A sphere at temperature 600 K is placed in environment of temperature 200 K, its cooling rate is H. If the temperature is reduced to 400 K, the cooloing is same environment will be

Question

A sphere at temperature 600 K is placed in environment of temperature 200 K, its cooling rate is H. If the temperature is reduced to 400 K, the cooloing is same environment will be (A)  (H/16)  (B)  ( (9/27) )H  (C)  ( (16/3) )H  (D)  ( (3/16) )H

Tardigrade Admin · Accepted Answer

From Stefans law, the total radiant energy emitted per second per unit surface area of a black body is proportional to the fourth power of the absolute temperature of the body. That is

E = σ T^{4}

where,

σ

is Stefans constant. When sphere cools from 600 K to 200 K, energy 400 K to 200 K then.

H = σ [(600)^{4} - (400)^{4}]

\frac{H}{H} = \frac{[ ( 600 ) ^{4} - ( 200 ) ^{4} ]}{[ ( 600 ) ^{4} - ( 400 ) ^{4} ]}

Using

a^{4} - b^{4} = (a^{2} - b^{2}) (a^{2} + b^{2}),

we have

\frac{H}{H} = \frac{[ ( 600 ) ^{2} - ( 200 ) ^{2} ]}{[ ( 600 ) ^{2} - ( 400 ) ^{2} ]} \times \frac{[ ( 600 ) ^{2} + ( 200 ) ^{2} ]}{[ ( 600 ) ^{2} + ( 400 ) ^{2} ]}

\frac{H}{H} = \frac{32}{12} \times \frac{40}{20} = \frac{16}{3}

H = \frac{3}{16} H

Q. A sphere at temperature 600 K is placed in environment of temperature 200 K, its cooling rate is H. If the temperature is reduced to 400 K, the cooloing is same environment will be

$\frac{H}{16}$

$(\frac{9}{27}) H$

$(\frac{16}{3}) H$

$(\frac{3}{16}) H$

Q. A sphere at temperature 600 K is placed in environment of temperature 200 K, its cooling rate is H. If the temperature is reduced to 400 K, the cooloing is same environment will be

16H​

(279​)H

(316​)H

(163​)H

$\frac{H}{16}$

$(\frac{9}{27}) H$

$(\frac{16}{3}) H$

$(\frac{3}{16}) H$