Question:
Two bodies A and B having temperatures \( 327^\circ C \) and \( 427^\circ C \) are radiating heat to the surrounding. The surrounding temperature is \( 27^\circ C \). The ratio of rates of heat radiation of A to that of B is
Two bodies A and B having temperatures \( 327^\circ C \) and \( 427^\circ C \) are radiating heat to the surrounding. The surrounding temperature is \( 27^\circ C \). The ratio of rates of heat radiation of A to that of B is
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Two bodies A and B having temperatures $327C$ are radiating heat to the surrounding. The surrounding temperature is $27C$. The ratio of rates of heat radiation of A to that of B is
Updated On: Apr 15, 2026
- 0.52
- 0.31
- 0.81
- 0.42
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The Correct Option is A
Solution and Explanation
Step 1: Concept
According to Stefan-Boltzmann law, the net loss of energy of a body by radiation is $Q \propto (T^{4} - T_{0}^{4})$.
Step 2: Conversion
Convert temperatures to Kelvin: $T_1 = 327 + 273 = 600 \text{ K}$. $T_2 = 427 + 273 = 700 \text{ K}$. $T_0 = 27 + 273 = 300 \text{ K}$.
Step 3: Calculation
$\frac{Q_{1}}{Q_{2}} = \frac{600^{4} - 300^{4}}{700^{4} - 300^{4}} = \frac{6^{4} - 3^{4}}{7^{4} - 3^{4}} = \frac{1296 - 81}{2401 - 81} = \frac{1215}{2320} \approx 0.52$.
Final Answer: (A)
According to Stefan-Boltzmann law, the net loss of energy of a body by radiation is $Q \propto (T^{4} - T_{0}^{4})$.
Step 2: Conversion
Convert temperatures to Kelvin: $T_1 = 327 + 273 = 600 \text{ K}$. $T_2 = 427 + 273 = 700 \text{ K}$. $T_0 = 27 + 273 = 300 \text{ K}$.
Step 3: Calculation
$\frac{Q_{1}}{Q_{2}} = \frac{600^{4} - 300^{4}}{700^{4} - 300^{4}} = \frac{6^{4} - 3^{4}}{7^{4} - 3^{4}} = \frac{1296 - 81}{2401 - 81} = \frac{1215}{2320} \approx 0.52$.
Final Answer: (A)
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