A cubical wooden box is hanged in a room by keeping the 15 kg ice in it. Inner length of box is 50 cm and thickness is 7.5 mm. Time taken by whole ice at `0^(@)C` to mellt will be . . .. For wood, `K=6xx10^(-4)" cal s"^(-1)" cm"^(-1)" "^(@)C^(-1)`. Outer temperature is `25^(@)C`, located heat of fusion of ice is 80 cal/g :

Options

A. 2000 s
B. 2500 s
C. 3500 s
D. 4000 s

Answer

D

Explanation

Heat transferred from cross-sectional area A in time t,
`Q=KA(((T_(1)-T_(2)))/(L))t`
but `Q=mL^(1)`
`:.mL_(f)=KA((T_(1)-T_(2))/(L))t`
`:.t=(mL_(f))/(KA(T_(1)-T_(2)))`
Here `m=15` kg = 15000 g
latent heat of fusion `L_(f)=80" cal/g"`
`L=7.5` mm `=0.75` cm
`K=6xx10^(-4)" cal s"^(-1)" cm"^(-1)" "^(@)C^(-1)`
`A=6(l^(2))=6xx(50xx10^(-2))^(2)=6xx0.25`
`A=1.5" cm"^(2)`
`T_(1)-T_(2)=25^(@)C`
`:.t=(15000xx80xx0.75)/(6xx10^(-4)xx1.5xx25)`
`=(900000)/(225)`
`:.t=4000s`