91Ó°ÊÓ

i)The thickness of slab, $L=5.0\text{cmor0.05m}$

ii) The temperature of the air above the ice,$T_C=−10°\text{C}$

iii) The thermal conductivity of ice,
$\begin{array}{c}k=0.0040\frac{\text{cal}}{\text{s}}×4.186\left(\frac{\text{J}}{\text{cal}}\right)×(1×{10}^{−2}\frac{\text{m}}{\text{cm}})\\ =1.674\frac{\text{W}}{\text{m.K}}\end{array}$

iv) The density of ice,$\mathrm{ÒÏ}=0.92\frac{\text{g}}{{\text{cm}}^{\text{3}}}\text{or}0.92×{10}^3\frac{\text{kg}}{{\text{m}}^{\text{3}}}$

We can use the concept of conduction through the slab.The heat transfer from a hotter body to a colder body is defined as the process of conduction and this heat transfer is known as thermal conduction. Here, a certain amount of heat is released by the body during the phase change process.The heat transformation involved in the phase change from liquid to solid is called heat of fusion. We are going to find the rate of ice formation in cm/h which is dL /dt.

Formulae:

The conduction rate at which heat energy is transferred by a body,

$P_{cond}=\frac{kA(T_H−T_C)}{L}$ …(¾±)

where,$k$is the thermal conductivity of the material,$A$is the surface area of radiation,$T_H$is the temperature at the hotter end,$T_L$is the temperature at the colder end,$L$is the length of the conduction.

The heat energy released by the body, $Q=L_{f}m$ …(¾±¾±)

Where, $L_f$is the latent heat of fusion, is the mass of the substance.

We know that,

$T_H$is the temperature of water,$T_H=0°\text{C}$

$T_C$is the temperature of the air above the ice,$T_C=−10°\text{C}$

The heat of fusion for water,$L_f=333×{10}^3\frac{\text{J}}{\text{kg}}$

The water leaves the heat and freezes it. There is phase change from liquid to solid, hence heat of fusion is given using equation (ii).

Differentiating equation (ii) with respect to the time, we get the rate of conduction that is

$P_{cond}=L_f\frac{dm}{dt}\text{(}âˆ\mu \frac{dQ}{dt}=P_{cond}\text{)}$ …(¾±¾±¾±)

Again, we know that the equation of mass in terms of density and volume is given as:

$\begin{array}{c}m=\mathrm{ÒÏ}V\\ =\mathrm{ÒÏ}AL\text{(}âˆ\mu \text{Volume}=\text{Area}×\text{Length)}\end{array}$ …(¾±±¹)

Substituting equation (iv) in equation (iii), it becomes:

$\begin{array}{c}{P}_{cond}=L_f\frac{d\left(\mathrm{ÒÏ}AL\right)}{dt}\\ =\mathrm{ÒÏ}A{L}_f\frac{d\left(L\right)}{dt}\end{array}$ …(±¹)

By equating equation (iii) with equation (v) and using the given values, we can get the rate of formation of ice as follows:

$\begin{array}{c}\mathrm{ÒÏ}A{L}_f\frac{d\left(L\right)}{dt}=\frac{kA(T_H−T_C)}{L}\\ \frac{dL}{dt}=\frac{kA(T_H−T_C)}{\mathrm{ÒÏ}A{L}_{f}L}\\ =\frac{k(T_H−T_C)}{\mathrm{ÒÏ}{L}_{f}L}\\ =\frac{(1.674\frac{\text{W}}{\text{m}}\text{.K})(0°\text{C}+10°\text{C})}{0.92×{10}^3\frac{\text{kg}}{{\text{m}}^{\text{3}}}×333×{10}^3\frac{\text{J}}{\text{kg}}×0.050\text{m}}\\ =1.1×{10}^{−6}\text{m/s}\\ =0.40\text{cm/h}\end{array}$

Hence, the required rate of the formation is $0.40\text{cm/h}$