Question

In a 10 m deep lake, the bottom is at a constant temperature of ${4^\circ C}$. The air temperature is constant at ${-4^\circ C}$. The thermal conductivity of ice is 3 times that of water. Neglecting the expansion of water on freezing, the maximum thickness of ice will be,
(a) 7.5 m
(b) 6 m
(c) 5 m
(d) 2.5 m

Answer 1

Heat is a form of energy which produces a change in the temperature of any substance. Heat produced is given as $\dfrac{{\text{Q}}}{{\text{t}}}{\text{ = }}\dfrac{{{{KA\Delta T}}}}{{\text{d}}}$
Where K = Thermal conductivity of medium,
ΔT = The difference in temperature between one side of the material and the other
t = Time required to transfer amount of heat
A = The cross sectional area of the material transferring heat
d = The thickness of the material

Complete Step by step answer: Given: Thermal conductivity of ice Kice is $3$ times that of Thermal conductivity of water Kwater.
As it is mention that bottom of the lake is maintained at temperature of $4^\circ C$ and upper portion is at ${-4^\circ C}$ and depth of lake is 10 m, So let us assume an intermediate temperature of $0^\circ C$ will lie at distance x from top end so it's distance from bottom end will be 10–x.
Heat is a form of energy which produces a change in the temperature of any substance
Now we know that heat produced is given as
${\text{Q = }}\dfrac{{{{KA\Delta TL}}}}{{\text{L}}}$
Where K = Thermal conductivity of medium,
ΔT = The difference in temperature between one side of the material and the other
t = Time required to transfer amount of heat
A = The cross sectional area of the material transferring heat
d = The thickness of the material
Now for(water) bottom end
${{\text{Q}}_{\text{1}}}{\text{ = }}\dfrac{{{\text{A(1 - 0)}}}}{{10 - x}}$ ………… (1)

And for(ice) top end -
${{\text{Q}}_{\text{2}}}{\text{ = }}\dfrac{{{\text{3A\\{ 0 - ( - 4)\\} }}}}{{\text{x}}}$ .…… (2)
it is given that Kice = 3 × Kwater
so equating (1) & (2) we get,
4x = 120 – 12x
$\Rightarrow$ 4x + 12x = 120
$\Rightarrow$ X = 7.5
So, the maximum thickness of ice will be: 7.5 m.

Hence option (A) is a correct answer.

Note: Materials with a high thermal conductivity constant k (like metals and stones) will conduct heat well both ways; into or out of the material. Here also we have to check which temperature is positive and which is negative and the temperatures for water and ice.