A lake is covered with ince 2 cm thick.t he 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

Options

A. 1.5cm/h
B. 0.75 cm/h
C. 3 cm/h
D. 2 cm/h

Answer

A

Explanation

Heat energy flowing per sec is given by
`H=(Q)/(t)=KA(Deltatheta)/(Deltax)` ..(i)
if dm mass of ice is increased ini time dt, then
`(dm)/(dt)=(Adxrho)/(dt)=A.rho.(dx)/(dt)`
Since, `H=((dm)/(dm))L`
`thereforeH=Arho(dx)/(dt)L` ..(ii)
From eq. (i) and (ii)
`ArhoL(dx)/(dt)=KA(Deltatheta)/(Deltax)`
Rate of thickening of ice `=dx//dt`
`therefore(dx)/(dt)=(KA)/(rhoAL)(Deltatheta)/(Deltax)=(K)/(rhoL)(Deltatheta)/(Deltax)=(4xx10^(-4))/(0.9xx10^(3)xx80)xx[(0-(-15))/(2xx10^(-2))]`
`=4.166xx10^(-6)m//s=1.5cm//hr`.