Question 11.20:
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 ....(i)
Where,
T1 = Temperature of the flame in contact with the boiler
T2 = Boiling point of water = 100°C
Heat required for boiling the water:
θ = mL … (ii)
Equating equations (i) and (ii), we get:
∴ mL = KA(T1 - T2) t / l
T1 - T2 = mLl / KAt
= 6 X 2256 X 103 X 0.01 / (109 X 0.15 X 60)
= 137.98 o C
Therefore, the temperature of the part of the flame in contact with the boiler is 237.98°C.
A
brass boiler has a base area of 0.15 m2
and thickness 1.0 cm. It boils water
at the rate of 6.0 kg/min when placed on a gas stove. Estimate the
temperature of the part of the flame in contact with the boiler.
Thermal conductivity of brass = 109 J s –1
m–1 K–1;
Heat of vaporisation of water = 2256 × 103
J kg–1.
Solution:
Base
area of the boiler, A =
0.15 m2Thickness 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 ....(i)
Where,
T1 = Temperature of the flame in contact with the boiler
T2 = Boiling point of water = 100°C
Heat required for boiling the water:
θ = mL … (ii)
Equating equations (i) and (ii), we get:
∴ mL = KA(T1 - T2) t / l
T1 - T2 = mLl / KAt
= 6 X 2256 X 103 X 0.01 / (109 X 0.15 X 60)
= 137.98 o C
Therefore, the temperature of the part of the flame in contact with the boiler is 237.98°C.
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