Question

A deep rectangular pond of surface are A, containing water (density = r, specific heat capacity = s), is located in a region where the outside air temperature is at a steady value of -26°C. The thickness of the frozen ice layer in this pond, at a certain instant is x. Taking the thermal conductivity of ice as K, and its specific latent heat of fusion as L, the rate of increase of the thickness of ice layer, at this instant would be given by :-

• A. \(\frac {26K}{\rho r\left(L-4s\right)}\)
• B. \(\frac {6K}{\left(\rho x^{2}-L\right)}\)
• C. \(\frac {26K}{\left(\rho xL\right)}\)
• D. \(\frac {26K}{\rho r\left(L+4s\right)}\)

Answer 1

The Correct Option is C

\(KA \frac{[0-(-26)]}{x} dt\)\(= A(dx) \rho L\)

\(\Rightarrow \frac{dx}{dt} = \frac{26K}{\rho K x}\)

So, the correct option is (C): \(\frac {26K}{\left(\rho xL\right)}\).

Answer 2

The rate of increase in the thickness of the ice layer can be calculated using the concept of heat transfer through the ice layer. Heat transfer occurs due to the temperature difference between the water in the pond and the surrounding air.

The rate of heat transfer per unit area (q) through the ice layer can be calculated using Fourier's law of heat conduction, which states that the rate of heat transfer is proportional to the temperature gradient and the thermal conductivity of the material.

\(q = K \times \frac{dT}{dx}\)

where K is the thermal conductivity of ice, \(\frac{dT}{dx}\) is the temperature gradient across the ice layer, and x is the thickness of the ice layer.

The rate of increase of the thickness of the ice layer (dx/dt) can be calculated using the heat transfer equation as:

\(\frac{dx}{dt} = \frac{L}{rAs} \times q\)

where L is the specific latent heat of fusion of ice, r is the density of water, A is the surface area of the pond, and s is the specific heat capacity of water.

Substituting the value of q from Fourier's law of heat conduction, we get:
\(\frac{dx}{dt} = \frac{L}{rAs} \times K \times \frac{dT}{dx}\)

Since the air temperature outside the pond is at a steady value of -26°C, the temperature gradient across the ice layer can be assumed to be constant, and equal to the temperature difference between the water and the air, which is:
\(\frac{dT}{dx} = \frac{0°C - (-26°C)}{x} = \frac{26°C}{x}\)

Substituting this value in the above equation, we get:
\(\frac{dx}{dt} = \frac{LK}{rAs} \times \frac{26°C}{x}\)

Therefore, the rate of increase of the thickness of the ice layer at a certain instant is given by:
\(\frac{dx}{dt} = \frac{LK}{rAs} \times \frac{26°C}{x}\)

Answer 3

Given:
Surface Area = A,
Density = ρ,
Specific heat capacity = s,
Thermal conductivity of ic = K,
Specific latent heat of fusion = L
Assuming Volume to be V.
The rate of increase in the thickness of the ice layer can be calculated using the concept of heat transfer through the ice layer. Heat transfer occurs due to the temperature difference between the water in the pond and the surrounding air.
We assume that at any instant the thickness of ice is (x) and time taken to form an additional thickness (dx) is (dt).

Let the mass formed by the additional thickness of ice be m.
The heat energy released by forming the additional thickness of ice (m) be (dq).
Therefore, \(dq = mL = ρ\times (A \times dx) \times L\)
We know, \(r = \frac mv\)
Therefore, \(m = ρ\times v\)
\(m= ρ \times (A \times dx)\)
Heat current (i_H) flows due to (m) which is calculated using Fourier's law of heat conduction, which states that the rate of heat transfer is proportional to the temperature gradient and the thermal conductivity of the material.
\(\frac {dq}{dt}=\frac {KA(0+26)}{x}\)

\(\frac {ρ(Adx)L}{dt}=\frac {KA26}{x}\)

\(\frac {dx}{dt}=\frac {26K}{ρxL}\)
Therefore, the rate of increase of the thickness of the ice layer, at this instant would be given by \(\frac {26K}{ρxL}\).

So, the correct option is (C): \(\frac {26K}{ρxL}\).