Find the time for which layer of ice 5 cm thick on the surface of a pond will increase its thickness by 0.1 cm when temperature of the surrounding air is `-20^(@) C` .

Explanation

Average thickness of ice block d = `d = ( 5 + 5.1)/(2) = 5.05` cm = `5.05 xx 10^(-2)` m
dx = `0.1 cm , theta_(2) = -20^(@) C , theta _(1) = 0^(@)C `
`K = 2.1 W m^(-1) K^(-1) , rho_("ice") = 900 kgm^(-3)`
`L = 3.36 xx 10^(5) J kg^(-1)` ,
We know that Q = `mL = (KA (theta_(1) - theta_(2))t)/(d)`
i.e., `Vrho L = (KA (theta_(1) - theta_(2))t)/(dx) therefore V = Ad`

We write `dt = (Adrho L dx)/(KA (theta_(1) - theta_(2)))`
Hence dt = `(5.05 xx 10^(-2) xx 10^(2) xx 3.36 xx 10^(5) xx 0.1 xx 10^(-2))/(2.1 xx (0 - (-20)))`
i.e. `dt = (5.05 xx 3.36 xx 10^(-2 + 3 +5 -1 = 2))/(2.1 xx 20)`
i.e. , `dt = 0.404 xx 10^(3)` s
i.e. dt = 404 s or 6 min 44 s .
It takes 6 min 44 s for ice layer to grow by a thickness 0.1 cm