Calculate the rate of increment of the thickness of ice layer on a lake when thickness of ice is 10 cm and the air temperature is -5^(@) C. If thermal conductivity of ice is 0.008 cal "cm"^(-1) "s"^(-1)°"C"^(-1), density of ice is 0.91 xx 10^(3) "kg m"^(-3) and latent heat is 79.8 cal "gm"^(-1). How long will it take the layer to become 10.1 cm ?

Calculate the rate of increment of the thickness of ice layer on a lak

1.	Calculate the rate of increment of the thickness of ice layer on a lake when thickness of ice is 10 cm and the air temperature is -5^(@) C. If thermal conductivity of ice is 0.008 cal "cm"^(-1) "s"^(-1)°"C"^(-1), density of ice is 0.91 xx 10^(3) "kg m"^(-3) and latent heat is 79.8 cal "gm"^(-1). How long will it take the layer to become 10.1 cm ?
Answer» Solution :The time taken for the layer of ice to increase its thickness from `x_1` to `x_2` is given by `int_(0)^(t) DT = (rho L)/( K THETA ) int_(x_1)^(x_2) X" "dx` `t = ( rho L)/( 2 K theta ) [ x_(2)^(2) - x_(1)^(2) ]` Given `theta = 5°C, rho = 0.91 "GM/c.c." L= 79.8 "cal gm"^(-1)` `x_1 = 10 "cm", x_2 = 10.1 "cm", K = 0.008 "cal cm"^(-1) s^(-1)°"C"^(-1)` Rate of increment of thickness`= (dx)/(dt) = (K ( theta_2 - theta_1) )/(x rho L) = (0.0008 xx 5)/( 10 xx 0.91 xx 79.8) = 5.5 xx 10^(-5) "cm/s"` `t = (0.91 xx 79.8)/( 2 xx 0.008 xx 5 ) [ 10.1^(2) - 10^(2) ] = 1824.527` S