The standard molar heat of fusion of ice is 6020 J/mol. Calculate q, w, and (delta) E for melting 1.00 mol of ice at 0 deg. C and 1.00 atm pressure.
E = q+w
q = 6020 J/mol x 1 mol = ?
w = -pdV.
p is 101.325
Look up te density of ice and calculate the volume of a mol at zero C. Then calculate the volume of a mol of water at zero C. The differecne is dV. I found densities ice/H2O here.
http://en.wikipedia.org/wiki/Ice
To calculate q (heat transfer), w (work), and ΔE (change in internal energy) for melting 1.00 mol of ice at 0°C and 1.00 atm pressure, we need to consider the enthalpy change, pressure, and volume involved in the process.
First, let's determine the heat transfer (q) using the equation:
q = n * ΔH
where n is the number of moles and ΔH is the standard molar heat of fusion.
q = 1.00 mol * 6020 J/mol
q = 6020 J
Now, let's calculate the work done (w). In this case, since the system is not performing any mechanical work, we can assume w = 0 J.
Finally, to find the change in internal energy (ΔE), we can use the first law of thermodynamics:
ΔE = q + w
Since w is zero, the change in internal energy will be equal to the heat transfer:
ΔE = 6020 J
Therefore, the values for q, w, and ΔE are as follows:
q = 6020 J
w = 0 J
ΔE = 6020 J
To calculate q, w, and ΔE for melting 1.00 mol of ice at 0°C and 1.00 atm pressure, we can use the following equations and information:
1. q = ΔH_fus × n
2. w = -PΔV
3. ΔE = q + w
where:
- q is the heat absorbed or released during the process (in J),
- ΔH_fus is the standard molar heat of fusion of ice (in J/mol),
- n is the number of moles of substance,
- w is the work done on or by the system (in J),
- P is the pressure (in atm),
- ΔV is the change in volume (in L),
- ΔE is the change in internal energy (in J).
Given:
ΔH_fus = 6020 J/mol
n = 1.00 mol
P = 1.00 atm
Now, let's calculate q:
q = ΔH_fus × n
q = 6020 J/mol × 1.00 mol
q = 6020 J
Next, let's calculate w:
Since the melting is occurring at constant pressure, there is no change in volume, so ΔV = 0. Therefore:
w = -PΔV
w = 0
Finally, let's calculate ΔE:
ΔE = q + w
ΔE = 6020 J + 0
ΔE = 6020 J
Therefore, for melting 1.00 mol of ice at 0°C and 1.00 atm pressure, q = 6020 J, w = 0 J, and ΔE = 6020 J.