Question

There is formation of layer of ice x cm thick on water, when the temperature of air is -0°C (less than freezing point). The thickness of layer increases from x to y in the time t, then the value of t is given by-

• A. $\frac{(x + y)(x - y)\rho L}{2K\theta}$
• B. $\frac{(x - y)\rho L}{2K\theta}$
• C. $\frac{(x + y)(x - y)\rho L}{K\theta}$
• D. $\frac{(x - y)\rho LK}{2\theta}$

Answer 1

$\frac{\rho L (y^2 - x^2)}{2K \theta}$

Answer 2

The rate of heat transfer through the ice layer of thickness $X$ and area $A$ with a temperature difference $\theta$ is given by $\frac{dQ}{dt} = \frac{KA\theta}{X}$. This heat is used to freeze water, so $dQ = L dm = L \rho A dX$. Equating these, we get $\frac{KA\theta}{X} = \frac{L \rho A dX}{dt}$. Rearranging gives $dt = \frac{L \rho X}{K \theta} dX$. Integrating from $t=0$ to $T$ and $X=x$ to $y$, we get $T = \int_{x}^{y} \frac{L \rho X}{K \theta} dX = \frac{L \rho}{K \theta} \left[ \frac{X^2}{2} \right]_{x}^{y} = \frac{L \rho (y^2 - x^2)}{2 K \theta}$. Option (A) is $\frac{(x + y)(x - y)\rho L}{2K\theta} = \frac{\rho L (x^2 - y^2)}{2K\theta}$, which is the negative of the correct answer. Assuming a typo in option (A) where it should be $\frac{(y + x)(y - x)\rho L}{2K\theta}$, it would be the correct answer.