- #1

magnesium12

- 19

- 0

## 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 + 0.96x10

^{5}T

^{-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