a.) 13.09 Consult Interactive Solution 13.9 to explore a model for solving this

Question

a.) 13.09 Consult Interactive Solution 13.9 to explore a model for solving this problem. One end of a brass bar is maintained at 306 oC, while the other end is kept at a constant, but lower, temperature. The cross-sectional area of the bar is 2.10 × 10-4 m2. Because of insulation, there is negligible heat loss through the sides of the bar. Heat flows through the bar, however, at a rate of 6.74 J/s. What is the temperature of the bar at a point 0.227 m from the hot end?

b.)13. 29 The amount of radiant power produced by the sun is approximately 3.9 × 1026 W. Assuming the sun to be a perfect blackbody sphere with a radius of 6.96 × 108 m, find its surface temperature (in kelvins).

Explanation / Answer

a)

from the theory we have the equation

   Q = (k A T) t / L

   L = 0.227 m

   Q / t = 6.74 J / s

   T = (Q / t) L / k A

         = (6.74 J / s)(0.227 m) / [110 J / (m.s.oC)](2.1 x 10-4m2)

306-T =66.23

T =239.76oC

b)

Radiant power produced by the sun ,P=3.9*10^26 W

Let A be the Surface area of the sphere
Here Radius ,r =6.96 x 10^8 m

Therefore Area ,A=4**r2 =4**(6.96 x 10^8)2 =6.08*1018 m2

We know that Power.P=*A*T4

Here is Stefan-Boltzmann constant=5.7*10-8 W m-2 K-4 (Approximately)

Therefore T4=P/*A=(3.9*10^26 )/(5.7*10^-8)*(6.08*10^18)=1.1255791 × 1015 K4

Therefore T=5791 K (approx)

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