Water is boiled in a container having a bottom of surface area `25cm^(2)` , thickness `1.0mm and thermal conductivity `50Wm^(-1)`^(@)C^(-1)` , 100g of water is converted into steam per minute in the steady state after the boiling starts.Assuming that no heat is lost to the atmosphere, calculate the temperature of the lower surface of the bottom. Latent heat of vaporrization of water `=2.26xx10^(6)Jkg^(-1)` .

Answer

A, C

Explanation

`A=25cm^(2)=25 XX10^_(-4)m^(2)`
`L=1 mm =10^(-3)m,K=50W//m^(@)C.`
(Q)/(T)=rate of conversion of water in to steam
`Q/T=(100xx10^(-3)xx2.26xx10^(6))/(1 min)`
`=(100xx10^(3)xx2.26)/(1 min)`
`2.26/6 xx10^(4)`
`=0.376xx10^(4) J//s`
`(Q)/(T)=(kA(T_1-T_2))/(l)`
`implies 0.376xx10^(4)=(50xx25xx10^(-4)xx(T-100))/(10^(-3))`
`implies(T-100)=(10^(-3)xx0.376xx10^(4))/(50xx25xx10^(-4))`
`=(10^(-3)xx0.376)/(50xx25)`
`implies(T-100)=3.008xx10^(-4)xx10xx(-3)xx10^(8)`
`implies(T-100)=3.008xx10`
`=30^(@)C=30`
`:. T=100+30=130^(@)C.`