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Question: Two spheres of the same material have radii 1 m and 4 m and temperatures 4000 K and 2000 K respectiv...

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

A

Greater than that by the second

B

Less than that by the second

C

Equal in both cases

D

The information is incomplete to draw any conclusion

Answer

Equal in both cases

Explanation

Solution

The energy radiated per second E, by a body of surface area A, at temperature TK, is given by

E=eσAT4E = e\sigma AT^{4}

Hence, E1=eσ4π(1)2×(4000)4E_{1} = e\sigma 4\pi(1)^{2} \times (4000)^{4}

=eσπ×1024×1012Js1= e\sigma\pi \times 1024 \times 10^{12}Js^{- 1}And E2=eσ4π(4)2(2000)4=eσπ×1024×1012Js1E_{2} = e\sigma 4\pi(4)^{2}(2000)^{4} = e\sigma\pi \times 1024 \times 10^{12}Js^{- 1} So, E1=E2E_{1} = E_{2}