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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) ] |
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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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