When the tungsten filament in a 60 W, 120 V light bulb reaches 2010degrees C, 60

Question

When the tungsten filament in a 60 W, 120 V light bulb reaches 2010degrees C, 60 W of electric power is used to produce light andheat.
a) Find the current through the filament under theseconditions which are established very shortly after the light isturned on.
b)Find the filament current when the light is first turned on,assuming an initial filament temperature of 19 degrees C.
c) Find the power used when the light is first turned on,assuming an initial filament temperature of 19 degrees C.
d) estimate the time needed to heat the filament to 2000degrees C. The filament has a msss of .038g and a specific heat of0.032 cal/g-C
a) Find the current through the filament under theseconditions which are established very shortly after the light isturned on.
b)Find the filament current when the light is first turned on,assuming an initial filament temperature of 19 degrees C.
c) Find the power used when the light is first turned on,assuming an initial filament temperature of 19 degrees C.
d) estimate the time needed to heat the filament to 2000degrees C. The filament has a msss of .038g and a specific heat of0.032 cal/g-C

Explanation / Answer

      Current   I   =   Power/ voltage    a.   I2010   =   60/ 120                   =   0.5   A    b.   Resistance varies withtemperature as          RT2   =   RT1* {1 + *(T2   -   T1)}          T2   =   2010   0C          T1      =   19   0C    temperature coefficient of resistivity oftungsten      =   0.0045 / 0C       And   R2010   =   V/ I2010                R19   =   V/ I19       =>   V /I2010   =   V /I19 * { 1 + 0.0045 * (2010 - 19)}             1/0.5   =   1/I19* 9.959       Currentat   190C      I19   =   0.5* 9.959                                              =   4.98   A     c.     Power   P19   =  V* I192                               =   120* 4.982

                                       =   2975.75   W     d.   Heatrequired   H   =   m *c * T       also   H   =   P* t       =>   2975.75* t   =   0.038 * (0.032 *4.2) * (2000 - 19)       ( sp. heat is convertedinto J/g - K by multiplying by 4.2)          time   t   =   10.12/ 2975.75                            =   3.40* 10-3   s                               =   3.40   ms

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