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Q11.17:
A ‘thermacole’ icebox is a cheap and an efficient method for storing small quantities of cooked food in summer in particular. A cubical icebox of side 30 cm has a thickness of 5.0 cm. If 4.0 kg of ice is put in the box, estimate the amount of ice remaining after 6 h. The outside temperature is 45 °C, and co-efficient of thermal conductivity of thermacole is $0.01 J s^{-1} m^{-1} K^{-1}$. [Heat of fusion of water = $335 \times 10^3 J kg^{-1}$]

Solution :

Given Variables & Initial Setup

To determine the mass of ice remaining, we must calculate the total heat energy ($Q$) that flows into the icebox through its walls via thermal conduction over the specified duration.

  • Side of the cubical box ($L$): $0.30 \, m$
  • Thickness of the walls ($\Delta x$): $0.05 \, m$
  • Initial mass of ice ($M_{initial}$): $4.0 \, kg$
  • Time duration ($t$): $6 \, hours = 6 \times 3600 \, s = 21600 \, s$
  • Outside temperature ($T_2$): $45 \, ^\circ C$
  • Inside temperature ($T_1$): $0 \, ^\circ C$ (since ice is present)
  • Temperature difference ($\Delta T$): $45 - 0 = 45 \, K$
  • Thermal conductivity ($K$): $0.01 \, J \, s^{-1} \, m^{-1} \, K^{-1}$
  • Latent heat of fusion of water ($L_f$): $335 \times 10^3 \, J \, kg^{-1}$

Step 1: Calculating Surface Area

A cube has 6 faces. The surface area ($A$) of each face is $L^2$. Therefore, the total surface area for heat transfer is:

$A = 6 \times L^2 = 6 \times (0.30 \, m)^2 = 6 \times 0.09 \, m^2 = 0.54 \, m^2$

Step 2: Determining Total Heat Influx

Per the Fourier Law of Thermal Conduction: $Q = \frac{K \cdot A \cdot \Delta T \cdot t}{\Delta x}$

Substituting the values:

$Q = \frac{0.01 \cdot 0.54 \cdot 45 \cdot 21600}{0.05}$

$Q = \frac{0.243 \times 21600}{0.05} = \frac{5248.8}{0.05} = 104976 \, J$

Step 3: Calculating Mass of Melted Ice

The heat required to melt a mass $m$ of ice is $Q = m \cdot L_f$. Thus, $m = \frac{Q}{L_f}$

$m = \frac{104976}{335 \times 10^3} \, kg$

$m \approx 0.313 \, kg$

Step 4: Final Mass Remaining

$M_{remaining} = M_{initial} - m$

$M_{remaining} = 4.0 - 0.313 = 3.687 \, kg$

30 cm 5 cm thick Heat Flux Q = KAΔT/Δx

Final Solution: The amount of ice remaining after 6 hours is approximately 3.69 kg.


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