Topic: Heat Transfer
Two bodies A and B having equal surface area are maintained at temperatures 10 C and 20 C. The energy of thermal radiations emitted in a given time by A and B are in the ratio:
- A. 1:1.15
- B. 1:2.5
- C. 1:4.2
- D. 1:5.6
Answer: A. 1:1.15
Explanation: According to the Stefan-Boltzmann law, the energy of thermal radiation emitted is proportional to the fourth power of the absolute temperature (E is proportional to T^4). The temperatures in Kelvin are T_A = 10 + 273 = 283 K and T_B = 20 + 273 = 293 K. The ratio of their emitted energies is (E_A / E_B) = (283 / 293)^4, which evaluates to approximately 1 / 1.149, or a ratio of 1:1.15.