Final answer:
By using the conservation of energy principle and the specific heat capacity and latent heat of fusion for water, it is calculated that approximately 136.58 g of ice would have melted when 421 g of water at 26°C is added to ice at 0°C.
Step-by-step explanation:
To determine how much ice melted when 421 g of water at 26°C is added to ice at 0°C, resulting in a final temperature of 0°C, we use the concept of conservation of energy. The heat lost by the water (cooling from 26°C to 0°C) will be used to melt a portion of the ice.
- Calculate the heat lost by the water: Q = mcΔT, where m is the mass of the water, c is the specific heat capacity, and ΔT is the change in temperature.
- Calculate the mass of ice melted: Use the heat of fusion for ice, Q = mL, where L is the latent heat of fusion and solve for m.
Given that the specific heat capacity of water is approximately 4.18 J/g°C and the latent heat of fusion of ice is about 334 J/g, we proceed with the calculations.
Heat lost by water: Q = (421 g)(4.18 J/g°C)(26°C) = 45615.56 J
Mass of ice melted: Q = (m)(334 J/g), solving for m we get m = 45615.56 J / 334 J/g = 136.58 g.
Hence, approximately 136.58 g of ice melted.