A thermocole cubical icebox of side 30 cm has a thickness of 5.0 cm if 4.0 kg of ice are put ini the box, estimate the amount of ice remaining after 6 h. The outside temperature is `45^@C` and coefficient of thermal conductivity of thermocole`=0.01 J//kg`.

Explanation

Let m = mass of ice melted in 6 hr
Amount of heat required to melt m grams of ice
`Q = m xx 335 xx 10^3 Jkg^(-1)`
` = 335 xx 10^3 m J kg^(-1)`
since , ` Q = (KA (theta_1 - theta_2) t)/(d)`
`335 xx 10^3 m = (0.01 xx 6 xx 30 xx 30 xx 10^(-4) xx 45 xx 6 xx 60 xx 60)/(5 xx 10^(-2) )`
` = (0.01 xx 6 xx 30 xx 30 xx 10^(-4) xx 45 xx 6 xx 60 xx 60)/(335 xx 10^3 xx 5 xx 10^(-2) )`
` m = (3 xx 3 xx 45 xx 6 xx 6 xx 6 xx 6)/(335 xx 5 xx 1000) = 0.0313 kg `
Amount of ice left = 4.00 - 0.313 = 3.68 kg.