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 \[{{0}^{0}}C\]. 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 \[{{K}_{thermocol}}=0.03W{{m}^{-1}}{{K}^{-1}},{{L}_{fusion(ice)}}=3.00\times {{10}^{5}}Jk{{g}^{-1}}\]) –
A) 1 kg
B) 2.88 kg
C) 25.992 kg
D) 8.64 kg

Answer 1

Hint: We need to understand the relation between the rate of cooling of ice and the given parameters such as the temperature of the surroundings, the latent heat of fusion of ice and the surface area of the container to solve this problem easily.

Complete answer:
We are given a thermocol box with the ice at \[{{0}^{0}}C\] in it which is closed and kept in order to avail maximum insulation from heat. We know that the thermocol is heat insulating material to certain extent, even then there is a possibility for the conduction which can transfer heat from the related hotter surrounding in a long time.
We know that the rate of conduction of heat is proportional to the surface area of the box, the length from the surrounding to the other end of the container and the temperature difference. This is given mathematically as –
\[\dfrac{H}{t}=K\dfrac{A\Delta T}{l}\]
Now, we know that at \[{{0}^{0}}C\], the ice is changing its state from solid to liquid. So, the heat involved in the conversion is given by the latent heat of fusion of as –
\[H=m{{L}_{fusion(ice)}}\]
Where, m is the mass that we are supposed to find from this solution. Now, we can apply this formula in the rate of heat transfer to find the mass converted from ice to water in 24 hours as –
\[\begin{align}
& \dfrac{H}{t}=K\dfrac{A\Delta T}{l} \\
& \text{but,} \\
& H=m{{L}_{fusion(ice)}} \\
& \Rightarrow \dfrac{m{{L}_{fusion(ice)}}}{t}=K\dfrac{A\Delta T}{l} \\
& \Rightarrow \dfrac{m(3.00\times {{10}^{5}}Jk{{g}^{-1}})}{24\times 60\times 60s}=(0.03W{{m}^{-1}}{{K}^{-1}})\dfrac{1.0{{m}^{2}}({{30}^{0}}C-{{0}^{0}}C)}{0.03m} \\
& \Rightarrow m=\dfrac{(0.03)(1)(30)(86400)}{(300000)(0.03} \\
& \therefore m=8.64kg \\
\end{align}\]
The mass of ice converted to water in one day is 8.64 kg.

The correct answer is option D.

Note:
We have found the mass of ice converted to water in this question. If the complete ice was converted to water at some point before 24 hours, then the temperature difference would have contributed to the rise in temperature of the water.