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A 'thermacole' icebox is cheap and efficient method for storing small quantities of cooked food in summer in particular. A cubical icebox of side 30 cm has a thickness of 5.0 cam if 4.0 kg of ice is put in the box, estimate the amount of ice remaining after 6 h. The outside temperature is 45^(@)C, and coefficient of thermal conductivity of thermacole is 0.01" Js"^(-1)m^(-1)K^(-1). [Heat of fusion of water =335xx10^(3)" Jkg"^(-1) ]

Answer»

Solution :Area of cube `A=6l^(2)=6XX(30xx10^(-2))^(2)`
`:.A=6xx900xx10^(-4)`
`:.A=54xx10^(-2)m^(2)`
`dx=5" cm"=5xx10^(-2)m`
`t=6` hours `=6xx3600s=21600s`
`T_(1)=45^(@)C` and `T_(2)=0^(@)C`
Latent heat `L.=335xx10^(3)"J kg"^(-1)`
Thermal conductivity `K=0.01" J s"^(-1)m^(-2)K^(-1)`
Heat required to melt ice of mass `m.Q.=mL.` br> and heat entering in ice box in time .t.,
`Q=KA(dT)/(dx)t`
`:.Q.=Q`
`:." mL.=KA(dT)/(dx)t`
`:.m=(KAdTxxt)/(L.xxdx)`
`:.m=(0.01xx54xx10^(-2)xx45xx2121600)/(335xx10^(3)xx5xx10^(-2))`
`:.m=313.36xx10^(-3)" kg"`
`:.m~~0.313" kg"`
`:.` Ice REMAINING in box `=M-m`
where M = mass of ice in box
`=3.687` kg
`~~3.7` kg


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