Question

A thermocol box has a total wall area (including the lid) of $1.0 m ^{2}$ and wall thickness of $3 cm$. It is filled with ice at $0^{\circ} C$. If the average temperature outside the box is $30^{\circ} C$ throughout the day, the amount of ice that melts in one day is [Use $K_{\text {thermocol }}=0.03 W / mK$ $.L_{\text {fusion (ice) }}=3.00 \times 10^{5} J / kg ]$

• A. 1 kg
• B. 2.88 kg
• C. 25.92 kg
• D. 8.64 kg

Answer 1

Answer: (D)

8.64 kg

Answer 2

Given,
Total wall area (including the lid) $(A)=1.0\, cm ^{2}$
Thickness of wall $(1)=3 cm =3 \times 10^{-2} m$
Average temperature outside the box $=30^{\circ} C$
$\Delta \theta =30-0=30^{\circ} C$
$L_{\text {fusion (ice) }} =3 \times 10^{5} J / kg$
$K_{\text {thermocol }}=0.03 W / m K$
We know that,
$\frac{Q}{t}=\frac{K A}{1} \Delta \theta$
For one day, $t=24 \times 60 \times 60 s$
$\frac{m \times L_{\text {fusion (ice) }}}{t} =\frac{K A}{1} \Delta Q$
$\frac{m \times 3 \times 10^{5}}{24 \times 60 \times 60} =\frac{0.03 \times 1}{3 \times 10^{-2}} \times 30$
$m =\frac{0.03 \times 1 \times 30 \times 24 \times 60 \times 60}{3 \times 10^{-2} \times 3 \times 10^{5}}$
$m =\frac{77760}{9000}=8.64 \,kg$