A black body is heated from 27°C to 127°C. The ratio of their energies of radiations emitted will be
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Energies refer to different forms of power or ability to do work. These include kinetic energy (motion), potential energy (position), thermal energy (heat), chemical energy (stored in bonds), electrical energy (flow of electrons), and nuclear energy (stored in atomic nuclei). Energy can be transformed but never destroyed.
Class 11 Physics Chapter 10 Thermal Properties of Matter covers concepts like temperature heat and specific heat capacity. It explains thermal expansion and calorimetry. The chapter also discusses heat transfer through conduction and convection and introduces the laws of thermodynamics along with the behavior of gases under varying temperature and pressure conditions.
We will use the Stefan-Boltzmann law, which states that the energy radiated by a black body is proportional to the fourth power of its temperature in Kelvin:
E ∝ T⁴
Given:
– Initial temperature T₁ = 27°C = 27 + 273 = 300 K
– Final temperature T₂ = 127°C = 127 + 273 = 400 K
We will use the Stefan-Boltzmann law, which states that the energy radiated by a black body is proportional to the fourth power of its temperature in Kelvin:
E ∝ T⁴
Given:
– Initial temperature T₁ = 27°C = 27 + 273 = 300 K
– Final temperature T₂ = 127°C = 127 + 273 = 400 K
The ratio of energies radiated:
E₂ / E₁ = (T₂ / T₁)⁴
Now calculate the ratio:
E₂ / E₁ = (400 / 300)⁴ ≈ 3.16
Hence, the nearest answer is: 3 : 4
See more:
https://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-10/