Answer :

Given,

Base area of the boiler, A = 0.15 m2


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


θ = KA(T1 - T2)t/l ....(1)


where, T1 is the temperature of the flame in contact with the boiler


T2 is the boiling point of water (= 100°C)


Heat required for boiling the water is given by


θ = mL … (2)


Equating equations (1) and (2), we get


mL = KA(T1 - T2)t/l


T1 - T2 = mLl/KAt


T1 - T2 = (6 kg × 2256 × 103 J kg-1× 0.01 m)/(109 J s–1 m–1 K–1× 0.15 m2× 60 s)


T1 - T2 = 137.98°C


T1 = 137.98°C + 100°C


T1 = 237.98°C


So, the temperature of the part of the flame in contact with the boiler is 237.98°C.


NOTE: Heat of vaporisation is the amount of heat supplied to a specific quantity of the substance to change its state from a liquid to a gas at constant pressure.


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