Question

two spheres p and q of same emissivity having radii 8cm and 2cm are maintained at temperature 127°C and 527°C respectively ther ratii of energy radiated by p and q in a given time interval is

Answer 1

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Answer 2

Answer:

The ratio of energy radiated by p and q in a given time interval is 1.

Explanation:

The energy radiated by a spherical blackbody is given by

$\rm E = \epsilon A\sigma T^4 t.$

where,

• $\epsilon$ = emissivity of the sphere.
• A = surface area of the sphere = $\rm \pi R^2$
• R = radius of the sphere.
• T = absolute temperature of the sphere.
• $\sigma$ = Stefan-Boltzmann constant.
• t = time interval.

Given that,

The radii of the sphere p and q are:

$\rm r_p = 8\ cm.\\r_q = 2\ cm.$

The temperature of the spheres are:

$\rm T_p = 127^\circ C = 127+273.15=400.15\ K.\\T_q=527^\circ C = 527+273.15=800.15\ K.$

The energy radiated by the two spheres in time interval t are:

$\rm E_p = \epsilon_p A_p\sigma T_p^4 t=\epsilon_p \pi r_p^2\sigma T_p^4 t.\\E_q = \epsilon_q A_q\sigma T_q^4 t=\epsilon_q \pi r_q^2\sigma T_q^4 t.$

The emissivity of both the sphere are same, therefore, the ratio of energy radiated by p and q in a given time interval is given by:

$\rm \dfrac{E_p}{E_q}=\dfrac{\epsilon \pi r_p^2\sigma T_p^4 t}{\epsilon \pi r_q^2\sigma T_q^4 t}=\dfrac{r_p^2T_p^4}{r_q^2T_q^4}=\dfrac{8^2\times 400.15^4}{2^2\times 800.15^4}=1.00075\approx 1.$