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Two spheres of the same material have radii 1 m and 4 m and temperatures 4000 K and 2000 K respectively. The energy radiated per second by the first sphere is
detailed solution
Correct option is C
The energy radiated per second E, by a body of surface area A, at temperature TK is given byE=kσAT4 [Stefan's law)where k = relative emissivityE1 = energy radiated per second by first sphere =kσ4π(1)2(4000)4 =kσπ1024×1012 joule Similarly, E2=kσ4π(4)2(2000)4 =kσπ1024×1012 joule ∴ E1=E2Talk to our academic expert!
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For a black body at temperature 727°C, its radiating power is 60W and temperature of surrounding is 227° C. If temperature of black body is changed to 1227°C, then its
radiating power will be :
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