    Question

# Water at 50∘C is filled in a cubical container of side 1 m. The thickness of the walls of the container is 1 mm. The container is surrounded by large amount of ice at 0∘C. The temperature of the water becomes 25∘C in 10 ln2 seconds. Choose the correct options. [Given, specific heat of water =1 cal/(gm∘C); latent heat of fusion of ice =80 cal/gm; density of water =1gm/cm3; heat capacity of the container =1 J/K]

A
Thermal conductivity of the material is 70 J/mC
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B
Thermal conductivity of the material is 60 J/mC
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C
Mass of the ice melted is 312.5 kg
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D
Mass of the ice melted is 252 kg
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Solution

## The correct option is C Mass of the ice melted is 312.5 kgLet temperature of water at any instant be T, then heat current i=kAx.(T−0)−(1) where A=6a2=6 m2;x= thickness =1 mm=10−3 m Rate of heat lost from water, dQdt=+msdTdt−(2) Ratw of heat lost from water = Rate of heat transfer through the wall So, we get from (1)&(2),−msdTdt=kATx⇒−∫25∘50∘dTT=kAmsx∫10ℓn20dt⇒ln(2)=kAmsx⋅10(ln2) So, kA(10)msx=1 On Putting values we get k=msx10A=(103kg)(4.2×103J/kg∘C)10−3m10×(6m3)=k=70J/m∘C As we know, total heat transferred through the container = total heat absorbed by ice. Q=∫dQ=∫kAxTdt Lost by water. Q=mSΔT=103×4200×25J=(10∘gm)(1 cal) (25)=miceL ⟹mice=(106×2580)gm=2500080kg=312.5kg mass of ice meited =312.5kg  Suggest Corrections  0      Similar questions
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