91Ó°ÊÓ

a) Temperature of air above ice,$T_L=−5.0°\text{ C}$

b)Temperature at the bottom of the pond,$T_H=4.0°\text{ C}$

c) Depth of ice + water,$d=1.4\text{″¾}$

d) Thermal conductivity of ice,$k_{ice}=0.40\text{ }\frac{\text{cal}}{\text{m}â‹\dots °\text{C}â‹\dots \text{s}}$

e) Thermal conductivity of water,data-custom-editor="chemistry" $k_{water}=0.12\text{ }\frac{\text{cal}}{\text{m}â‹\dots °\text{C}â‹\dots \text{s}}$

The rate of heat transfer by conduction, a temperature gradient that exits perpendicular to the area through the unit cross-section area of a material is referred to as the thermal conductivity of a material. Considering that the heat transfer at the interface of ice and water is equal, we can get the required value of the thickness of the ice.

Formula:

The heat transferred by the body due to thermal conductivity of a material,

$Q=\frac{kA\mathrm{Δ}T}{L}$ …… (i)

Here, $k$is the thermal conductivity of the material, $A$is the area of the cross section, i$\mathrm{Δ}T$s the temperature difference between two surfaces of the material and$L$ is the thickness of the material.

The interface between the ice and the water is at$T_X=0.0°\text{C}$.

The heat transferred from bottom of the water to the interface of ice-water is equal to the heat transferred to the ice as the heat transfers from high to low temperature points.

Now, assuming the area through which heat transferred to be equal, the thickness of the ice can be calculated using equation (i) as follows:

$\begin{array}{l}\frac{k_{water}A(T_H−T_X)}{L−L_{ice}}=\frac{k_{ice}A(T_X−T_C)}{L_{ice}}\\ \frac{\left(0.12\right)A({4.0}^0−{0.0}^0)}{1.4−L_{ice}}=\frac{\left(0.40\right)A({0.0}^0+{5.0}^0)}{L_{ice}}\\ \frac{0.48}{1.4−L_{ice}}=\frac{2}{L_{ice}}\\ \frac{0.24}{1.4−L_{ice}}=\frac{1}{L_{ice}}\end{array}$

Solve further as:

$\begin{array}{l}0.24L_{ice}=1.4−L_{ice}\\ 1.24L_{ice}=1.4\\ L_{ice}=\frac{1.4}{1.24}\text{m}\\ L_{ice}=1.13\text{″¾}\end{array}$

Hence, the thickness of the ice is .$1.13\text{″¾}$