To calculate the mass of ice required to lower the temperature of 420g of water at 45 °C to water at 0 °C, use the equations for heat transfer and combine the specific heat capacity of water and the latent heat of ice. The mass of ice required is approximately 262.6 grams.
To calculate the mass of ice required to lower the temperature of 420g of water at 45 °C to water at 0 °C, we need to consider the heat transferred from the water to the ice. First, we calculate the heat transferred from the water to cool it from 45 °C to 0 °C using the specific heat capacity of water:
Q1 = mcΔT
Q1 = (420g)(4.186 J/g°C)(45 °C)
Q1 = 88,117.4 J
Then, we calculate the heat required to melt the ice at 0 °C using the latent heat of ice:
Q2 = mL
Q2 = (m)(336 J/g)
Now, since the heat lost by the water is gained by the ice, we can equate Q1 and Q2:
Q1 = Q2
88,117.4 J = (m)(336 J/g)
Solving for m, the mass of ice:
m = 88,117.4 J / 336 J/g
m ≈ 262.6 g
Therefore, the mass of ice required to lower the temperature of 420g of water at 45 °C to water at 0 °C is approximately 262.6 grams.