A quantity of ice at 0.0°C was added to 25.6 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).
I don't understand how to set this problem up (what the formula is and where to plug in which qualities). Any help would be very much appreciated.
duplicate post. Answered below.
To solve this problem, we can use the principles of heat transfer and the specific heat capacity of water to calculate the amount of ice that was added.
First, let's determine the heat gained by the ice and water during the process.
The heat gained or lost by a substance can be calculated using the formula:
Q = m * c * ΔT
Where:
Q is the heat gained or lost (in joules),
m is the mass of the substance (in grams),
c is the specific heat capacity of the substance (in J/(g · °C)), and
ΔT is the change in temperature (in °C).
In this case, we want to find the amount of ice that was added, so we'll let x be the mass of the ice (in grams).
The heat gained by the ice is given by:
Q_ice = x * (6.01 kJ/mol) * (1 mol/18 g) (since the heat of fusion is given in kJ/mol)
The heat gained by the water is given by:
Q_water = (25.6 g) * (4.18 J/(g · °C)) * (0.0°C - 21.0°C)
Since the final water temperature is 0.0°C, we can set up an equation by equating the heat gained by the ice and the heat gained by the water:
x * (6.01 kJ/mol) * (1 mol/18 g) = (25.6 g) * (4.18 J/(g · °C)) * (0.0°C - 21.0°C)
Now, let's solve for x, the mass of the ice:
x = (25.6 g) * (4.18 J/(g · °C)) * (0.0°C - 21.0°C) / (6.01 kJ/mol) * (1 mol/18 g)
Simplifying the equation and performing the calculations will give us the mass of the ice that was added.
Note: Make sure to convert kilojoules to joules and keep the units consistent throughout the calculation.
I hope this helps! Let me know if you have any further questions.