A metallic sphere is heated to a very high temperature T 1 . Its peak emission wavelength is λ 1 . When its temperature is reduced to T 2 , the total radiated power decreases by a factor of 16.

Question

A metallic sphere is heated to a very high temperature T 1 ​ . Its peak emission wavelength is λ 1 ​ . When its temperature is reduced to T 2 ​ , the total radiated power decreases by a factor of 16. If the emissivity of the sphere is 0.8, what is the ratio of the peak emission frequencies, f 1 ​ / f 2 ​ ?

TestHub · Accepted Answer

According to the Stefan-Boltzmann law, the total radiated power $P = ϵ σ A T^{4}$ . Given that the total radiated power decreases by a factor of 16, we have $\frac{P _{1}}{P _{2}} = 16$ .

Since $P \propto T^{4}$ , we have $(\frac{T _{1}}{T _{2}})^{4} = 16$ . This implies $\frac{T _{1}}{T _{2}} = 2$ . The emissivity (0.8) and surface area $A$ cancel out when taking the ratio of powers. According to Wien's Displacement Law, $λ_{m} T = b$ (where $b$ is Wien's constant). Thus, $λ_{1} T_{1} = λ_{2} T_{2}$ . Substituting $T_{1} = 2 T_{2}$ , we get $λ_{1} (2 T_{2}) = λ_{2} T_{2}$ , which simplifies to $\frac{λ _{1}}{λ _{2}} = \frac{1}{2}$ . The relationship between frequency $f$ and wavelength $λ$ is $f = c / λ$ . Therefore, the ratio of peak emission frequencies is $\frac{f _{1}}{f _{2}} = \frac{c / λ _{1}}{c / λ _{2}} = \frac{λ _{2}}{λ _{1}}$ . Using the ratio of wavelengths, $\frac{f _{1}}{f _{2}} = \frac{1}{1/2} = 2$ .

Physics - Heat & Thermo Question with Solution | TestHub

Options:

Doubts & Discussion