Discussion Forum

1)

A thermocol box has a total wall area (including the lid)  of 1.0 m² and wall thickness of 3 cm. It is filled with ice at $0^{0}C$. If the average temperature outside the box is $30^{0}C$ throughout  the day, the amount of ice that melts in one day is 

[Use $K_{termocol}= 0.03 W/mK,$

  $L_{fusion(ice)}=300 \times 10^{5} J/kg$]

Answer:

Option D

Explanation:

 Given,

 Total wall area (including the lid) (A)=1.0 cm²

 Thickness of wall(l)=3 cm= $3 \times 10^{-2} m$

 Average temperature outside  the box = $30^{0}C$

 $\triangle \theta =30-0= 30^{0}C$

 $L_{fusion (ice)}=3 \times 10^{5} J/kg$

 $K_{thermocol}=0.03 W/m K$

 We know that,

    $\frac{Q}{t}=\frac{KA}{l} \triangle Q$

 For one day, t=24 x 60 x 60 s

$\frac{ m \times L_{fusion (ice)}}{t}=\frac{KA}{l} \triangle Q$

  $\frac{m\times3\times 10^{5}}{24\times60\times60}=\frac{0.03\times1}{3\times 10^{-2}}\times30$

   $m=\frac{0.03\times1 \times30\times24\times 60\times60}{3\times 10^{-2}\times3\times10^{5}}$

   $m= \frac{77760}{9000}$= 8.64 kg