A brass boiler has a base area of 0.15; m^{2} | vedclass

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(B) Thickness of the boiler,$l = 1.0\; cm = 0.01\; m$.Boiling rate of water,$R = 6.0\; kg/min$.Mass,$m = 6.0\; kg$.Time,$t = 1\; min = 60\; s$.Thermal

Vedclass · Accepted Answer

$238$

Vedclass · Answer

(B) Thickness of the boiler,$l = 1.0\; cm = 0.01\; m$.Boiling rate of water,$R = 6.0\; kg/min$.Mass,$m = 6.0\; kg$.Time,$t = 1\; min = 60\; s$.Thermal conductivity of brass,$K = 109\; J s^{-1} m^{-1} K^{-1}$.Heat of vaporisation,$L = 2256 	imes 10^{3}\; J kg^{-1}$.The amount of heat flowing into water through the brass base of the boiler is given by:$Q = \frac{K A (T_{1} - T_{2}) t}{l} \dots (i)$Where $T_{1}$ is the temperature of the flame and $T_{2} = 100^{\circ}C$ is the boiling point of water.Heat required for boiling the water is $Q = m L \dots (ii)$.Equating $(i)$ and $(ii)$:$m L = \frac{K A (T_{1} - T_{2}) t}{l}$$T_{1} - T_{2} = \frac{m L l}{K A t} = \frac{6.0 	imes 2256 	imes 10^{3} 	imes 0.01}{109 	imes 0.15 	imes 60} \approx 137.98^{\circ}C$.Thus,$T_{1} = 137.98 + 100 = 237.98^{\circ}C \approx 238^{\circ}C$.

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