A lake is freezing in a region where the steady outside air temperature is -θ^ C . The density of ice is and its specific latent heat of fusion is L . A graph is plotted showing the square of the ice thickness ( x^2 )

Question

Quantrex Academy · Accepted Answer

Correct answer: A. (m L)/(2θ)

The rate of heat conduction through the ice layer is equal to the rate of heat released during freezing.

\dfrac{KA\theta}{x} = \dfrac{dm}{dt} L

Since

dm = \rho A dx

, we have:

\dfrac{KA\theta}{x} = \rho A L \dfrac{dx}{dt}

\Rightarrow \dfrac{dx}{dt} = \dfrac{K\theta}{\rho L x}

Integrating both sides with respect to time:

\int x \, dx = \int \dfrac{K\theta}{\rho L} \, dt

\dfrac{x^2}{2} = \dfrac{K\theta}{\rho L} t

\Rightarrow x^2 = \left(\dfrac{2K\theta}{\rho L}\right) t

Comparing this with the equation of a straight line passing through the origin,

y = mt

, the slope

m

of the

x^2

versus

t

graph is:

m = \dfrac{2K\theta}{\rho L}

Rearranging for the thermal conductivity

K

, we get:

K = \dfrac{m\rho L}{2\theta}

Answer:

\dfrac{m\rho L}{2\theta}

A lake is freezing in a region where the steady outside air...

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