A cubical ice box of side 50 cm has a thickness of 5.0 cm. if 5 kg of ice is put in the box, estimate the amount of ice remaining after 4 hours. The outside temperature is `40^(@)C` and coefficient of thermal conductivity of the material of the box = `0.01 Js^(-1) m^(-1) .^(@)C^(-1)`. Heat of fusion of ice` = 335 Jg^(-1)`.

Explanation

Here, `l=50 cm = 0.50 m` ,
`Delta x = 5 cm = 5 xx 10^(-2)m`
As cubical box has six faces, so that area
`A = 6 xx l^(2) = 6 xx (0.5)^(2) = 1.50 m^(2)`
`Delta T = T_(1) -T_(2) = 40^(@)-0^(@) = 40^(@)C`,
`Delta t = 4 hr = 4 xx 60 xx 60 s`,
`L = 335 jg^(-1) = 335 xx 10^(3) J kg^(-1)`,
`K=0.01 J s^(-1)m^(-1) .^(@)C^(-1)`
Amount of heat conducted into the box is
`Q = (KA(Delta T)t)/(Delta x)`
`=90.01 xx (1.5) xx 940 xx (4 xx 60 xx 60)/(5 xx 10^(-2))`
`=(172800)/(5) = 34560 J`
Let m be the mass of ice melted in 4 hr, then heat spent, `Q = mL = m xx (335 xx 10^(3))J`
`:. m xx 335 xx 10^(3) = 34560`
or `m = (34560)/(335 xx 10^(3)) = 0.103 kg`
Mass of the ice left in box after 4 hr
`=5 - 0.103 = 4.897 kg`.