A ‘thermacole’ icebox is a 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 cm. 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 co-efficient of thermal conductivity of thermacole is `0.01 J s^(–1) m^(–1) K^(–1)`. [Heat of fusion of water `= 335xx 103 J kg^(–1)]`
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Latent heat of fusion of ice `= L = 335 xx 10^(3) "Jkg"^(-1)` Area through which heat is lost `= 6xx` area of one surface `= 6 xx 0.3^(2) = 0.54 "m"^(2)` `x=5` cm `=0.05`m Mass of ice melted in 6 hours `= m=?` heat absorbed by the ice `=Q="mL" = m xx 335 xx 10^(3)` `theta_1 - theta_2 = 45 -0 = 45` Time `= t = 6 h = 6 xx 3600 s` `K= 0.01 "Wm"^(-1)°"C"^(-1)` `Q= "mL" = ( KA ( theta_1 - theta_2 ) t) /( x L)` `= ( 0.01 xx 0.54 xx 45 xx 6 xx 3600)/( 0.05 xx 335 xx 10^(3) ) = 0.313 "kg"` Remaining ice = Total ice - melted ice `= 4 - 0.313 = 3.687` kg
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