Question

The ratio of emissive power of perfectly blackbody at 1327°C and 527°C is...​

Answer 1

Answer:

$\boxed{\mathfrak{Ratio \ of \ emissive \ power = 16:1}}$

Explanation:

As per Stefan's Law the radiant energy emitted by a perfectly black body per unit area per second i.e. emissive power is directly proportional to the fourth power of its absolute temperature.

$\boxed{ \bold{E \propto T^4}}$

According to the question:

Absolute temperature of $\sf 1^{st}$ body = 1327°C = 1600 K

Absolute temperature of $\sf 2^{nd}$ body = 527°C = 800 K

So,

Emissive power of $\sf 1^{st}$ body = $\sf E_1 \propto 1600^4$

Emissive power of $\sf 2^{nd}$ body = $\sf E_2 \propto 800^4$

Ratio of emissive power:

$\sf \implies \dfrac{E_1}{E_2} = \dfrac{ {1600}^{4} }{ {800}^{4} } \\ \\ \sf \implies \dfrac{E_1}{E_2} = \dfrac{1600 \times 1600 \times 1600 \times 1600}{800 \times 800 \times 800 \times 800} \\ \\ \sf \implies \frac{E_1}{E_2} = \frac{2 \times 2 \times 2 \times 2}{1 \times 1 \times 1 \times 1} \\ \\ \sf \implies \frac{E_1}{E_2} = \frac{16}{1} \\ \\ \sf \implies E_1 : E_2 = 16 : 1$

Answer 2

Answer:

T1 = 1600 K

T2 = 800 K

Q1:Q2 = T1:T2

= 1600^4:800^4

= 16:1