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NEET-XII-Physics

28: Heat Transfer

with Solutions - page 4
Qstn# iv-14 Prvs-QstnNext-Qstn
  • #14
    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-1 and thermal conductivity of ice = 1.7 W m-1°C-1. Neglect the expansion of water of freezing. (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-1 and thermal conductivity of ice = 1.7 W m-1°C-1. Neglect the expansion of water of freezing.
    Ans : 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
    `` \frac{\,\mathrm{\,\Delta Q\,}}{\,\mathrm{\,\Delta \,}t}=\frac{\left({\,\mathrm{\,T\,}}_{1}-{\,\mathrm{\,T\,}}_{2}\right)·\,\mathrm{\,KA\,}}{l}``
    `` \Rightarrow \frac{l}{\,\mathrm{\,\Delta \,}t}=\frac{\left({T}_{1}-{T}_{2}\right)\mathit{·}KA}{\,\mathrm{\,\Delta Q\,}}``
    `` =\frac{\,\mathrm{\,K\,}·\,\mathrm{\,A\,}\left({\,\mathrm{\,T\,}}_{1}-{\,\mathrm{\,T\,}}_{2}\right)}{\,\mathrm{\,m\,}·\,\mathrm{\,L\,}}``
    `` =\frac{\,\mathrm{\,KA\,}\left({\,\mathrm{\,T\,}}_{1}-{\,\mathrm{\,T\,}}_{2}\right)}{\left(\,\mathrm{\,A\,}l·{\,\mathrm{\,\rho \,}}_{\,\mathrm{\,\omega \,}}\right)\,\mathrm{\,L\,}}``
    `` =\frac{\left(1.7\right)\left(0-10\right)}{\left(10\times {10}^{-2}\right)\times {10}^{3}\times 3.36\times {10}^{5}}``
    `` =\frac{17}{3.36}\times {10}^{-7}``
    `` =5.059\times {10}^{-7}\,\mathrm{\,m\,}/\,\mathrm{\,s\,}``
    `` =5\times {10}^{-7}\,\mathrm{\,m\,}/\,\mathrm{\,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
    `` \frac{d\,\mathrm{\,Q\,}}{dt}=\frac{K·A\left(∆T\right)}{x}``
    `` \frac{L\,\mathrm{\,d\,}m}{\,\mathrm{\,d\,}t}=\frac{KA\left(∆T\right)}{x}``
    `` \frac{\left(\,\mathrm{\,A\,}dx{\,\mathrm{\,\rho \,}}_{w}\right)L}{\,\mathrm{\,d\,}t}=\frac{KA\mathit{∆}T}{x}``
    `` \Rightarrow \underset{0}{\overset{t}{\int }}\,\mathrm{\,d\,}t=\frac{{\rho }_{w}L}{K\left(\Delta T\right)}\underset{0}{\overset{0.1}{\int }}xdx``
    `` \Rightarrow t=\frac{{\rho }_{w}L}{K\left(\Delta T\right)}{\left[\frac{{x}^{2}}{2}\right]}_{0}^{0.1}``
    `` t=\frac{{\rho }_{w}L}{K\left(\Delta T\right)}\times \left(\frac{{\left(0.1\right)}^{2}}{2}\right)``
    `` t=\frac{{10}^{3}\times 3.36\times {10}^{5}\times 0.01}{1.7\times 10\times 2}``
    `` t=\frac{3.36}{2\times 17}\times {10}^{6}\,\mathrm{\,sec\,}``
    `` t=\frac{3.36}{2\times 17}\times \frac{{10}^{6}}{3600}\,\mathrm{\,hours\,}``
    `` t=27.45\,\mathrm{\,hours\,}``
    Page No 99: (a) Rate of flow of heat is given by
    `` \frac{\,\mathrm{\,\Delta Q\,}}{\,\mathrm{\,\Delta \,}t}=\frac{\left({\,\mathrm{\,T\,}}_{1}-{\,\mathrm{\,T\,}}_{2}\right)·\,\mathrm{\,KA\,}}{l}``
    `` \Rightarrow \frac{l}{\,\mathrm{\,\Delta \,}t}=\frac{\left({T}_{1}-{T}_{2}\right)\mathit{·}KA}{\,\mathrm{\,\Delta Q\,}}``
    `` =\frac{\,\mathrm{\,K\,}·\,\mathrm{\,A\,}\left({\,\mathrm{\,T\,}}_{1}-{\,\mathrm{\,T\,}}_{2}\right)}{\,\mathrm{\,m\,}·\,\mathrm{\,L\,}}``
    `` =\frac{\,\mathrm{\,KA\,}\left({\,\mathrm{\,T\,}}_{1}-{\,\mathrm{\,T\,}}_{2}\right)}{\left(\,\mathrm{\,A\,}l·{\,\mathrm{\,\rho \,}}_{\,\mathrm{\,\omega \,}}\right)\,\mathrm{\,L\,}}``
    `` =\frac{\left(1.7\right)\left(0-10\right)}{\left(10\times {10}^{-2}\right)\times {10}^{3}\times 3.36\times {10}^{5}}``
    `` =\frac{17}{3.36}\times {10}^{-7}``
    `` =5.059\times {10}^{-7}\,\mathrm{\,m\,}/\,\mathrm{\,s\,}``
    `` =5\times {10}^{-7}\,\mathrm{\,m\,}/\,\mathrm{\,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
    `` \frac{d\,\mathrm{\,Q\,}}{dt}=\frac{K·A\left(∆T\right)}{x}``
    `` \frac{L\,\mathrm{\,d\,}m}{\,\mathrm{\,d\,}t}=\frac{KA\left(∆T\right)}{x}``
    `` \frac{\left(\,\mathrm{\,A\,}dx{\,\mathrm{\,\rho \,}}_{w}\right)L}{\,\mathrm{\,d\,}t}=\frac{KA\mathit{∆}T}{x}``
    `` \Rightarrow \underset{0}{\overset{t}{\int }}\,\mathrm{\,d\,}t=\frac{{\rho }_{w}L}{K\left(\Delta T\right)}\underset{0}{\overset{0.1}{\int }}xdx``
    `` \Rightarrow t=\frac{{\rho }_{w}L}{K\left(\Delta T\right)}{\left[\frac{{x}^{2}}{2}\right]}_{0}^{0.1}``
    `` t=\frac{{\rho }_{w}L}{K\left(\Delta T\right)}\times \left(\frac{{\left(0.1\right)}^{2}}{2}\right)``
    `` t=\frac{{10}^{3}\times 3.36\times {10}^{5}\times 0.01}{1.7\times 10\times 2}``
    `` t=\frac{3.36}{2\times 17}\times {10}^{6}\,\mathrm{\,sec\,}``
    `` t=\frac{3.36}{2\times 17}\times \frac{{10}^{6}}{3600}\,\mathrm{\,hours\,}``
    `` t=27.45\,\mathrm{\,hours\,}``
    Page No 99:
  • #14-a
    Calculate the rate of increase of thickness of the ice when 10 cm of the ice is already formed.
    Ans : Rate of flow of heat is given by
    `` \frac{\,\mathrm{\,\Delta Q\,}}{\,\mathrm{\,\Delta \,}t}=\frac{\left({\,\mathrm{\,T\,}}_{1}-{\,\mathrm{\,T\,}}_{2}\right)·\,\mathrm{\,KA\,}}{l}``
    `` \Rightarrow \frac{l}{\,\mathrm{\,\Delta \,}t}=\frac{\left({T}_{1}-{T}_{2}\right)\mathit{·}KA}{\,\mathrm{\,\Delta Q\,}}``
    `` =\frac{\,\mathrm{\,K\,}·\,\mathrm{\,A\,}\left({\,\mathrm{\,T\,}}_{1}-{\,\mathrm{\,T\,}}_{2}\right)}{\,\mathrm{\,m\,}·\,\mathrm{\,L\,}}``
    `` =\frac{\,\mathrm{\,KA\,}\left({\,\mathrm{\,T\,}}_{1}-{\,\mathrm{\,T\,}}_{2}\right)}{\left(\,\mathrm{\,A\,}l·{\,\mathrm{\,\rho \,}}_{\,\mathrm{\,\omega \,}}\right)\,\mathrm{\,L\,}}``
    `` =\frac{\left(1.7\right)\left(0-10\right)}{\left(10\times {10}^{-2}\right)\times {10}^{3}\times 3.36\times {10}^{5}}``
    `` =\frac{17}{3.36}\times {10}^{-7}``
    `` =5.059\times {10}^{-7}\,\mathrm{\,m\,}/\,\mathrm{\,s\,}``
    `` =5\times {10}^{-7}\,\mathrm{\,m\,}/\,\mathrm{\,s\,}``
  • #14-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-1 and thermal conductivity of ice = 1.7 W m-1°C-1. Neglect the expansion of water of freezing.
    Ans : 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
    `` \frac{d\,\mathrm{\,Q\,}}{dt}=\frac{K·A\left(∆T\right)}{x}``
    `` \frac{L\,\mathrm{\,d\,}m}{\,\mathrm{\,d\,}t}=\frac{KA\left(∆T\right)}{x}``
    `` \frac{\left(\,\mathrm{\,A\,}dx{\,\mathrm{\,\rho \,}}_{w}\right)L}{\,\mathrm{\,d\,}t}=\frac{KA\mathit{∆}T}{x}``
    `` \Rightarrow \underset{0}{\overset{t}{\int }}\,\mathrm{\,d\,}t=\frac{{\rho }_{w}L}{K\left(\Delta T\right)}\underset{0}{\overset{0.1}{\int }}xdx``
    `` \Rightarrow t=\frac{{\rho }_{w}L}{K\left(\Delta T\right)}{\left[\frac{{x}^{2}}{2}\right]}_{0}^{0.1}``
    `` t=\frac{{\rho }_{w}L}{K\left(\Delta T\right)}\times \left(\frac{{\left(0.1\right)}^{2}}{2}\right)``
    `` t=\frac{{10}^{3}\times 3.36\times {10}^{5}\times 0.01}{1.7\times 10\times 2}``
    `` t=\frac{3.36}{2\times 17}\times {10}^{6}\,\mathrm{\,sec\,}``
    `` t=\frac{3.36}{2\times 17}\times \frac{{10}^{6}}{3600}\,\mathrm{\,hours\,}``
    `` t=27.45\,\mathrm{\,hours\,}``
    Page No 99: