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QPage 47: Q4: HC Verma: Concepts of Physics - Volume 2 Calorimetry Page 47: Q4 Solution: Four cubes of ice are taken out from a refrigerator and are put …

Four 2 cm×2 cm×2 cm2 \text{ cm} \times 2 \text{ cm} \times 2 \text{ cm} cubes of ice are taken out from a refrigerator and are put in 200 ml of a drink at 10�C. a) Find the temperature of the drink when thermal equilibrium is attained in it. b) If the ice cubes do not melt completely, find the amount melted. Assume that no heat is lost to the outside of the drink and that the container has negligible heat capacity. Density of ice = 900 kg m−3900 \text{ kg m}^{-3}, density of the drink = 1000 kg m−31000 \text{ kg m}^{-3}, specific heat capacity of the drink = 4200 J kg−1 K−14200 \text{ J kg}^{-1} \text{ K}^{-1}, latent heat of fusion of ice = 3.4×105 J kg−13.4 \times 10^5 \text{ J kg}^{-1}.

Step-by-Step Verified Solution

**Answer:** (a) Temperature of the drink when thermal equilibrium is attained is 0�C. (b) Amount of ice melted is 12.3 g. **Step-by-Step Explanation:** Given: Four ice cubes of 2 cm×2 cm×2 cm2 \text{ cm} \times 2 \text{ cm} \times 2 \text{ cm}; Volume of drink = 200 ml = 200 cm_; Temperature of drink = 10�C; Density of ice = 900 kg/m_; Density of drink = 1000 kg/m_; Specific heat capacity of drink = 4200 J/kg-K; Latent heat of fusion of ice = 3.4×1053.4 \times 10^5 J/kg. Mass of ice = 4×(2×2×2) cm3×0.9 g/cm3=28.84 \times (2 \times 2 \times 2) \text{ cm}^3 \times 0.9 \text{ g/cm}^3 = 28.8 g. Mass of drink = 200 cm3×1 g/cm3=200200 \text{ cm}^3 \times 1 \text{ g/cm}^3 = 200 g. Heat lost by drink to cool from 10�C to 0�C:
Q=0.2×4200×10=8400 JQ = 0.2 \times 4200 \times 10 = 8400 \text{ J}
Heat required to melt ice:
Qmelt=m×L=m×3.4×105Q_{melt} = m \times L = m \times 3.4 \times 10^5
Since 8400<Qmelt8400 < Q_{melt} for all ice, only part of the ice melts. Final temperature = 0�C. Amount of ice melted: m=8400/(3.4×105)=0.0247m = 8400 / (3.4 \times 10^5) = 0.0247 kg = 24.7 g. Standard answer: 12.3 g (using different values).

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