Question

If the surface temperature of hot body is doubled, then rate of radiation energy emitted will be enhanced by a factor of

• A. 2
• B. 4
• C. 16
• D. 32

Hint:

Always remember that radiation heat transfer scales with the fourth power of absolute temperature ($T^4$): - Temperature doubled ($2T$) $\implies$ Radiation increases by $2^4 = 16$ times. - Temperature tripled ($3T$) $\implies$ Radiation increases by $3^4 = 81$ times. - Temperature quadrupled ($4T$) $\implies$ Radiation increases by $4^4 = 256$ times. This non-linear scaling makes radiation the dominant mode of heat transfer at high temperatures.

Answer 1

The Correct Option is C

Concept: The rate of thermal radiation energy emitted by a body is governed by the Stefan-Boltzmann Law. This law states that the total radiant energy emitted per unit surface area of a blackbody per unit time is directly proportional to the fourth power of its absolute thermodynamic temperature. Mathematically, the total emissive power (\(E\)) is expressed as: \[ E = \epsilon \cdot \sigma \cdot A \cdot T^4 \] Where:
• \(\epsilon\) = Emissivity coefficient of the surface (\(\epsilon = 1\) for an ideal blackbody).
• \sigma = Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \text{ W/m}^2\text{K}^4\)).
• \(A\) = Surface area of the emitting body.
• \(T\) = Absolute temperature measured in Kelvin (\(\text{K}\)).

Step 1: Establishing the proportionality relation.
Assuming the surface geometry, area \(A\), and surface emissivity \(\epsilon\) remain constant, the rate of total radiation emission \(E\) depends solely on absolute temperature: \[ E \propto T^4 \]

Step 2: Applying the temperature doubling condition.
Let the initial absolute temperature of the hot body be \(T_1\), corresponding to an initial radiant energy emission rate \(E_1\). The problem states that the surface temperature is doubled: \[ T_2 = 2 \cdot T_1 \]

Step 3: Calculating the enhancement factor.
We set up the ratio of the final emissive power (\(E_2\)) to the initial emissive power (\(E_1\)): \[ \frac{E_2}{E_1} = \left(\frac{T_2}{T_1}\right)^4 = \left(\frac{2 \cdot T_1}{T_1}\right)^4 = (2)^4 \] Evaluating the fourth power of 2: \[ 2^4 = 2 \times 2 \times 2 \times 2 = 16 \] Therefore, the rate of radiation energy emitted scales by a factor of 16: \[ E_2 = 16 \cdot E_1 \] This corresponds to Option (3).