A quantity of ice at 0°C is added to 65.0 g of water in a glass at 55°C. After the ice melted, the temperature of the water in the glass was 15°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).
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To solve this problem, we can use the concept of heat transfer. The heat lost by the water at 55°C will be equal to the heat gained by the ice that melts and raises the temperature of the water to 15°C.
First, let's calculate the heat lost by the water:
q1 = m1 * c1 * ΔT1
Where:
- q1 is the heat lost by the water
- m1 is the mass of water in the glass (65.0 g)
- c1 is the specific heat capacity of water (4.18 J/(g·°C))
- ΔT1 is the change in temperature of the water (55°C - 15°C = 40°C)
q1 = 65.0 g * 4.18 J/(g·°C) * 40°C
q1 = 109,480 J
This heat lost by the water will be equal to the heat gained by the ice:
q2 = m2 * ΔHf
Where:
- q2 is the heat gained by the ice
- m2 is the mass of ice that melted (which is what we need to find)
- ΔHf is the heat of fusion of water (6.01 kJ/mol)
However, we need to convert ΔHf to J/g. The molar mass of water is approximately 18.02 g/mol, so:
ΔHf = 6.01 kJ/mol * 1000 J/kJ / (18.02 g/mol)
ΔHf = 334,184 J/g
Now, we can calculate the mass of ice:
q1 = q2
109,480 J = m2 * 334,184 J/g
m2 = 109,480 J / 334,184 J/g
m2 ≈ 0.3279 g
Therefore, approximately 0.3279 g of ice was added to the glass.