Question

A cubical thermocole ice box of side $20cm$ has a thickness $5cm$. If $5kg$ of ice is put in the box, the amount of ice remaining after $10hours$ is
(The outside temperature is $50$ and the coefficient of thermal conductivity of thermocole $=0.01J{s}^{-1}{m}^{-1}{}^{-1}$, latent heat of fusion of ice $=335\times {10}^{3}J{kg}^{-1})$

• A. $3.7kg$
• B. $3.9kg$
• C. $4.7kg$
• D. $4.9kg$

Answer 1

The correct option is C $4.7kg$

Here,

Length of each side, $l=20=20\times {10}^{-2}m$

Thickness, $x=5=5\times {10}^{-2}m$

Total surface area through which heat enters into box,

$a={6l}^2=6\times {(20\times {10}^{-2}m)}^2=24\times {10}^{-2}{m}^2$

Temperature difference, $\Delta T=50-0=50$

Thermal conductivity, $K=0.01J{s}^{-1}{m}^{-1}{}^{-1}$

Time, $t=10h=10\times 60\times 60s=36\times {10}^{3}s$

$L_f=335\times {10}^{3}J{kg}^{-1}$

Total heat entering the box through all the six faces is

$Q=\frac{KA\Delta Tt}{x}$

$=\frac{0.01J{s}^{-1}{m}^{-1}{}^{-1}\times 24\times {10}^{-2}{m}^2\times 50\times 36\times {10}^{3}s}{5\times {10}^{-2}m}$

$=86400J$ .....(i)

let $m$ kg of ice melts in this time.

$\therefore Q=m{L}_{f}4$ ......(ii)

From (i) and (ii), we get

$m=\frac{86400J}{335\times {10}^{3}J{kg}^{-1}}=0.258kg$

$\therefore$ Amount of ice left $=(5-0.258)kg=4.742kg=4.7kg$