Q. 205.0( 7 Votes )

# A brass boiler ha

Answer :

Given,

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 × 10^{3} J kg^{–1}

The amount of heat flowing into water through the brass base of the boiler is given by

θ = KA(T_{1} - T_{2})t/l ....(1)

where, T_{1} is the temperature of the flame in contact with the boiler

T_{2} 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(T_{1} - T_{2})t/l

⇒ T_{1} - T_{2} = mLl/KAt

⇒ T_{1} - T_{2} = (6 kg × 2256 × 10^{3} J kg^{-1}× 0.01 m)/(109 J s^{–1} m^{–1} K^{–1}× 0.15 m^{2}× 60 s)

⇒ T_{1} - T_{2} = 137.98°C

⇒ T_{1} = 137.98°C + 100°C

⇒ T_{1} = 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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