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Heat Transfer. — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsHeat Transfer.5 Marks
Question
An icebox almost completely filled with ice at $0^\circ C$ is dipped into a large volume of water at $20^\circ C$. The box has walls of surface area $2400cm^2$, thickness $2.0mm$ and thermal conductivity $0.06\text{Wm}^{-1}{^{\circ}}\text{C}^{-1}.$ Calculate the rate at which the ice melts in the box. Latent heat of fusion of ice $= 3.4\times10^5\text{Jkg}^{-1}.$
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Answer

$\text{A}=2400\text{cm}^2=2400\times10^{-4}\text{m}^2$$\ell=2\text{mm}=2\times10^{-3}\text{m}$
$\text{K}=0.06\text{w/m-}^\circ\text{C}$
$\theta_1=20^\circ\text{C}$
$\theta_2=0^\circ\text{C}$
$\frac{\text{Q}}{\text{t}}=\frac{\text{KA}(\theta_1-\theta_2)}{\ell}$
$=\frac{0.06\times2400\times10^{-4}\times20}{2\times10^{-3}}$
$=24\times6\times10^{-1}\times10$
$=24\times6=144\text{J/sec}.$
Rate in which ice melts $=\frac{\text{mL}_\text{f}}{\text{t}}$
$\Rightarrow\frac{\Delta\text{Q}}{\Delta\text{t}}=\Big(\frac{\text{m}}{\text{t}}\Big)\text{L}_\text{f}$
$\Rightarrow144=\Big(\frac{\text{m}}{\text{t}}\Big)\times3.4\times10^5$
$\Rightarrow\frac{\text{m}}{\text{t}}=\frac{144}{3.4\times10^{5}}\text{kg/s}$
$\Rightarrow\frac{\text{m}}{\text{t}}=\frac{144\times60\times60}{3.4\times10^5}\text{kg/h}$
$\Rightarrow\frac{\text{m}}{\text{t}}=1.52\text{kg/h}$

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