Question

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 $0.05\, Wm ^{-10} C ^{-1}$ )

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

Answer 1

Answer: (B)

$ \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 $