A layer of ice of `0^(@)C` of thickness `x_(1)` is floating on a pond. If the atmospheric temperature is `-7^(@)C`. Show that the time taken for thickness of the layer of ice to increase from `x_(1)` to `x_(2)` is given by
`t=(pL)/(2kT)(x_(2)^(2)-x_(1)^(2))`
where p is the density of ice, k its thermal conductivity and L is the latent heat of fusion of ice.

Explanation

When the temperature of the air is less than `0^(@)C`, the cold air near the surface ofthe pond takes heat (latent) from the water which freezes in the forms of layers. Figure-4.16. Consequently, the thickness of the ice layer keeps increasing with time. Letx be the thickness of the ice layer at a certain time. If the thickness is increased by dx in time dt, then the amount of heat flowing through the slab in time dt is givenby(see figure-4.16)
`Q=(kA[0-(-T)]dt)/(x) = (kATdt)/(x)` .....(4.23)
Where A is the area of the layer of ice and `- T^(@)C` is the temperature of the surrounding air. If dm is the mass of water frozen into ice, then `Q= dm xx L`. But `dm=Apdx`, where p is the density of ice. Hence
`Q = ApLdx` ....(4.24)
Equating (4.23) and (4.24), we have
`(kATdt)/(x) = ApLdx or dt =(rho L)/(kT).xdx`
Integrating, we have
`int_(0)^(t) dt =(rhoL)/(kT)int_(x_(1))^(x_(2))xdx or t=(rhoL)/(kT)|(x^(2))/(2)|_(x_(1))^(x_(2))= (rhoL)/(2kT)(x_(2)^(2)-x_(1)^(2))`