How many grams of ice at -6.1C can be completely converted to liquid at 27.4C if the available heat for this process is 4.60×10^3 kJ ?
For ice, use a specific heat of 2.01J/(g*C) and traingle Hfus = 6.01kJ/mol.
4.60E6 J = [mass ice x specific heat ice x (Tfinal-Tinitial)] + [mass ice x heat fusion] + [mass water from ice x specific heat water x (Tfinal-Tinitial)]
Solve for mass ice/water
Watch that first (Tfinal-Tinitial) because it can be tricky. Tfinal = 0 and Tinitital = -6.1; therefore, it is 0 -(-6.1) = 0+6.1
To solve this problem, we need to calculate the amount of heat required to convert the ice to liquid, and then use that heat to determine how much ice can be converted.
The first step is to calculate the heat required to raise the temperature of the ice from -6.1°C to 0°C:
Heat1 = (Mass of ice) × (Specific heat of ice) × (Change in temperature)
= (Mass of ice) × (2.01 J/(g°C)) × (0 - (-6.1)°C)
= (Mass of ice) × (2.01 J/(g°C)) × (6.1°C)
Next, we calculate the heat required to melt the ice at 0°C:
Heat2 = (Mass of ice) × (Triangle Hfus)
Finally, we calculate the heat required to raise the temperature of the liquid water from 0°C to 27.4°C:
Heat3 = (Mass of water) × (Specific heat of water) × (Change in temperature)
= (Mass of water) × (4.18 J/(g°C)) × (27.4°C - 0°C)
Since the total available heat is given as 4.60×10^3 kJ, we convert it to joules:
Total heat = (4.60×10^3 kJ) × (10^3 J/kJ)
= 4.60×10^6 J
Now, we can set up the equation to solve for the mass of ice:
Total heat = Heat1 + Heat2 + Heat3
Substituting the previously calculated values:
4.60×10^6 J = (Mass of ice) × (2.01 J/(g°C)) × (6.1°C) + (Mass of ice) × (6.01 kJ/mol) + (Mass of water) × (4.18 J/(g°C)) × (27.4°C - 0°C)
To continue solving the equation, we need additional information: the molar mass of water.