calculate q , w , and ∆ E for melting 1.00 mol of ice at 0 ◦ C and 1.00 atm pressure
q = mass ice x dHfusion.
w = -p*delta V.
Calculate volume ice at zero C. I would look up the density of ice at zero C and liquid H2O at 0C, then solve for volume for 18.015g ice and water. I found that here of (0.9167 g/cc for ice @ 0C vs 0.9998 @ 0Cg/cc for liquid water).
http://en.wikipedia.org/wiki/Ice
That will give you delta V from which you can solve for w (in L*atm). Multiply by 101.325 to change to Joules.
Then E = q + w.
NOTE: The volume of ice is greater than the volume of liquid water. Watch the sign.
Calculate the enthalpy change upon converting 1.00 mol of ice at -25°C to water vapor (steam) at 125°C under a constant pressure of 1 atmosphere.
To calculate q, w, and ∆E for melting 1.00 mol of ice at 0°C and 1.00 atm pressure, we need to use the equations related to heat, work, and change in energy.
1. q (heat):
To melt ice, we need to provide heat to raise its temperature to the melting point (0°C) and then provide additional heat to convert it from a solid to a liquid state. The heat required to melt the ice can be calculated using the equation q = n∆H.
- ∆H is the enthalpy change per mole, which represents the heat required to convert the ice at 0°C to liquid water at 0°C. For the melting of ice at its melting point, the enthalpy change (∆H) is 6.01 kJ/mol.
Substituting the values into the equation, we get:
q = (1.00 mol) * (6.01 kJ/mol)
q = 6.01 kJ
Therefore, the heat required is 6.01 kJ.
2. w (work):
In this case, no work is being done because the system (the ice) is not expanding or compressing, and there is no external force acting on the system. Therefore, w = 0.
3. ∆E (change in internal energy):
The change in internal energy (∆E) is the sum of the heat (q) and the work (w):
∆E = q + w
∆E = 6.01 kJ + 0
∆E = 6.01 kJ
Therefore, the change in internal energy (∆E) is 6.01 kJ.
In summary:
- q = 6.01 kJ (heat required)
- w = 0 (work)
- ∆E = 6.01 kJ (change in internal energy)
To calculate q (heat transferred), w (work done), and ΔE (change in internal energy) for the process of melting 1.00 mol of ice at 0 °C and 1.00 atm pressure, we need to use the following equations:
1. q = m * ΔH_fusion
2. w = -P * ΔV
3. ΔE = q + w
Where:
- q is the heat transferred.
- m is the mass of the substance.
- ΔH_fusion is the enthalpy of fusion (heat of fusion) of the substance.
- w is the work done.
- P is the pressure.
- ΔV is the change in volume.
- ΔE is the change in internal energy.
First, let's calculate q using equation 1:
q = m * ΔH_fusion
The molar mass of water is approximately 18.02 g/mol, so the mass of 1.00 mol of ice is 18.02 g. The enthalpy of fusion of ice is close to 6.01 kJ/mol.
q = (18.02 g) * (6.01 kJ/mol)
q ≈ 108.36 kJ
Now, let's calculate w using equation 2:
w = -P * ΔV
Since the process is occurring at constant pressure (1.00 atm) and the volume does not change during melting, the work done can be assumed to be zero:
w = - (1.00 atm) * 0
w = 0
Next, we can calculate ΔE using equation 3:
ΔE = q + w
ΔE = 108.36 kJ + 0
ΔE = 108.36 kJ
Therefore, the values for q, w, and ΔE for melting 1.00 mol of ice at 0 °C and 1.00 atm pressure are approximately:
- q = 108.36 kJ
- w = 0
- ΔE = 108.36 kJ