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Q10.17: NCERT Physics - Part 2 THERMAL PROPERTIES OF MATTER 10.17 Solution: A 'thermacole' icebox is a cheap and an efficient method for…

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__ m__ K__. [Heat of fusion of water = 335 _ 10_ J kg__]

Step-by-Step Verified Solution

**Answer:** Mass of ice remaining = 3.69 kg **Step-by-Step Explanation:** Side of cube a=0.30a = 0.30 m. Total surface area A=6a2=6(0.30)2=0.54A = 6a^2 = 6(0.30)^2 = 0.54 m_. Thickness l=0.05l = 0.05 m. Temperature difference ΔT=45−0=45∘C\Delta T = 45 - 0 = 45^\circ\text{C}. Thermal conductivity K=0.01K = 0.01 J s__ m__ K__. Time t=6t = 6 h =21600= 21600 s. Heat entering the box:
Q=KAΔTl×t=0.01×0.54×450.05×21600=104976 JQ = \frac{KA\Delta T}{l} \times t = \frac{0.01 \times 0.54 \times 45}{0.05} \times 21600 = 104976 \text{ J}
Mass of ice melted:
m=QL=104976335×103=0.313 kgm = \frac{Q}{L} = \frac{104976}{335 \times 10^3} = 0.313 \text{ kg}
Mass of ice remaining:
m′=4.0−0.313=3.687 kg≈3.69 kgm' = 4.0 - 0.313 = 3.687 \text{ kg} \approx 3.69 \text{ kg}
Standard answer: 3.69 kg .

All Questions in this Chapter

Q10.1 Q10.2 Q10.3 Q10.4 Q10.5 Q10.6 Q10.7 Q10.8 Q10.9 Q10.10 Q10.11 Q10.12 Q10.13 Q10.14 Q10.15 Q10.16 Q10.17 Q10.18 Q10.19 Q10.20

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