Question 17 of 20intermediate🔧 ApplyNumerical4 marks

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^{-1} m−1^{-1} K−1^{-1}. [Heat of fusion of water = 335×103335 \times 10^3 J kg−1^{-1}]

Correct Answer

Approximately 3.69 kg of ice remains after 6 hours.

Exercise: EXERCISES | Q: 10.17 | (Page 24)
For More Understanding

Explanation

This problem applies heat conduction through the thermacole walls to find how much heat enters the icebox, then uses the latent heat of fusion to determine how much ice melts. The thermacole acts as an insulator with low thermal conductivity, minimizing heat transfer.

Solution Steps

  1. Step 1: Convert all quantities to SI units. Side of cube = 30 cm = 0.3 m; Thickness (d) = 5.0 cm = 0.05 m; Time (t) = 6 h = 6×36006 \times 3600 s = 21600 s.

  2. Step 2: Calculate total surface area of cubical icebox. Total Area (A) = 6 × (side)² = 6 × (0.3)² = 6×0.096 \times 0.09 = 0.54 m².

  3. Step 3: Calculate heat transferred through thermacole using conduction formula. Q = (K × A × ΔT × t) / d. Given: K = 0.01 J s⁻¹ m⁻¹ K⁻¹, ΔT = 45°C - 0°C = 45 K. Q = (0.01×0.540.01 \times 0.54 × 45×2160045 \times 21600) / 0.05 = 104976 J ≈\approx 1.05×101.05 \times 10⁵ J.

  4. Step 4: Calculate mass of ice melted using latent heat. Mass melted (m) = Q / L. Given: L = 335×10335 \times 10³ J kg⁻¹. m = 104976 / (335×10335 \times 10³) = 0.313 kg.

  5. Step 5: Calculate remaining ice. Remaining ice = Initial mass - Mass melted = 4.0 - 0.313 = 3.69 kg.

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