Consider the process where 1.3 moles of ice, initially at -30.0 C, is heated to 140.0 C at constant pressure of 1.00 atm. The molar heat capacities (Cp) for solid, liquid, and gaseous water (37.5 J K-1 mol-1, 75.3 J K-1 mol-1, 36.4 J K-1 mol-1, respectively) are assumed to be temperature independant. The enthalpies of fusion and vaporization are 6.01 kj/mol and 40.7 kj/mol, respectively. Assume ideal gas behavior.
Calculate the heat, q
calculate the work
q is computed in steps.
Within a phase q = mass x specific heat x (Tfinal-Tinitial). For example, to heat 20 g H2O from zero to 100 C (within the liquid phase) it requires
q = 20g x 4.184 J/g*C x (100C-0C) = ?
At a phase change (solid to liquid or liquid to vapor) it is q = mass x heat fusion to melt or mass x heat vaporization to change to vapor.
For example, q to change 20 g liuqid H2O to steam is
q = 20 x 2260 kJ/g = ?kJ.
Then added the qs together.
You have a solid phase from -30 to zero. It changes to a liquid at zero, is a liquid to 100, changes to a vapor at 100, is steam from there to 140 C.
For part 2.
work is p* delta V.
Calculate the volume at -30 and the volume at 140.
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To calculate the heat (q), we need to consider the following steps in the process:
1. Heating the ice from -30.0°C to 0°C.
2. Melting the ice at 0°C.
3. Heating the liquid water from 0°C to 100°C.
4. Boiling the liquid water at 100°C.
5. Heating the steam from 100°C to 140°C.
Step 1: Heating the ice from -30.0°C to 0°C
To calculate the heat for this step, we use the equation:
q = n * Cp * ΔT
where n is the number of moles of ice, Cp is the molar heat capacity for the solid water, and ΔT is the temperature change.
q1 = 1.3 mol * 37.5 J K-1 mol-1 * (0 - (-30.0)°C)
q1 = 1.3 mol * 37.5 J K-1 mol-1 * 30.0°C
Step 2: Melting the ice at 0°C
To calculate the heat for this step, we use the equation:
q = n * ΔHfusion
where n is the number of moles of ice, and ΔHfusion is the enthalpy of fusion.
q2 = 1.3 mol * 6.01 kJ/mol
Step 3: Heating the liquid water from 0°C to 100°C
To calculate the heat for this step, we use the equation:
q = n * Cp * ΔT
where n is the number of moles of water, Cp is the molar heat capacity for liquid water, and ΔT is the temperature change.
q3 = 1.3 mol * 75.3 J K-1 mol-1 * (100 - 0)°C
Step 4: Boiling the liquid water at 100°C
To calculate the heat for this step, we use the equation:
q = n * ΔHvaporization
where n is the number of moles of water, and ΔHvaporization is the enthalpy of vaporization.
q4 = 1.3 mol * 40.7 kJ/mol
Step 5: Heating the steam from 100°C to 140°C
To calculate the heat for this step, we use the equation:
q = n * Cp * ΔT
where n is the number of moles of steam, Cp is the molar heat capacity for gaseous water, and ΔT is the temperature change.
q5 = 1.3 mol * 36.4 J K-1 mol-1 * (140 - 100)°C
Now, to calculate the total heat (q), we sum up all the heats from each step:
q = q1 + q2 + q3 + q4 + q5
To calculate the work, we need to determine if the volume changes during the process. If the volume remains constant, the work done will be zero (w = 0). If the volume changes, we will need additional information to calculate the work.
To calculate the heat, q, you need to consider the stages involved in the process:
1. Heating the ice from -30.0°C to 0°C:
q1 = n * ΔHfus
Here, n is the number of moles of ice, which is given as 1.3 moles.
ΔHfus is the enthalpy of fusion, which is given as 6.01 kJ/mol.
Calculate q1: q1 = 1.3 moles * 6.01 kJ/mol
2. Melting the ice at 0°C to liquid water at 0°C:
q2 = n * Cp_liquid * ΔT
Cp_liquid is the molar heat capacity of liquid water, which is given as 75.3 J/K/mol.
ΔT is the change in temperature, which is 0°C-(-30.0°C) = 30.0°C.
Convert ΔT to Kelvin: ΔT = 30.0°C + 273.15
Calculate q2: q2 = 1.3 moles * 75.3 J/K/mol * ΔT
3. Heating the liquid water from 0°C to 100°C:
q3 = n * Cp_liquid * ΔT
Use the same values for n, Cp_liquid, and ΔT as in step 2.
4. Vaporizing the liquid water at 100°C to gaseous water at 100°C:
q4 = n * ΔHvap
ΔHvap is the enthalpy of vaporization, which is given as 40.7 kJ/mol.
Calculate q4: q4 = 1.3 moles * 40.7 kJ/mol
5. Heating the gaseous water from 100°C to 140°C:
q5 = n * Cp_gas * ΔT
Cp_gas is the molar heat capacity of gaseous water, which is given as 36.4 J/K/mol.
ΔT is the change in temperature, which is 140°C - 100°C.
Convert ΔT to Kelvin: ΔT = 140°C + 273.15
Calculate q5: q5 = 1.3 moles * 36.4 J/K/mol * ΔT
Now, sum up all the q values to find the total heat, q:
q = q1 + q2 + q3 + q4 + q5
To calculate the work, you need to know if the process is reversible or irreversible. If it is reversible, the work can be calculated using the equation:
w = -PΔV
Since the process is assumed to be at constant pressure (1.00 atm), the equation simplifies to:
w = -PΔV = -nRΔT
Here, R is the ideal gas constant, which is 8.314 J/(K*mol).
ΔT is the change in temperature, which is 140°C - (-30.0°C).
Convert ΔT to Kelvin: ΔT = 140°C + 273.15
Calculate work: w = -1.3 moles * 8.314 J/(K*mol) * ΔT