An ice cube of dimensions $60\,cm \times 50\,cm \times 20\,cm$ is placed in an insulation box of wall thickness $1\,cm$. 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 $0.05\,Wm ^{-10} C ^{-1}$ )
JEE MAIN 2022, Diffcult
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$\frac{ dQ }{ dt }=\frac{ KA \Delta T }{\ell}$
$A =2(0.6 \times 0.5+0.5 \times 0.2+0.2 \times 0.6)$
$=2(0.3+0.1+0.12)$
$=2(0.4+0.12)$
$=2(0.52)$
$=1.04\,m ^{2}$
$R _{ th }=\frac{\ell}{ KA } \Rightarrow \frac{1 \times 10^{-2}}{0.05 \times 1.04} \Rightarrow \frac{10^{-2}}{0.052}$
$\frac{ dQ }{ dt }=\frac{\Delta T }{ R _{ th }} \Rightarrow \frac{40 \times 0.052}{10^{-2}} \Rightarrow 2.08 \times 10^{2}\,J / s$
$2.08 \times 10^{2}= m \times 3.4 \times 10^{5}$
$m =\frac{2.08}{3.4 \times 10^{3}} \Rightarrow 0.61 \times 10^{-3}\,kg / s$
$=61 \times 10^{-5}\,Kg / s$
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