A 'thermacole' icebox is a cheap and an 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 (in $kg$) remaining after $6 \;h$. The outside temperature is $45\,^{\circ} C ,$ and co-efficient of thermal conductivity of thermacole is $0.01\; J s ^{-1} m ^{-1} K ^{-1} .$ Heat of fuston of water $=335 \times 10^{3}\;J kg ^{-1} $
A 'thermacole' icebox is a cheap and an 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 (in $kg$) remaining after $6 \;h$. The outside temperature is $45\,^{\circ} C ,$ and co-efficient of thermal conductivity of thermacole is $0.01\; J s ^{-1} m ^{-1} K ^{-1} .$ Heat of fuston of water $=335 \times 10^{3}\;J kg ^{-1} $
Side of the given cubical ice box, $s=30 cm =0.3 m$
Thickness of the ice box, $l=5.0 cm =0.05 m$
Mass of ice kept in the ice box, $m=4 kg$
Time gap, $t=6 h =6 \times 60 \times 60 s$
Outside temperature, $T=45^{\circ} C$
Coefficient of thermal conductivity of thermacole, $K=0.01 J s ^{-1} m ^{-1} K ^{-1}$
Heat of fusion of water, $L=335 \times 10^{3} J kg ^{-1}$ Let
$m$ be the total amount of ice that melts in $6 h$.
The amount of heat lost by the food:
$\theta=\frac{K A(T-0) t}{l}$
Where,
$A=$ Surface area of the box $=6 s^{2}=6 \times(0.3)^{2}=0.54 m ^{3}$
$\theta=\frac{0.01 \times 0.54 \times(45) \times 6 \times 60 \times 60}{0.05}=104976 J$
But $\theta=m^{\prime} L$
$\therefore m^{\prime}=\frac{\theta}{L}$
$=\frac{104976}{335 \times 10^{3}}=0.313 kg$
Mass of ice left $=4-0.313=3.687 kg$
Hence, the amount of ice remaining after $6 h$ is $3.687\, kg .$
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- [AIEEE 2006]