Final answer:
Approximately 0.125 g of ice was added to cool the water from 20 °C to 10 °C.
Step-by-step explanation:
First, we need to calculate how much heat is needed to cool the water from 20 °C to 10 °C. We can use the formula:
Q = mcΔT
where Q is the heat transfer, m is the mass of water, c is the specific heat capacity of water, and ΔT is the change in temperature.
Assuming the specific heat capacity of water is approximately 4.18 J/g°C, the mass of water is 1.0 kg, and the change in temperature is 20 °C - 10 °C = 10 °C, we can calculate the heat transfer as:
Q = (1.0 kg)(4.18 J/g°C)(10 °C) = 41.8 J
Now, we need to calculate the amount of heat released by the ice to cool the water. We can use the formula:
Q = mL
where Q is the heat transfer, m is the mass of ice, and L is the latent heat of fusion of ice.
Assuming the latent heat of fusion of ice is approximately 334 J/g, we can calculate the mass of ice as:
m = Q / L = 41.8 J / 334 J/g ≈ 0.125 g
Therefore, approximately 0.125 g of ice was added to cool the water from 20 °C to 10 °C.