Problem 16.63 6 of6 Part A An electric current passes through a metal strip 041b

Question

Problem 16.63 6 of6 Part A An electric current passes through a metal strip 041by43 cm by 0.19 mm , heating it at the rate of 44 w . The strip has emissivity e 1.0 and its surroundings are at 250 K. What will be the temperature of the strip if it's enclosed in (a) a vacuum bottle transparent to all radiation and (b) an insulating box with thermal resistance R- 8.7 K/W that blocks all radiation? Express your answer using two significant figures. Submit Part B Express your answer using two significant figures. Submit

Explanation / Answer

Solution:

Let us go to the basics first.

(a)The strip is in energy balance between the input power and the net power radiated (the only transfer mechanism available).

Then, Pin = Prad = A(T14 - T24)

=> T14 = T24 + Pin /A

The total surface area of the strip is 2[0.41*4.3 + 0.019*4.71] = 3.705 cm2 = 3.705*10-4 m2

(the edges contributing only 2%).

Making the numerical substitutions, we find:

T14  = (250)4  + 44 /(5.67*10-8)(3.705*10-4)

=>T1 = 1203.573 K = 1.2*103 K (Two significant figures) (Answer a)

(b) Now, all the input power is transferred by conduction through the box,

Pin = H = - (T2 - T1) / R

=> T1 = T2 + RPin

=> T1 = 250 + 8.7*44 = 632.8 K = 630 K   (Two significant figures) (Answer b)

Thanks!!!

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