A cylinderical brass boiler of radius 15 cm and thickness 1.0 cm is filled with water and placed on an elerctric heater. If the If the water boils at the rate of `200 g//s`, estimate the temperature of the heater filament. Thermal conductivity of `brass=109 J//s//m^@C` and heat of vapourization of water `=2.256xx10^3 J//g`.

Explanation

Rate of boiling of water `=200 g//s`. Since the heat of vapourization of water is `2.256xx10^3 J//g`, the amount of heat energy required to boil 200 g of water is
`2.256xx10^3J//gxx200g=4.512xx10^5J`
Since water is boiling at the rate of 200 g/s, the rate at which heat energy is supplied by the heater to water is
`(Q)/(t)=4.512xx10^5J//s`
Now, radius of the boiler
`r=15cm=0.15m`
`:.` Base area of the boiler
`A=pir^2=3.142xx(0.15)^2=0.0707m^2`
Thickness of brass
`d=1.0cm=1.0xx10^-2m`
Thermal conductivity of brass`k=109 J//s//m^(@)C`
Temperature of boiling water `T_W=100^@C`
If `T_f` is the temperature of the filament the, rate at which heat energy is transmitted through the base is given by
`(Q)/(t)=(kA(T_f-100))/(1.0xx10^-2)=4.512xx10^5`
or `T_f-100=585.5`
or `T_f=685.5^@C`