Standard entropy of liquid lead at 500C?

[magnesium12]

Homework Statement

The standard entropy of lead at 25C is S(298)=64.80 J/Kmol.
The heat capacity of solid lead is Cp(s) = 22.13 + .01172T + 0.96x10⁵T^-2.
The heat capacity of liquid lead is Cp(l) = 32.51 - 0.00301T
Melting point is 327.4C
Heat of fusion is 4770J/mol.
Calculate the standard entropy of liquid lead at 500C.

Homework Equations

ΔS(500) = S(298) + ∫(Cp(s)/T)dT + ΔHfus/T + ∫(Cp(l)/T)dT
ΔS(500) = ∫(Cp(s)/T)dT + ΔHfus/T + ∫(Cp(l)/T)dT

The Attempt at a Solution

I'm just wondering which equation to use?
At first I thought the equation was just ΔS(500) = ∫(Cp(s)/T)dT + ΔHfus/T + ∫(Cp(l)/T)dT but the question includes the standard entropy at 25C, so do I have to account for that too by using ΔS(500) = S(298) + ∫(Cp(s)/T)dT + ΔHfus/T + ∫(Cp(l)/T)dT? But shouldn't the initial entropy not matter since entropy is a state function?
And if I do have to account for entropy at 25C, does that entropy stay the same up to the melting point so I wouldn't need to account for the ∫(Cp(s)/T)dT

 

[Chestermiller]

The Attempt at a Solution

I'm just wondering which equation to use?
At first I thought the equation was just ΔS(500) = ∫(Cp(s)/T)dT + ΔHfus/T + ∫(Cp(l)/T)dT but the question includes the standard entropy at 25C, so do I have to account for that too by using ΔS(500) = S(298) + ∫(Cp(s)/T)dT + ΔHfus/T + ∫(Cp(l)/T)dT?

Yes.

But shouldn't the initial entropy not matter since entropy is a state function?

When you are talking about the "standard entropy," you are talking about the change in entropy from the state of pure elements comprising the compound at 25 C.

And if I do have to account for entropy at 25C, does that entropy stay the same up to the melting point so I wouldn't need to account for the ∫(Cp(s)/T)dT

What would make you think this?