Question 60 of 126

An ice cube of dimensions $60cm\times 50cm\times 20cm$ is placed in an insulation box of wall thickness $1cm$. The box keeping the ice cube at $0^{\circ}C$ of temperature is brought to a room of temperature $40^{\circ}C$. The rate of melting of ice is approximately: (Latent heat of fusion of ice is $3.4\times {10}^{5}J{kg}^{-1}$ and thermal conducting of insulation wall is 0.05W{m}^{-1}^{\circ}{C}^{-1})

A$61\times {10}^{-1}kg{s}^{-1}$B$61\times {10}^{-5}kg{s}^{-1}$C$208kg{s}^{-1}$D$30\times {10}^{-5}kg{s}^{-1}$

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Answer: (B) $61\times {10}^{-5}kg{s}^{-1}$

Using the equation of conduction,
$\dfrac{dQ}{dt}=\dfrac{KA\Delta T}{l}$
Now the total area of the box will be,
$A=2\left(0.6\times 0.5+0.5\times 0.2+0.2\times 0.6\right)$
$A=2\left(0.3+0.1+0.12\right)$
$A=2\left(0.52\right)=1.04{m}^{2}$
Therefore,
$\dfrac{dQ}{dt}=\dfrac{KA\Delta T}{l}=\dfrac{0.05\times 1.04\times 40}{0.01}=208J{s}^{-1}$
This heat will melt the ice. Therefore,
$\dfrac{dQ}{dt}=\dfrac{dm}{dt}L \Rightarrow 208=\dfrac{dm}{dt}\times 3.4\times {10}^{5} \Rightarrow \dfrac{dm}{dt}=61\times {10}^{-5}kg{s}^{-1}$