Physics

Physics

Problem

The specific heat of water = 4200 J kg−1 K−14200 \text{ J kg}^{-1} \text{ K}^{-1} and the latent heat of ice = 3.4×105 J kg−13.4 \times 10^5 \text{ J kg}^{-1} . 100 grams of ice at 0∘C0^\circ\text{C} is placed in 200 g of water at 25∘C25^\circ\text{C} . The amount of ice that will melt as the temperature of water reaches 0∘C0^\circ\text{C} is close to (in grams): a. 61.7 b. 63.8 c. 69.3 d. 64.6

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