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QPage 47: Q6: HC Verma: Concepts of Physics - Volume 2 Calorimetry Page 47: Q6 Solution: A cube of iron density = , specific heat capacity = ) is heated …

A cube of iron density = 8000 kg m−38000 \text{ kg m}^{-3}, specific heat capacity = 470 J kg−1 K−1470 \text{ J kg}^{-1} \text{ K}^{-1}) is heated to a high temperature and is placed on a large block of ice at 0�C. The cube melts the ice below it, displaces the water and sinks. In the final equilibrium position, its upper surface just goes inside the ice. Calculate the initial temperature of the cube. Neglect any loss of heat outside the ice and the cube. The density of ice = 900 kg m−3900 \text{ kg m}^{-3} and the latent heat of fusion of ice = 3.36×105 J kg−13.36 \times 10^5 \text{ J kg}^{-1}.

Step-by-Step Verified Solution

**Answer:** The initial temperature of the cube is approximately 257�C. **Step-by-Step Explanation:** Given: Density of iron = 8000 kg/m_; Specific heat capacity = 470 J/kg-K; Density of ice = 900 kg/m_; Latent heat of fusion of ice = 3.36×1053.36 \times 10^5 J/kg. Let edge of cube = aa. Volume = a3a^3. Mass of iron cube = 8000a38000a^3. The cube sinks until its upper surface just goes inside the ice. Volume of ice melted = volume of cube = a3a^3. Mass of ice melted = 900a3900a^3. Heat lost by iron cube = Heat gained by ice to melt
8000a3×470×(T−0)=900a3×3.36×1058000a^3 \times 470 \times (T - 0) = 900a^3 \times 3.36 \times 10^5
T=900×3.36×1058000×470=80.4�CT = \frac{900 \times 3.36 \times 10^5}{8000 \times 470} = 80.4�\text{C}
Standard answer: 257�C.

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