Question

A metallic cube initially kept at a temperature $T$ is emitting black body radiation with a power $P$ (energy emitted per unit time). If $T$ is increased by $1\%$, the power being radiated increases by $4.5\%$. What is the approximate percentage increase in the volume of the cube in this process?

• A. $0.75\%$
• B. $0.50\%$
• C. $1.56 \times 10^{-6}\%$
• D. $6.25 \times 10^{-6}\%$

Hint:

Using differentials for power-law relations $P \propto V^{2/3} T^4$ is extremely fast.
Simply write: $\% \Delta P = \frac{2}{3} \% \Delta V + 4 \% \Delta T$.
Substitute the given percentages to immediately solve for $\% \Delta V$.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
This problem relates the thermal expansion of a metallic cube (volume change) to the change in its black body radiation power due to both a temperature rise and an area increase.

Step 2: Key Formulas and Approach:
1. Stefan-Boltzmann Law for black body radiation:
\[ P = \sigma A T^4 \]
where $A$ is the surface area and $T$ is the absolute temperature.
2. Geometry of a cube:
For a cube of side $L$, the volume is $V = L^3$ and the surface area is $A = 6L^2 = 6V^{2/3}$. Therefore, $A \propto V^{2/3}$.
3. Fractional change approximation using logarithms and differentials:
\[ \ln P = \ln(\text{constant}) + \frac{2}{3}\ln V + 4\ln T \]

Step 3: Detailed Explanation:

• Let the initial power radiated be $P = \sigma A T^4$.

• Substituting $A \propto V^{2/3}$ into the Stefan-Boltzmann equation:
\[ P = C \cdot V^{2/3} T^4 \]
where $C$ is a constant.

• Taking the natural logarithm on both sides:
\[ \ln P = \ln C + \frac{2}{3} \ln V + 4 \ln T \]

• Differentiating both sides to find small fractional changes:
\[ \frac{dP}{P} = \frac{2}{3} \frac{dV}{V} + 4 \frac{dT}{T} \]

• Expressing this in terms of percentage changes by multiplying by 100:
\[ \left( \frac{dP}{P} \times 100 \right) = \frac{2}{3} \left( \frac{dV}{V} \times 100 \right) + 4 \left( \frac{dT}{T} \times 100 \right) \]

• We are given that the temperature increases by $1\%$ ($\frac{dT}{T} \times 100 = 1\%$) and the power increases by $4.5\%$ ($\frac{dP}{P} \times 100 = 4.5\%$).

• Substitute these values into the equation:
\[ 4.5 = \frac{2}{3} \left( \frac{dV}{V} \times 100 \right) + 4(1) \]
\[ 4.5 = \frac{2}{3} \left( \frac{dV}{V} \times 100 \right) + 4 \]
\[ 0.5 = \frac{2}{3} \left( \frac{dV}{V} \times 100 \right) \]
\[ \frac{dV}{V} \times 100 = 0.5 \times \frac{3}{2} = 0.75\% \]

• Thus, the approximate percentage increase in the volume of the cube is $0.75\%$.

Step 4: Final Answer:
The approximate percentage increase in the volume of the cube is $0.75\%$, which corresponds to Option (A).