100 grams of ice at 0°C is put in 240 grams of water at 5°C. The mixture finally comes to equilibrium at 0°C with m grams of ice melted in the process. Value of m is close to : (latent heat of ice = 3.33 × 105 J/kg and specific heat of water = 4.2 J g-1°C)
0 g
10 g
5 g
15 g
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UPSC Geoscientist – 2024
The heat lost by the water is Q_water = m_water * c_water * ΔT_water. Given m_water = 240 g, c_water = 4.2 J g⁻¹°C⁻¹, and ΔT_water = 5°C – 0°C = 5°C. Q_water = (240 g) * (4.2 J g⁻¹°C⁻¹) * (5°C) = 5040 J. This heat melts a mass m of ice. The heat required to melt ice is Q_melt = m * L_f. Given L_f = 3.33 × 10⁵ J/kg = 3.33 × 10⁵ J / 1000 g = 333 J/g.