Key Idea If area of cross-section of a surface is not uniform or if the steady state condition is not reached, the heat flow equation can be applied to a thin layer of material perpendicular to direction of heat flow. The rate of heat flow by conduction for growth of ice is given by, dtdθ=xKA(θ0−θ0)] where, dθ=ρAdxL,θ0=0 and θ1=−θ Given, θ0=0∘C,θ1=−26∘C The rate of increase of thickness can be calculated from Eq. dtdθ⇒dtρAdxL⇒dtdx=xKA(θ0−θ1)=xKA(θ0−θ1)=ρAxLKA(θ0−θ1)=ρ×LK[0−(−26)]=ρ×L26 K
A deep rectangular pond of surface area A, containing water (density =ρ, specific heat capacity =s ), is located in a region where the outside air…
NEET UG 2019 — Physics Thermodynamics
A deep rectangular pond of surface area A, containing water (density =ρ, specific heat capacity =s ), is located in a region where the outside air temperature is a steady value at the −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
Held on 30 Apr 2019 · Verified 9 Jul 2026.
Options
- A
26 K/ρr( L−4 s)
- B
26 K/(ρx2−L)
- C
26K/(ρxL)
- D
26K/ρr(L+4s)
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