The rate at which radiant energy from the sun reaches the earth’s upper atmosphere is about 1.50 kW/m2. The distance from the earth to the sun is 1.50 X 1011 m, and the radius of the sun is 6.96 X 108 m. (a) What is the rate of radiation of energy per unit area from the sun’s surface? (b) If the sun radiates as an ideal blackbody, what is the temperature of its surface?

Solution 114P Step 1 : In this question , we need to find rate of radiation energy per unit area from surface of sun’s surface Temperature on surface of sun Data given Rate of radiation dQ = 1.50 kW/m 2 dt 11 Distance from earth to Sun = 1.50 × 10 m Radius of Sun r = 6.96 × 10 m 8 Let us find the area of the Sun’s surface It is given by A = 4 r 2 Substituting the values we get 8 A = 4 × 6.96 × 10 m 18 A = 4 × 1.5218 × 10 m A = 6.0873 × 10 m 18 Step 2 : To find the temperature on the surface of the sun dQ 4 dt = A(T ) 8 We know = 5.67 × 10 Emissivity of blackbody = 1.0 Substituting the values we get 2 18 8 4 1.50 kW/m = 6.0873 × 10 m × 1.0 × 5.67 × 10 (T ) 1.50 kW/m2 = T 4 3.4518 ×10 T = 1.7404 × 10 11kW/m 3 T = 1.4438 × 10 kw/m In terms of kelvin we obtain as T = 1.4438 × 10 kw/m × 4 T = 5775 K Hence the temperature on the surface of the Sun will be 5775 K