Question

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 $3.4\times 10^{5}Jk{g}^{-1}$ and thermal conducting of insulation wall is $0.05W{m}^{-10}{C}^{-1}$ )

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

Answer 1

Answer: (B)

$\frac{dQ}{dt}=\frac{KA\Delta T}{\ell }$
$A=2(0.6\times 0.5+0.5\times 0.2+0.2\times 0.6)$
$=2(0.3+0.1+0.12)$
$=2(0.4+0.12)$
$=2(0.52)$
$=1.04m^2$
$R_{th}=\frac{\ell }{KA}\Rightarrow \frac{1\times 10^{-2}}{0.05\times 1.04}\Rightarrow \frac{10^{-2}}{0.052}$
$\frac{dQ}{dt}=\frac{\Delta T}{R_{th}}\Rightarrow \frac{40\times 0.052}{10^{-2}}\Rightarrow 2.08\times 10^{2}J/s$
$2.08\times 10^2=m\times 3.4\times 10^5$
$m=\frac{2.08}{3.4\times 10^3}\Rightarrow 0.61\times 10^{-3}kg/s$
$=61\times 10^{-5}Kg/s$