A brass boiler has a base area of `0.15m^2` and thickness 1.0 cm it boils water at the rate of `6.0 kg//min`, When placed on a gas. Estimate the temperature of the part of the flame in contact with the boiler. Thermal conductivity of brass`=109 J//s-.^@C` ) and heat of vapourization of water `=2256 J//g`.

Explanation

Base area of the boiler, `A = 0.15 m^(2)`
Thickness of the boiler, l = 1.0 cm = 0.01 m
Boiling rate of water, R = 6.0 kg/min
Mass, m = 6 kg
Time, t = 1 min = 60 s
Thermal conductivity of brass, `K = 109 J s^( –1) m^(–1) K^(–1)`
Heat of vaporisation,` L = 2256 × 103 J kg^(–1)`
The amount of heat flowing into water through the brass base of the boiler is given by:
`theta=(KA(T_(1)-T_(2)t)/(t))`...(i)
Where,
`T_(1)` = Temperature of the flame in contact with the boiler
`T_(2)` = Boiling point of water `= 100^(@)C`
Heat required for boiling the water:
`theta = mL` … (ii)
Equating equations (i) and (ii), we get:
`:. mL=(KA(T_(1)-T_(2)t)/(t))`
`T_(1)-T_(2)=(mLl)/(KAt)`
`=(6xx2256xx10^(3)xx0.01)/(109xx0.15xx60)`
`=137.98^(@)C`
Therefore, the temperature of the part of the flame in contact with the boiler is `237.98^(@)C`.