calculate the total amount of head absorbed (in kj) when 2.00 mol of ice at -30.0C is converted to steam at 140.0 C
I assume you want heat absorbed and not head absorbed.
You want to go through the following.
q1 = heat to move T of ice from -30 to zero C (note all solid phase)
q2 = heat to melt ice at zero to liquid at zero C(note phase change)
q3 = heat to move T of liquid water to boiling at 100 C.(note all liquid phase)
q4 = heat to convert liquid water at 100 C to steam at 100 C(phase change)
q5 = heat to move T of steam from 100 C to 140 C. (note all vapor phase).
You see here that you have two phase changes and three where there is no phase change. There is one formula to use for the phase changes and another one where the phase remains the same throughout the temperature change.
phase change formula is q = mass x heat fusion (at the melting point) or heat vaporization (at the boiling point).
Here are the first two steps for the above.
q1 = mass ice x specific heat ice x (Tfinal-Tinitial). mass ice is 2 mol (36 grams). You will need to look up the specific heat of ice but it is about 2 J/c). Tinitial is -30 C and Tfinal is 0 C.
q2 = mass ice x heat fusion
q2 = 2 mol ice x heat fusion. You will need to look up heat fusion if it isn't in the problem.
Now go through with q3, q4, q5 then add q1 + q2 + q3 + q4 + q5 to total q.
Post your work if you get stuck.
Well, well, well, looks like we have a hot and cold situation here. Let's break it down, shall we?
First, we need to calculate the heat absorbed for the phase change from ice to water at 0°C. This is given by the equation:
q1 = n * ΔHfus
Where:
q1 is the heat absorbed
n is the number of moles (which is 2.00 mol)
ΔHfus is the molar heat of fusion for ice
Then, we need to calculate the heat absorbed for the temperature change of the water from 0°C to 100°C. This is given by the equation:
q2 = n * Cp * ΔT
Where:
q2 is the heat absorbed
n is the number of moles (which is still 2.00 mol)
Cp is the molar heat capacity of water
ΔT is the change in temperature (100°C - 0°C)
Finally, we need to calculate the heat absorbed for the phase change from water to steam at 100°C. This is given by the equation:
q3 = n * ΔHvap
Where:
q3 is the heat absorbed
n is the number of moles (yes, still 2.00 mol)
ΔHvap is the molar heat of vaporization for water
Now you have all the equations, but do you have all the variable values?
To calculate the total heat absorbed during this process, we need to consider the heat required for three separate steps:
1. Heating the ice from -30.0°C to 0°C
2. Melting the ice at 0°C
3. Heating the water from 0°C to 140.0°C
4. Evaporating the water at 100°C
5. Heating the steam from 100°C to 140.0°C
Step 1: Heating the ice from -30.0°C to 0°C
The specific heat capacity of ice is 2.09 J/g°C, and the molar mass of water is 18.015 g/mol. The heat absorbed in this step can be calculated using the formula:
Q1 = m * c * ΔT
Where:
m = mass of the ice = moles of ice * molar mass of water
c = specific heat capacity of ice
ΔT = change in temperature
m = 2.00 mol * 18.015 g/mol = 36.03 g
ΔT = 0°C - (-30.0°C) = 30.0°C
Q1 = 36.03 g * 2.09 J/g°C * 30.0°C = 2,373.447 J
Step 2: Melting the ice at 0°C
The heat required to melt ice can be calculated using the formula:
Q2 = m * ΔH_fusion
Where:
ΔH_fusion = enthalpy of fusion of ice = 6.01 kJ/mol
Q2 = 2.00 mol * 6.01 kJ/mol = 12.02 kJ
Step 3: Heating the water from 0°C to 100°C
The specific heat capacity of water is 4.18 J/g°C. The heat absorbed in this step can be calculated using the formula:
Q3 = m * c * ΔT
Where:
m = mass of water
c = specific heat capacity of water
ΔT = change in temperature
m = 2.00 mol * 18.015 g/mol = 36.03 g
ΔT = 100.0°C - 0°C = 100.0°C
Q3 = 36.03 g * 4.18 J/g°C * 100.0°C = 15,066.054 J
Step 4: Evaporating the water at 100°C
The heat required to vaporize water can be calculated using the formula:
Q4 = m * ΔH_vaporization
Where:
ΔH_vaporization = enthalpy of vaporization of water = 40.7 kJ/mol
Q4 = 2.00 mol * 40.7 kJ/mol = 81.4 kJ
Step 5: Heating the steam from 100°C to 140.0°C
The specific heat capacity of steam is 2.03 J/g°C. The heat absorbed in this step can be calculated using the formula:
Q5 = m * c * ΔT
Where:
m = mass of steam
c = specific heat capacity of steam
ΔT = change in temperature
m = 2.00 mol * 18.015 g/mol = 36.03 g
ΔT = 140.0°C - 100.0°C = 40.0°C
Q5 = 36.03 g * 2.03 J/g°C * 40.0°C = 2,927.052 J
Now, let's calculate the total heat absorbed:
Total Q = Q1 + Q2 + Q3 + Q4 + Q5
Total Q = 2,373.447 J + 12,020 J + 15,066.054 J + 81,400 J + 2,927.052 J
Total Q = 113,786.553 J
To convert this value to kJ, divide by 1000:
Total Q = 113,786.553 J / 1000
Total Q ≈ 113.8 kJ
Therefore, the total amount of heat absorbed during this process is approximately 113.8 kJ.
To calculate the total amount of heat absorbed (in kJ) during the conversion of ice to steam, we need to consider the following steps and the respective heat exchange:
1. Heating the ice from -30.0°C to its melting point at 0.0°C.
2. Melting the ice at 0.0°C to form water at 0.0°C.
3. Heating the water from 0.0°C to its boiling point at 100.0°C.
4. Vaporizing the water at 100.0°C to form steam at 100.0°C.
5. Heating the steam from 100.0°C to 140.0°C.
Let's break down each step and calculate the amount of heat absorbed in each:
1. Heating the ice from -30.0°C to 0.0°C:
The specific heat capacity of ice is approximately 2.09 J/g°C.
The molar mass of water is 18.015 g/mol.
The molar amount of ice is 2.00 mol.
The temperature change is 0.0°C - (-30.0°C) = 30.0°C.
The amount of heat absorbed (Q1) can be calculated as:
Q1 = molar mass × molar amount × specific heat capacity × temperature change
= 18.015 g/mol × 2.00 mol × 2.09 J/g°C × 30.0°C
= 3616.54 J
Therefore, the amount of heat absorbed in step 1 is 3616.54 J.
2. Melting the ice at 0.0°C:
The heat of fusion (enthalpy of fusion) of water is approximately 6.02 kJ/mol.
The molar amount of ice melted is 2.00 mol.
The amount of heat absorbed (Q2) can be calculated as:
Q2 = molar amount × heat of fusion
= 2.00 mol × 6.02 kJ/mol
= 12.04 kJ
Therefore, the amount of heat absorbed in step 2 is 12.04 kJ.
3. Heating the water from 0.0°C to 100.0°C:
The specific heat capacity of water is approximately 4.18 J/g°C.
The molar mass of water is 18.015 g/mol.
The molar amount of water is 2.00 mol.
The temperature change is 100.0°C - 0.0°C = 100.0°C.
The amount of heat absorbed (Q3) can be calculated as:
Q3 = molar mass × molar amount × specific heat capacity × temperature change
= 18.015 g/mol × 2.00 mol × 4.18 J/g°C × 100.0°C
= 15036.72 J
Therefore, the amount of heat absorbed in step 3 is 15036.72 J.
4. Vaporizing the water at 100.0°C:
The heat of vaporization (enthalpy of vaporization) of water is approximately 40.7 kJ/mol.
The molar amount of water vaporized is 2.00 mol.
The amount of heat absorbed (Q4) can be calculated as:
Q4 = molar amount × heat of vaporization
= 2.00 mol × 40.7 kJ/mol
= 81.4 kJ
Therefore, the amount of heat absorbed in step 4 is 81.4 kJ.
5. Heating the steam from 100.0°C to 140.0°C:
The specific heat capacity of steam is approximately 2.03 J/g°C.
The molar mass of water is 18.015 g/mol.
The molar amount of steam is 2.00 mol.
The temperature change is 140.0°C - 100.0°C = 40.0°C.
The amount of heat absorbed (Q5) can be calculated as:
Q5 = molar mass × molar amount × specific heat capacity × temperature change
= 18.015 g/mol × 2.00 mol × 2.03 J/g°C × 40.0°C
= 2918.32 J
Therefore, the amount of heat absorbed in step 5 is 2918.32 J.
To calculate the total amount of heat absorbed, we sum up the heat absorbed in each step:
Total heat absorbed = Q1 + Q2 + Q3 + Q4 + Q5
= 3616.54 J + 12.04 kJ + 15036.72 J + 81.4 kJ + 2918.32 J
Finally, we convert the combined units to kJ:
Total heat absorbed = (3616.54 J + 12040 J + 15036720 J + 81400 J + 2918320 J) / 1000
= 1.86 kJ + 12.04 kJ + 15.04 kJ + 81.4 kJ + 2.92 kJ
= 113.26 kJ
Therefore, the total amount of heat absorbed during the conversion of 2.00 mol of ice at -30.0°C to steam at 140.0°C is approximately 113.26 kJ.