A layer of ice 29 cm thick has formed on a pond. The temperature of air is `-10^(@)C`. Find how long will it take for another 0.1 cm layer of water to freeze? Conductivity of ice = `2.1 J s^(-1)m^(-1)K^(-1)`. Latent heat capacity of ice `=3.36 xx 10^(5) J kg^(-1)` and density of ice =`1000 kg m^(-3)`.

Explanation

Here, initial thicknes of ice layer = 20 cm
Temp. of one side of layer = `10^(@)C` ,
Temp. on other side of layer `= 0^(@)C`,
`:. Delta T = 0-(-10)=10^(@)C`
time , `Delta t = ?`
`K = 2.1 Js^(-1) m^(-1) K^(-1)`
`L = 3.36 xx 10^(5) J kg^(-1) , rho = 10^(3)kgm^(-3)`
When another 0.1 cm of water layer is frozen, final thickness of ice layer = `20 + 0.1 = 20.1 cm`
`:.` Average thickness through which heat is conducted `= (20+20.1)/(2) = 20.05 cm`
i.e., `Delta x = 20.05 xx 10^(-2)m`
Let `A m^(2)` be the surface area of fresh ice layer formed. as thickness of fresh ice layer `=0.1 cm = 10^(-3) m`.
`:.` Volume of water frozen =`A xx 10^(-3)m^(3)`
Mass of ice formed = `volume xx density`
`m= A xx 10^(-3) xx 10^(3) = A kg`
Amount of heat to be conducted
`Delta Q = m xx L = A xx 3.36 xx 10^(5) J`
From , `(Delta Q)/(Delta T) =KA (Delta T)/(Delta x)`
`Delta T = (3.36 xx 10^(5) xx 20.05 xx 10^(-2))/(2.1 xx 10) = 3208s`.