Thermal Properties of Matter - Result Question 10
10. A deep rectangular pond of surface area A, containing water $($ density $=\rho)$, specific heat $($ capacity $=s)$, is located in a region where the outside air temperature is at a steady value of $-26^{\circ} C$. 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
[NEET Odisha 2019]
(a) $26 K / \rho x(L+4 s)$
(b) $26 K / \rho x(L-4 s)$
(c) $26 K /(\rho x^{2} L)$
(d) $26 K /(\rho x L)$
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Answer:
Correct Answer: 10. (d)
Solution:
- (d) We assume that at any instant thickness of ice is $x$. And time taken to form additional thickness $(d x)$ is $d t$.
$ \begin{aligned} & \text{ 1010101010101010101 } _{-2}{ }^{x} \\ & mL=\frac{KA[26-0] d t}{x} \\ & \Rightarrow(A d x) \rho L=\frac{KA(26) d t}{x} \\ & \text{ So, } \frac{d x}{d t}=\frac{26 K}{x \rho L} \end{aligned} $
