Question

Ice has formed on a shallow pond and a steady state is reached where the temperature of air above the ic e is - 8 . 0 0 ° Cand the bottom of the pond has a temperature of 6.00° C. The depth of the ice and water combined is 1.80 m; what isthe depth of the ice alone? Use a thermal conductivity o f 1 . 6 7 W / m · K fo r ic e a nd 0 . 5 0 2 W / m · K fo r water.A.

Answer 1

Answer:

The depth of the ice = 1.47 m

Explanation:

Thermal conductivity of water, $K_{w} = 0.502 W/m.K$

Thermal conductivity of ice, $K_{i} = 1.67 W/m.k$

Temperature above ice, $T_{i} = -8^{0} C$

Temperature at the bottom of the pond, $T_{w} = 6^{0} C$

total depth of ice and water = 1.80 m

$d_{total} = d_{i} + d_{w} = 1.8 m\\ d_{w} = = 1.8 - d_{i}$

Let, transition between ice and water,  $T_{x} = 0$

At steady state;

$K_{w} A \frac{T_{w}-T_{x} }{d_{w} } = K_{i} A \frac{T_{x}-T_{i} }{d_{i} }$

$K_{w} A \frac{T_{w}-T_{x} }{1.8 -d_{i} } = K_{i} A \frac{T_{x}-T_{i} }{d_{i} }$

Inserting all the parameters

$0.502 A \frac{6-0 }{1.8 -d_{i} } = 1.67 A \frac{0+8 }{d_{i} }$

$0.301\frac{6}{1.8-d_{i} } = \frac{8}{d_{i} }$

$1.804 d_{i} = 8(1.8 - d_{i} )\\1.804 d_{i} = 14.4 - 8d_{i}\\1.804 d_{i} +8d_{i} = 14.4\\9.804 d_{i} = 14.4\\d_{i} = 14.4/9.804\\d_{i} = 1.47m$