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Question
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. The thermal conductivity of brass = 109 J s –1 m–1 K–1; Heat of vaporisation of water = 2256 × 103 J kg–1.
Solution 1
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:
`theta = (KA(T_1-T_2)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(T_1-T_2)t)/l`
`T_1 - T_2 = (mLl)/(KAt)`
`= (6xx2256xx10^3xx0.01)/(109xx0.15xx60)`
`= 137.98 ""^@C`
Therefore, the temperature of the part of the flame in contact with the boiler is 237.98°C.
Solution 2
Here `K = 109 Js^(-1) m^(-1) K^(-1)`
`A = 0.15 m^2`
`d = 1.0 cm = 10^(-2) m`
`T_2 = 100^@ C`
Let `T_1` = temperature of the part of the boiler in contact with the stove.
if Q be the amount of heat flowing per second throught the base of the boiler, then
`Q = (109 xx 0.15 xx (T_1 - 100))/10^(-2)` ..(i)
`= 1635 (T_1 - 100 )Js^(-1)`
Also heat of vaporisation of water `L = 2256 xx 10^3 J kg^(-1)`
Rate of boiling of water in the boiler, M = 6.0 kg` min^(-1) = 6.0/60 = 0.1 kg^(-1) s`
:. Heat received by water per second, Q = ML
`=> Q = 0.1 xx 2256 xx 10^3 Js^(-1)` ... ii
:. From equation i and ii we get
`1635 (T_1 - 100) = 2256 xx 10^2`
or `T_1 - 100 = (2256xx10^2)/1653 = 138`
`T_1 = 138 + 100 = 238^@C`
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