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 `=200gm//s`. Since the heat of vaporisation of water is `2.256 xx 10^(3)j//gm`, the amount of heat energy required to boil 200 g of water is
`2.256 xx 10^(3)Jg^(-1) xx 200g = 4.512 xx 10^(5)J`
Since water is boiling at the rate of `200 gm//s`, the rate at which heat energy is supplied by the heater to water is
`(Q)/(t) = 4.512 xx 10^(5) js^(-1)` .. (4.9)
Now , Radius of the boiler
`(r) = 15 cm = 0.15 m`
Base area of the boiler
`(A)=pi r^(2) = 3.14 xx (0.15)^(2) = 0.0707 m^(2)`
Thickness of brass
`(d) = 1.0 cm == 1.0 xx 10^(-2) m`
Thermal conductivity of brass
`(k)=109 Js^(-1) m^(-1)``(2)C^(-1)`
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))-T_(w))/(d)` ... (4.10)
Substuting the values of `k,A,T_(w) ` and d in (4.10) and equating with (4.9) , we get
`(109 xx 0.0707(T_(f)-100))/(1.0 xx 10^(-2)) = 4.512 xx 10^(5)`
or `T_(f) - 100 = 585.5`
or `T_(f) = 685.5^(@)C`