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 (in $^oC$) 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 $=2256 \times 10^{3}\; J kg ^{-1}$
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 (in $^oC$) 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 $=2256 \times 10^{3}\; J kg ^{-1}$
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, $K=109 Js ^{-1} m ^{-1} K ^{-1}$
Heat of vaporisation, $L=2256 \times 10^{3} J kg ^{-1}$
The amount of heat flowing into water through the brass base of the boiler is given by
$\theta=\frac{K A\left(T_{1}-T_{2}\right) t}{l}\dots (i)$
Where,
$T_{1}=$ Temperature of the flame in contact with the boiler
$T_{2}=$ Boiling point of water $=100^{\circ} C$
Heat required for boiling the water
$\theta=m L \ldots(i i)$
Equating equations $(i)$ and $(i i),$ we get:
$\therefore m L=\frac{K A\left(T_{1}-T_{2}\right) t}{l}$
$T_{1}-T_{2}=\frac{m L l}{K A t}$
$=\frac{6 \times 2256 \times 10^{3} \times 0.01}{109 \times 0.15 \times 60}$
$=137.98^{\circ} C$
Therefore, the temperature of the part of the flame in contact with the boiler is $237.98\,^{\circ} C$
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