An ice cube of dimensions $60cm×50cm×20cm$ is placed in an insulation box of wall thickness 1cm. The box keeping the ice cube at $0^{0} C$ of temperature is brought to a room of temperature $30^{0} C.$ The rate of melting of ice is approximately: (Latent heat of fusion of ice is ${3.4×10}^{5} {J kg}^{-1}$ and thermal conducting of insulation wall is ${0.05 Wm}^{-1}^{o} C^{-1}$ )

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${280 kg s}^{-1}$

${30×10}^{-5} {kg s}^{-1}$

${61×10}^{-5} {kg s}^{-1}$

${46×10}^{-5} {kg s}^{-1}$

answer is A.

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Detailed Solution

Given that latent heat of fusion of ice L, $= 3.4 \times 10^{5} J k g^{- 1}$

Thermal conductivity of insulation wall, $K = 0.05 W m^{- 10} C^{- 1}$

Change in temperature, $Δ T = 40 ° C$

Area of ice cube, $A = 2 (0.6 \times 0.5 + 0.5 \times 0.2 + 0.2 \times 0.6) = 2 (0.3 + 0.1 + 0.12) = 1.04 m^{2}$

We have $\frac{d Q}{d t} = \frac{K A Δ T}{l} = \frac{0.05×1.04×30}{10^{-2}} {=1.56×10}^{2} J/s$

$\frac{dQ}{dt} =mL$

${1.56×10}^{2} {=m×3.4×10}^{5}$

$m= \frac{1.56}{{3.4×10}^{3}} \Rightarrow {0.46×10}^{-3} {kg/s=46×10}^{-5} kg/s$

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