A thermacole cubical icebox of side 20 cm has a thickness of 4.0 cm. If 5.0 kg of ice is put in the box, estimate the amount of ice remaining after 5h. The outside temperature is 35°C, and coefficient of thermal conductivity of thermacole is 0.01 J $$\text s^{-1}\space\text m^{-1}\space\text k^{-1}$$​. (Heat of fusion of water = 335 × $$10^3$$​ J $$\text {kg}^{-1}$$​)

Correct option is C

​$$\begin{aligned}&\textbf{Given:} \\&\quad \text{Time } t = 5\, \text{hours} = 5 \times 60 \times 60\, \text{seconds} = 18000\, \text{seconds} \\&\quad \text{Thickness of the box } t_b = 0.04\, \text{m} \\&\quad \text{Latent heat of fusion } L = 335 \times 10^3\, \text{J/kg} \\&\quad \text{Coefficient of thermal conductivity of thermacole } K = 0.01\, \text{W/m} \cdot \text{K} \\&\quad \text{Outside temperature } T_a = 35^\circ \text{C} \\&\quad \text{Mass of ice placed in the box } m = 5\, \text{kg} \\&\quad \text{Side of the cubical box } s = 20\, \text{cm} = 0.2\, \text{m} \\&\quad \text{Surface area of the ice box } A = 6s^2 = 6 \times (0.2)^2 = 0.24\, \text{m}^2 \\[10pt]&\textbf{Formula:} \\&\text{The amount of heat transferred through the thermacole material is given by Fourier's law:} \\&\quad \dot{Q} = \frac{KA(T_a - T_b)}{t_b} \\[10pt]&\text{Where:} \\&\quad \dot{Q} \text{ is the heat transferred per second,} \\&\quad A \text{ is the surface area of the box,} \\&\quad T_a \text{ and } T_b \text{ are the temperatures of the outside and the box respectively.} \\[10pt]&\text{The total heat gained by the box in time } t = 5\, \text{hours is:} \\&\quad Q = \dot{Q} \times t\end{aligned}$$​$$\begin{aligned}&\textbf{ Heat gained per second:} \\&\quad \dot{Q} = \frac{KA(T_a - T_b)}{t_b} \\&\quad \dot{Q} = \frac{0.01 \times 0.24 \times (35 - 0)}{0.04} \\&\quad \dot{Q} = \frac{0.01 \times 0.24 \times 35}{0.04} = 1.68\, \text{J/s} \\[10pt]&\textbf{ Total heat gained in 5 hours:} \\&\quad Q = \dot{Q} \times t = 1.68 \times 18000 = 37800\, \text{J} \\[10pt]&\textbf{Mass of ice melted:} \\&\text{The energy required to melt } m \text{ kg of ice is given by:} \\&\quad Q = mL \\[5pt]&\text{Rearranging to solve for } m: \\&\quad m = \frac{Q}{L} \\[5pt]&\text{Substitute the values:} \\&\quad m = \frac{37800}{335 \times 10^3} = 0.1128\, \text{kg} \\[10pt]&\textbf{ Mass of ice left after 5 hours:} \\&\text{The remaining mass of ice is:} \\&\quad m_1 = m - 0.1128 = 5 - 0.1128 = 4.8\, \text{kg}\end{aligned}$$​​