An iron boiler is 1 cm thick and has a heating area `2.5 m^(2)`. The two surface of the boiler are at `230^(@)C` and `100^(@)C` respectively. If the latent heat of the steam is `540 kcal kg^(-1)` and thermal conductivity of iron is `1.6xx10^(-2)K cal s^(-1) m^(-1)K^(-1)`, then how much water will be evaporated into steam per minute ?

Explanation

Here, thickness of iron boiler, `Deltax=1`cm = 0.001 m
Heating area, `A=2.5 m^(2)`
Temperature `DeltaT= T_(1)-T_(2) = 230-100 =130^(@)`C
Latent heat of steam = 540 kcal`kg^(-1)`
Thermal conductivity of iron, K =`1.6 xx 10^(-2)` kcals^(-1)m^(-1)K^(-1)`
`Deltat`=1min = 60s
Let m kg of the mass be observed in 1 min, then
Heat conducted, `DeltaQ = (KA(T_(1)-T_(2))Deltat)/(Deltax) = mL`
`rArrm=(KA(T_(1)-T_(2)Deltat))/(DeltaxL)`
`=(1.6 xx 10^(-2)xx (2.5) xx 130 xx 60)/(0.01 xx 540)` = 57.78kg