The entropy of melting for fcc metals is nearly constant according to the interstitialcy theory of condensed matter and the empirical Richard’s rule.

Granato A V 1992 Phys. Rev. Lett. 68 974

Granato A V 2006 J. Non-Cryst. Solids 352 4821

M. Polak and L. Rubinovich 2019 J. Phys.: Condens. Matter in press https://doi.org/10.1088/1361-648X/ab0865

“that entropy of vaporisation is almost same (85 J/ (K.mol)) for lot of liquids near melting point” – no, you should use the normal boiling temperature.

Concerning the melting entropy: This is very variable, as it depends on the crystal structure. For a series of substances with the same crystal structure (e.g., metals with fcc structure, as Leonid Rubinovich pointed out), a Trouton-like rule may apply. But otherwise you cannot use this rule. Consider, for example, N2, which forms a rotator phase in which the molecules can rotate at the lattice sites and which therefore has a small melting entropy, in constrast to HCl, where the rotation is hindered because of hydrogen bonds.

The so-called "entropy " doesn't exist at all.

During the process of deriving the so-called entropy, in fact, ΔQ/T can not be turned into dQ/T. That is, the so-called "entropy " doesn't exist at all.

The so-called entropy was such a concept that was derived by mistake in history.

It is well known that calculus has a definition,

any theory should follow the same principle of calculus; thermodynamics, of course, is no exception, for there's no other calculus at all, this is common sense.

Based on the definition of calculus, we know:

to the definite integral ∫T f(T)dQ, only when Q=F(T), ∫T f(T)dQ=∫T f(T)dF(T) is meaningful.

As long as Q is not a single-valued function of T, namely, Q=F( T, X, …), then,

∫T f(T)dQ=∫T f(T)dF(T, X, …) is meaningless.

1) Now, on the one hand, we all know that Q is not a single-valued function of T, this alone is enough to determine that the definite integral ∫T f(T)dQ=∫T 1/TdQ is meaningless.

2) On the other hand, In fact, Q=f(P, V, T), then

∫T 1/TdQ = ∫T 1/Tdf(T, V, P)= ∫T dF(T, V, P) is certainly meaningless. ( in ∫T , T is subscript ).

We know that dQ/T is used for the definite integral ∫T 1/TdQ, while ∫T 1/TdQ is meaningless, so, ΔQ/T can not be turned into dQ/T at all.

that is, the so-called "entropy " doesn't exist at all.

Why did the wrong "entropy" appear ?

In summary , this was due to the following two reasons:

1) Physically, people didn't know Q=f(P, V, T).

2) Mathematically, people didn't know AΔB couldn‘t become AdB directely .

If people knew any one of them, the mistake of entropy would not happen in history.

Please read my paper and those answers of the questions related to my paper in my Projects.

https://www.researchgate.net/publication/230554936_Entropy_A_concept_that_is_not_a_physical_quantity

“As long as Q is not a single-valued function of T, namely, Q=F( T, X, …), then,

∫T f(T)dQ=∫T f(T)dF(T, X, …) is meaningless.”

I am not quite sure. First, even if Q is not a state function, dQ (reversibly exchanged heat), is a well-defined property if the path of the process is well-defined. Second, 1/T is the integrating factor that turns dQ into dS, the differential of the state function entropy.

The premise of differential is the existence of a differentiable function !

The premise of integral is the existence of a function !

To a single integral, first of all, there must be a Function y=f(x), then, ∫ ydx=∫x f(x)dx may be meaningful, similarly,

∫ xdy=∫x xdf(x) =∫x xf'(x)dx=∫x G(x)df(x)

=∫x dF(x) is meaningful.

( in detail, to review the definition of the integral)

But, as we know, Q is NOT a single valued function of T, this alone already results in ∫1/TdQ is meaningless. And, in fact, as I pointed out in my paper, Q=f(T, V, P), so, 1/TdQ=1/Tdf(T, V, P) is meaningless, that is ∫T 1/TdQ = ∫T 1/Tdf(T, V, P) is not a meaningful integral, or say, it is not a integral at all.

When A and B are any two quantities, you CAN NOT turn a product of any two quantities AΔB into AdB directely, then said B=f(A) !

e.g., when L is the Cable Length, I is the current intensity of the Cable, could you turn LΔI into LdI ?

NO, you certainly can’t ! for there is not the function relationship as I=f(L) or L=f(I), so, Lid is meaningless.

Similely, could you turn 1/TΔQ into 1/TdQ directely then said Q=f(T) ?

NO, you certainly can’t, for there is not the function relationship as Q=f(T) at all, in fact we know Q≠f(T).

And once we know Q≠f(T), then we immediately know that 1/TΔQ can’t be turned into 1/TdQ.

In fact, Q=f(P, V, T), generally speaking, two of them(P, V, T) are independent.

“when L is the Cable Length, I is the current intensity of the Cable, could you turn LΔI into LdI ?”

You can, by using partial integration.

You said "You can, by using partial integration.", and, then, could you said the current intensity of the Cable I is the single value function of the Cable Length L ?

That is, do you mean I=f(L) ?

Sorry, I may have misunderstood your remark, as the ResearchGate website uses a font in which upper-case “i” and lower-case “L“ look the same.

Coming back to the entropy question: The premise of a differential is not a function, but a variable with a continuous range of values. Of course one can define dQ as as infinitesimal amount of heat, and then Delta Q = integral dQ is the loss or absorption of heat during a process.