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

To solve the problem of determining the rate of increase of the thickness of the ice layer in the pond, we need to employ concepts related to thermal conduction and phase change. Here’s a step-by-step explanation:

1. First, understand the conditions:
   • The outside air temperature is -26°C.
   • The thermal conductivity of ice is \(K\).
   • The specific latent heat of fusion of ice is \(L\).
   • The density of water is \(\rho\), and its specific heat capacity is \(s\).
   • The current thickness of the ice layer is \(x\).
2. We need to determine the rate of change in the thickness of the ice layer. For a given small time interval, the layer of water turning to ice must lose heat as given by the rate of heat conduction through the existing ice.
3. The rate of heat conduction through the ice is described by Fourier’s law of heat conduction: q = \frac{K \cdot A \cdot (T_i - T_o)}{x} where:
   • \(A\) is the surface area of the ice.
   • \(T_i\) is the temperature at the bottom of the ice, approximately 0°C, where water turns to ice.
   • \(T_o\) is the outside air temperature, -26°C.
   • \(x\) is the current thickness of the ice.
4. The rate at which the thickness of ice increases, \(\frac{dx}{dt}\), is equated to the rate of heat conduction divided by the latent heat required to freeze a layer of water: \rho A L \frac{dx}{dt} = \frac{26KA}{x}
5. Solving for \(\frac{dx}{dt}\), we get: \frac{dx}{dt} = \frac{26K}{\rho xL}
6. Thus, the correct expression for the rate of increase of the ice thickness is \(\frac{26K}{\rho xL}\).

This solution matches the correct option provided in the question: \(\frac{26K}{\rho xL}\).

Conclusion: The reasoning here uses the balance between the rate of heat conducted through the ice and the heat required to freeze additional water into ice, integrating thermal and physical properties of the substances involved.