Verified Solution • 2025–26 Session moderate

QJEEM22Ph52ec: JEE Main 2022 Physics - Class 11 JEEM22Ph52ec Solution: An ice cube of dimensions is placed in an insulation box of…

An ice cube of dimensions 60 cm×50 cm×20 cm60 \mathrm{~cm} \times 50 \mathrm{~cm} \times 20 \mathrm{~cm} is placed in an insulation box of wall thickness 1 cm1 \mathrm{~cm} . The box keeping the ice cube at 0∘C0^{\circ} \mathrm{C} of temperature is brought to a room of temperature 40∘C40^{\circ} \mathrm{C} . The rate of melting of ice is approximately :

(Latent heat of fusion of ice is 3.4×105 J kg−13.4 \times 10^{5} \mathrm{~J} \mathrm{~kg}^{-1} and thermal conducting of insulation wall is 0.05 Wm−1∘C−10.05 \,\mathrm{Wm}^{-1 \circ} \mathrm{C}^{-1} )

Multiple Choice Options

  • A 61×10−3 kg s−161 \times 10^{-3} \mathrm{~kg} \mathrm{~s}^{-1}
  • B 61×10−5 kg s−161 \times 10^{-5} \mathrm{~kg} \mathrm{~s}^{-1}
  • C 208 kg s−1208 \mathrm{~kg} \mathrm{~s}^{-1}
  • D 30×10−5 kg s−130 \times 10^{-5} \mathrm{~kg} \mathrm{~s}^{-1}

Step-by-Step Verified Solution

ΔQΔt=kA(T1−T2)l{{\Delta Q} \over {\Delta t}} = {{kA({T_1} - {T_2})} \over l}

⇒mLΔt=kA(T1−T2)l\Rightarrow {{mL} \over {\Delta t}} = {{kA({T_1} - {T_2})} \over l}

⇒mΔt=kA(T1−T2)Ll\Rightarrow {m \over {\Delta t}} = {{kA({T_1} - {T_2})} \over {Ll}}

≃61.1×10−5\simeq 61.1 \times {10^{ - 5}} kg/s

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