Saved Bookmarks
| 1. |
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/kgA. 1.5cm/hB. 0.75 cm/hC. 3 cm/hD. 2 cm/h |
|
Answer» Correct Answer - A 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`. |
|