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)`.

Options

A. `3.7 kg`
B. `3.9 kg`
C. `4.7 kg`
D. `4.9 kg`

Answer

C

Explanation

Here,
Length of each side, `l = 20 cm = 20 xx10^(-2)m`
Thickness, `x = 5 cm = 5 xx 10^(-2) m`
Total surface area through which heat enters into the box,
`A = 6l^(2) = 6 xx (20 xx 10^(-2) m)^(2) = 24 xx 10^(-2) m^(2)`
Temperature difference , `DeltaT = 50^(@)C - 0^(@)C = 50^(@)C`
Thermal conductivity, `K = 0.01 j s^(-1) m^(-1) .^(@)C^(-1)`
Time, `t = 10 h = 10 xx 60 xx 60 s = 36 xx 10^(3) s`
`L_(f) = 335 xx 10^(3) J kg^(-1)`
Total heat energy the box through all the six faces is
`Q = (KADeltaTt)/(x)`
`= (0.01 J s^(-1) m^(-1) .^(@)C^(-1) xx 24 xx 10^(-2) m^(2) xx 50^(@)C xx 36 xx 10^(3)s )/(5 xx 10^(-2) m)`
`= 86400`
Let m kg of ice melts in this time.
`:. Q = mL_(f)`
`:. Q = mL_(f)`
From (i) and (ii), we get
`m = (86400 J)/(335 xx 10^(3) J kg^(-1)) = 0.258 kg`
`:.` Amount of ice left `= (5-0.258) kg = 4.7442 kg = 4.7 kg`