Question

An electric current passes through a metal strip 0.60 cm by 6.0 cm by 0.18 mm , heating it at the rate of 52 W . The strip has emissivity e=1.0and 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 = 9.0 K/W that blocks all radiation?

(A and B answers should be in K)

Answer 1

given

metal strip of dimensions 0.60 cm , 6 cm,0.18 mm = 0.018 cm

here the mode of heat transfer is only the radiation

we knwo that the radiation rate of heat transfer is P = A*e sigma*T^4 = 52 W

hewre total net power (energy rate) is  

a)

P = p_surface - P_envi

P = A*e sigma*T_s^4 - A*e sigma*T^4 = A*e*sigma(T_s^4 -T_envi^4)

here A is area of the strip is A = (0.6*6+0.018*0.6+ 0.018*6) cm^2 = 3.7188 2 cm^2 = 3.7188*10^-4 m^2

a

substituting the values

52 W = 3.7188*10^-4*1*5.67*10^-8(T_s^4 -250^4)

Ts = 1253.65 k

when the strip is enclosed in vacuum the temperature is T = 1253.6 k

b)

wooden box with thermal resistance R = 9.0 k/W

the temperature is T = ?

from the formula  

P = -k*A dT/t = -DT/R

here t is thickness fo the strip

P = T_envi - Ts/R

52 W *9 = T_envi - T_s

- T_s = P*R-P_envi

-T_S = 52*9 - 250 k

T_S = 218 k

when the strip is enclosed in insulating box with R = 9 k/W , the temperature is T_S = 218 k