5.5. A steelworker peers into a blast furnace through a window, using a small ha

Question

5.5. A steelworker peers into a blast furnace through a window, using a small hand-held spectroscope. He estimates that the brightest part of the broad visible emission from the molten steel is at 520 nm. The total area of the sample is 4 m^2.
(a) What does he deduce as the initial temperature of the steel?
(b) Assuming the efficiency of emission (the emissivity) is unity and that the efficiency of heating is also unity, how much power is being fed to the furnace from the factory, which is maintained at room temperature (293 K)?
(c) If the electrical power to the furnace is doubled, at what wavelength would the brightest emission be observed?
(b) What total power would be radiated by the molten liquid if the power to the furnace were abruptly turned off, allowing the steel to cool from its initial temperature to room temperature (T=293K)?

Explanation / Answer

Given     wavelength of light   ? = 520 nm    area of the sample A = 4 m^2       A) initial temperature of the steel                    ? = b / T                     T = b / ? where   b = 2.89*10^ -3     mK                   T = 2.89*10^ -3   / 520 nm                       = 5557 K = 5284 degrees      b) assume   emissivity e = 1                        efficiency of heating    = 1                  power P = s e A T ^4         where   s = 5.67*10^ -8   J / s                 P = ( 5.67*10^ -8   J / s )(1)( 4 m^2 )(293 K )^4                     = 1671.52 W     d)    P = s e A( T_i - T _f ) ^4                   = ( 5.67*10^ -8   J / s )(1)( 4 m^2 )(5557-293 K )^4   = 1.741*10^ 8 W      c)     ?_max = b / T                 T = b / ?               Power P   =s e A T ^4                               = s e A (b / ? ) ^4                        P ower is inversely proportional to ?^4                       P _1 / P _2 = ( ?_2/?_1)^4                            P / 2 P     = ?^4 / ?' ^4                            ?' =  ? / 2 ^ 1/4                                        ?'   =   0.84?                       

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