Question

A deep rectangular pond of surface area A
containing water (density = p), 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. [NEET-2019 (Odisha)]
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
(1) 26K/px(L + 4s) (2) 26K/px(L - 4s)
(3) 26K/(pxL)
(4) 26K (PxL)​

Answer 1

Answer:

26K/rho x L

Explanation:dQ/dt={KA(T1-T2)}/L

dQ={KA(theta)/x}dt

mL=(KA×26×dt)/x

{m=rho × v

=Rho × dxA}

Rho ×dxA L=(26KAdt)/x

dx/dt=26K/rho × x×L