A lake is covered with ice 2 cm thick. The temperature of ambient air is `-15^(@)C`. Find the rate of thickening of ice. For ice `k=4xx10^(-4)k-cal-m^(-1)s^(_1)(.^(@)C)^(-1)`, density `=0.9xx10^(3)kg//m^(3)` and latent heat `L=80` kilo cal/kg.

Explanation

Heat energy flowing per sec is given by
`H=(dQ)/(dt)=KA(Delta theta)/x`……(i)
If dm is the mass of ice formed in time dt, then
`(dm)/(dt)=(Adxx rho)/(dt)=A.rho. (dx)/(dt)`
Since `H=((dm)/(dt))L`
`H=A rho(dx)/(dt)L`………ii
From eq. (i) and (ii)
`A rho L (dx)/(dt)=-kA(Delta theta)/x`
Rate of thickening ofice `=(dx)/(dt)`
`:.(dx)/(dt)=-(KA)/(rhoAL) (Delta theta)/x=(-K)/(rhoL) (Delta theta)/(rhoL x)=(4xx10^(-4))/(0.9xx10^(3)xx80)xX((0-(-15))/(2xx10^(2)))=4.166xx10^(-6)m//s`
`=1.45` cm /hour