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The Radius of a Star Is 6.95×108 m6.95 \times 10 ^ { 8 } \mathrm {~m}

Question 4

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The radius of a star is 6.95×108 m6.95 \times 10 ^ { 8 } \mathrm {~m} and its rate of radiation has been measured to be 5.32×10265.32 \times 10 ^ { 26 } W. Assuming that it is a perfect emitter, what is the temperature of the surface of this star? (σ=( \sigma = 5.67×108 W/m2K4) \left. 5.67 \times 10 ^ { - 8 } \mathrm {~W} / \mathrm { m } ^ { 2 } \cdot \mathrm { K } ^ { 4 } \right)


A) 3.93×107 K3.93 \times 10 ^ { 7 } \mathrm {~K}
B) 8.87×103 K8.87 \times 10 ^ { 3 } \mathrm {~K}
C) 8.25×103 K8.25 \times 10 ^ { 3 } \mathrm {~K}
D) 5.78×107 K5.78 \times 10 ^ { 7 } \mathrm {~K}
E) 6.27×103 K6.27 \times 10 ^ { 3 } \mathrm {~K}

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