4 moles of water are heated from 240 K to 310. find delta entropy and graph entropy against K . Info : delta H fusion of water is 5 kJ/mol

delta H vaporization of water is 41.4 kJ/mol Cp of ice is 36 j/mol - k
Cp of water in liters is 74 j/mol-k

From the information given, I think it is the enthalpy, ∆H, you need to find for the overall process.

You have the following calculations:
1. Heating of ice from 240 to 273 K:
∆H1 = (36 j/mol.ºK)(4.0 mol)(273K-240K)
2. Melting of ice:
∆H2 = (4.0mol.)(5.0kJ/mol
3. Heating of water:
∆H3 = (74 J/mol.ºK)(4.0mol)(310ºK-273ºK)
Calculate and add the 3 enthalpies and add them up.
NOTE: 310 K is below the boiling of water. No significant evaporation takes place.

To find the delta entropy (ΔS), we can use the equation:

ΔS = ΔH / T

Where:
ΔS = Change in entropy
ΔH = Change in enthalpy
T = Temperature

In this case, we have two temperature points, 240 K and 310 K. We need to calculate the change in entropy for each temperature point, and then plot the entropy values against temperature.

1. ΔS for ice (240 K):
ΔS_ice = ΔH_fusion / T
= 5 kJ/mol / (240 K)

2. ΔS for water (310 K):
ΔS_water = (ΔH_fusion + ΔH_vaporization) / T
= (5 kJ/mol + 41.4 kJ/mol) / (310 K)

Now, let's calculate the values:

1. ΔS_ice = 5 kJ/mol / 240 K = 0.0208 kJ/(mol·K)

2. ΔS_water = (5 kJ/mol + 41.4 kJ/mol) / 310 K = 0.548 kJ/(mol·K)

Now, we can plot the entropy values against the temperature:

Temperature (K) | Entropy (ΔS) (kJ/(mol·K))
-------------------------------------------
240 | 0.0208
310 | 0.548

Plotting these two points on a graph will give us a linear relationship between entropy and temperature.