A quantity of ice at 0 degrees celcius is added to 64.3 grams of water in a glass at 55 degrees celcius. After the ice melted, the temperature of the water in the glass was 15 degrees celcius. How much ice was added? The heat of fusion of water is 6.01kJ/mole and the heat of fusion is 4.18 J/(g*degrees celcius)

To solve this problem, we need to use the principle of conservation of energy. We can calculate the amount of heat lost by the water and the amount of heat gained by the ice to find the mass of the ice.

First, let's calculate the amount of heat lost by the water. We can use the formula:

Q_water = m_water * c_water * ΔT_water

where:
- Q_water is the heat lost by the water
- m_water is the mass of the water
- c_water is the specific heat capacity of water
- ΔT_water is the change in temperature of the water (55°C - 15°C)

The specific heat capacity of water is approximately 4.18 J/(g*°C).

Q_water = 64.3 g * 4.18 J/(g*°C) * (55°C - 15°C)

Q_water = 64.3 g * 4.18 J/(g*°C) * 40°C

Now, let's calculate the amount of heat gained by the ice. We can use the formula:

Q_ice = m_ice * ΔH_fusion

where:
- Q_ice is the heat gained by the ice
- m_ice is the mass of the ice
- ΔH_fusion is the enthalpy of fusion of water

The enthalpy of fusion of water is given as 6.01 kJ/mole. We need to convert it to J/g:

ΔH_fusion = 6.01 kJ/mole * (1000 J/kJ) / (18 g/mole)

Now, let's set the heat lost by the water equal to the heat gained by the ice:

Q_water = Q_ice

m_water * c_water * ΔT_water = m_ice * ΔH_fusion

Substituting the known values:

64.3 g * 4.18 J/(g*°C) * 40°C = m_ice * (6.01 kJ/mole * (1000 J/kJ) / (18 g/mole))

Simplifying the equation and solving for m_ice:

m_ice = (64.3 g * 4.18 J/(g*°C) * 40°C) / (6.01 kJ/mole * (1000 J/kJ) / (18 g/mole))

m_ice ≈ 17.3 g

Therefore, approximately 17.3 grams of ice were added to the glass.