The standard molar heat of fusion of ice is 6020 J/mol. Calculate q, w, and ∆E for

melting 1.00 mol of ice at 0◦C and 1.00 atm pressure.

The standard molar heat of fusion of ice is

6020 J/mol. Calculate q, w, and ∆E for
melting 1.00 mol of ice at 0◦C and 1.00 atm
pressure.
1. q = 6020 J/mol, w = −6020 J/mol,
and ∆E = 0
2. q = 6020 J/mol, w = 0,
and ∆E = 6020 J/mol
3. q = −6020 J/mol, w = 0,
and ∆E = −6020 J/mol
4. q = −6020 J/mol, w = 6020 J/mol,
and ∆E = 0
5. q = 6020 J/mol, w = 6020 J/mol,

To calculate q, w, and ∆E for melting 1.00 mol of ice at 0°C and 1.00 atm pressure, we need to consider the energy exchanged during the phase change.

Here are the steps to calculate q, w, and ∆E:

1. Calculate q (heat) for the phase change using the formula:

q = n * ΔH_fusion

where n is the number of moles and ΔH_fusion is the standard molar heat of fusion of ice.

Substitute the values:

q = 1.00 mol * 6020 J/mol
q = 6020 J

Therefore, q = 6020 J

2. Calculate w (work). In this case, since the system is not doing any work and only undergoing a phase change, w = 0.

Therefore, w = 0 J

3. Calculate ∆E (change in internal energy) using the formula:

∆E = q + w

Substitute the values:

∆E = 6020 J + 0 J
∆E = 6020 J

Therefore, ∆E = 6020 J

To summarize, for melting 1.00 mol of ice at 0°C and 1.00 atm pressure:
- q (heat) is 6020 J
- w (work) is 0 J
- ∆E (change in internal energy) is 6020 J.

To calculate q (heat), w (work), and ∆E (change in internal energy) for melting 1.00 mol of ice at 0◦C and 1.00 atm pressure, we can use the equations:

q = ∆H
w = P∆V
∆E = q + w

Where:
∆H is the enthalpy change (in this case, the standard molar heat of fusion of ice)
P is the pressure
∆V is the change in volume (in this case, the change in the volume of ice as it melts)

Given:
∆H = 6020 J/mol (standard molar heat of fusion of ice)
P = 1.00 atm

Now, let's calculate q, w, and ∆E step by step:

Step 1: Calculate q (heat)
q = ∆H = 6020 J/mol

Since we are melting 1.00 mol of ice, q would be:
q = 6020 J/mol * 1.00 mol = 6020 J

So, q = 6020 J

Step 2: Calculate w (work)
To calculate work, we need to know the change in volume (∆V). In this case, as ice melts, there is no significant change in volume because the process occurs at constant pressure. Therefore, ∆V is approximately zero.

w = P∆V = 1.00 atm * ∆V

Since ∆V is negligible, w will also be close to zero.

w ≈ 0 J

Step 3: Calculate ∆E (change in internal energy)
∆E = q + w
∆E = 6020 J + 0 J

∆E = 6020 J

So, the values are:
q = 6020 J
w ≈ 0 J
∆E = 6020 J

Therefore, the heat required (q) to melt 1.00 mol of ice at 0◦C and 1.00 atm pressure is 6020 J, no work (w) is done as there is negligible volume change, and the change in internal energy (∆E) is 6020 J.