A brass boiler has a base area of 0.15 m^{2} 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 × 10^{3} J kg^{–1}.

#### Solution 1

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:

`theta = (KA(T_1-T_2)t)/l` .....(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:

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`