A quantity of ice at 0.0°C was added to 25.0 g of water at 21.0°C to give water at 0.0°C. How much ice was added? The heat of fusion of water is 6.01 kJ/mol and the specific heat is 4.18 J/(g · °C).
_g
(g ice x heat fusion) +[g melted ice x specific heat H2O x (Tfinal-Tinitial)] + [g H2O x specific heat water x (Tfinal-Tinitial)] = 0
You have only one unknown namely g ice. Solve for that. I estimated the answer to be about 7g ice at zero. Note that I have three terms above; i.e., ice + melted ice + water at 25. The middle term (melted ice) is multiplied by (Tfinal-Tinitial) but since Tfinal is zero and Tinitial is zero then Tf-Ti = 0 and anything times zero is zero so the middle term is zero.
To find out how much ice was added, we can use the principles of heat transfer.
First, let's calculate the heat gained by the ice as it warms up from 0.0°C to 21.0°C.
The formula for heat gained (or lost) is Q = mcΔT, where Q is the heat transfer in joules, m is the mass in grams, c is the specific heat capacity in J/(g·°C), and ΔT is the change in temperature in °C.
For the ice, the change in temperature (ΔT) is 21.0°C - 0.0°C = 21.0°C. The specific heat capacity (c) for ice is 2.09 J/(g·°C). We don't know the mass (m) of the ice, so we'll keep it as a variable.
Q_ice = m_ice × c_ice × ΔT_ice
Next, let's calculate the heat lost by the water as it cools down from 21.0°C to 0.0°C.
The change in temperature (ΔT) for the water is 21.0°C - 0.0°C = 21.0°C. The specific heat capacity (c) for water is 4.18 J/(g·°C), and the mass (m) of water is 25.0 g.
Q_water = m_water × c_water × ΔT_water
Since the ice and water exchange heat until they reach the same temperature (0.0°C), the heat lost by the water is equal to the heat gained by the ice:
Q_ice = Q_water
m_ice × c_ice × ΔT_ice = m_water × c_water × ΔT_water
Since the final temperature is 0.0°C, ΔT_ice = ΔT_water = 21.0°C.
m_ice × (2.09 J/(g·°C)) × (21.0°C) = (25.0 g) × (4.18 J/(g·°C)) × (21.0°C)
Now, we can solve for m_ice (mass of ice):
m_ice = (25.0 g) × (4.18 J/(g·°C)) × (21.0°C) / (2.09 J/(g·°C))
m_ice = 210 g
Therefore, 210 grams of ice were added to the water.