ Substitute this ratio into our percentage increase equation:
$$\text{\% Increase} = \left( 5.0625 - 1 \right) \times 100\% = 4.0625 \times 100\% = 406.25\%$$ Rounding to the nearest multiple provided in the options gives an approximate value of 400

Step 4: Final Answer:
The percentage increase in the rate of radiation is approximately 400%, which corresponds to option (C).