Medium MCQ +4 / -1 PYQ · JEE Mains 2022

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

  1. A $61 \times 10^{-3} \mathrm{~kg} \mathrm{~s}^{-1}$
  2. B $61 \times 10^{-5} \mathrm{~kg} \mathrm{~s}^{-1}$ Correct answer
  3. C $208 \mathrm{~kg} \mathrm{~s}^{-1}$
  4. D $30 \times 10^{-5} \mathrm{~kg} \mathrm{~s}^{-1}$

Solution

<p>${{\Delta Q} \over {\Delta t}} = {{kA({T_1} - {T_2})} \over l}$</p> <p>$\Rightarrow {{mL} \over {\Delta t}} = {{kA({T_1} - {T_2})} \over l}$</p> <p>$\Rightarrow {m \over {\Delta t}} = {{kA({T_1} - {T_2})} \over {Ll}}$</p> <p>$\simeq 61.1 \times {10^{ - 5}}$ kg/s</p>

About this question

Subject: Physics · Chapter: Thermodynamics · Topic: Zeroth and First Law

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