Question

A 'thermacole' icebox is a cheap and 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 remaining after 6 h. The outside temperature is ${45}^{o}C$, and co-efficient of thermal conductivity of thermacole is $0.01J{s}^{-1}{m}^{-1}{K}^{-1}$. [Heat of fusion of water $=335\times {10}^{3}Jk{g}^{-1}$]

Answer 1

Side of the given cubical ice box, $s=30cm=0.3m$
Thickness of the ice box, $l=5.0cm=0.05m$
Mass of ice kept in the ice box, $m=4kg$

Time gap, $t=6h=6\times 60\times 60s$
Outside temperature, $T={45}^{o}C$
Coefficient of thermal conductivity of thermacole, $K=0.01J{s}^{-1}{m}^{-1}{k}^{-1}$

Heat of fusion of water, $L=335\times {10}^{3}Jk{g}^{-1}$

Let m be the total amount of ice that melts in 6 h.

The amount of heat lost by the food:

$\theta =KA(T-0)t/l$
Where,
$A=$ Surface area of the box $=6s^2=6\times (0.3{)}^2=0.54m^3$

$\theta =0.01\times 0.54\times 45\times 6\times 60/0.05=104976J$
But $\theta =m^{\prime }L$

$\therefore m^{\prime }=\theta /L$
$=104976/(335\times {10}^3)=0.313kg$
Mass of ice left $=4–0.313=3.687kg$
Hence, the amount of ice remaining after 6 h is 3.687 kg.