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JEE MAIN - Physics (2022 - 26th July Evening Shift - No. 10)

An ice cube of dimensions $$60 \mathrm{~cm} \times 50 \mathrm{~cm} \times 20 \mathrm{~cm}$$ is placed in an insulation box of wall thickness $$1 \mathrm{~cm}$$. The box keeping the ice cube at $$0^{\circ} \mathrm{C}$$ of temperature is brought to a room of temperature $$40^{\circ} \mathrm{C}$$. The rate of melting of ice is approximately :

(Latent heat of fusion of ice is $$3.4 \times 10^{5} \mathrm{~J} \mathrm{~kg}^{-1}$$ and thermal conducting of insulation wall is $$0.05 \,\mathrm{Wm}^{-1 \circ} \mathrm{C}^{-1}$$ )

$$61 \times 10^{-3} \mathrm{~kg} \mathrm{~s}^{-1}$$
$$61 \times 10^{-5} \mathrm{~kg} \mathrm{~s}^{-1}$$
$$208 \mathrm{~kg} \mathrm{~s}^{-1}$$
$$30 \times 10^{-5} \mathrm{~kg} \mathrm{~s}^{-1}$$

Explanation

$${{\Delta Q} \over {\Delta t}} = {{kA({T_1} - {T_2})} \over l}$$

$$ \Rightarrow {{mL} \over {\Delta t}} = {{kA({T_1} - {T_2})} \over l}$$

$$ \Rightarrow {m \over {\Delta t}} = {{kA({T_1} - {T_2})} \over {Ll}}$$

$$ \simeq 61.1 \times {10^{ - 5}}$$ kg/s

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