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On a Winter Day When the Atmospheric Temperature Drops to −10°C, Ice Forms on the Surface of a Lake. (A) Calculate the Rate of Increase of Thickness - Physics

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Question

On a winter day when the atmospheric temperature drops to −10°C, ice forms on the surface of a lake. (a) Calculate the rate of increase of thickness of the ice when 10 cm of the ice is already formed. (b) Calculate the total time taken in forming 10 cm of ice. Assume that the temperature of the entire water reaches 0°C before the ice starts forming. Density of water = 1000 kg m−3, latent heat of fusion of ice = 3.36 × 105 J kg−1and thermal conductivity of ice = 1.7 W m−1°C−1. Neglect the expansion of water of freezing.

Solution

Thermal conductivity, K = 1.7 W/m°C

Density of water, ρω = 102 kg/m3

Latent heat of fusion of ice, Lice = 3.36 × 105 J/kg

Length, l = 10 × 10−2 m

(a) Rate of flow of heat is given by

`(DeltaQ)/(Deltat) = (( T_1 - T_2).KA)/l`

`l/Deltat = ((T_1 - T_2).KA)/(DeltaQ)`

`=(K.A (T_1 - T_2))/(m.l)`

`= (K.A(T_1- T_2))/( Al.pw.l)`

`= ((1.7)(0-10))/((10xx10^-2)xx10xx3.36xx10^5)`

`= 17/3.36 xx 10^-7`

`=5.059 xx 10^-7` m/s

= 5 × 10-7 m/s

(b) To form a thin ice layer of thickness dx, let the required be dt.
Mass of that thin layer, dm = A dx ρω
Heat absorbed by that thin layer, dQ = Ldm

`"dQ"/"dt"= (K.A (DeltaT))/ x`

`"ldm"/"dt" = (KA(DeltaT))/x`

`((Adxpw)L)/dt = (KA DeltaT)/x`
\[\int\limits_0^t\]   `dt = (rho_wL)/(K(DeltaT)` \[\int\limits_0 ^{0.1}\] `x  dx`

`⇒ t = (rho_wL)/ K(DeltaT) xx (0.1)^2/2`

`t =( 10^3xx3.36xx10^5xx0.01)/(1.7xx10xx2)`

`t = (3.36)/(2xx17) xx 10^6 sec`

`t=3.36/(2xx17)xx10^6 sec`

`t=3.36/(2xx17) xx 10^6/3600  hours`

`t = 27.45  hours`

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Solution On a Winter Day When the Atmospheric Temperature Drops to −10°C, Ice Forms on the Surface of a Lake. (A) Calculate the Rate of Increase of Thickness Concept: Heat Transfer - Conduction.
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