Reference for a folklore result about $T^1(B/k;M)$ If $B$ is a $k$-algebra, let $T^1(B/k;M)$ denote the first cotangent functor. It classifies first order deformations of the scheme $\mathrm{Spec} B$. 
Now, if $X \subset \mathbb P^n$ is a smooth projective scheme, we can look at its homogeneous coordinate ring $B$. Then $T^1(B/k;B)$ is a graded module, an under certain conditions it is true that its degree zero part classifies first order  deformations of the projective scheme $X$.
For example, I've computed in Macaulay2 that for the quintic, $T^1(B/k;B)_0=k^{101}$ which agrees with $\dim_k H^1(X,\mathcal T_X)$.
It appears that it is a folklore result that $T^1(B/k;B)_0=H^1(X,\mathcal T_X)$, when $X$ is a smooth projective scheme, but I haven't been able to find a reference for this fact.
I'm grateful for a reference or a proof (and also under what restrictions this statement is true).
 A: This is almost true. Indeed $T^1(B )$ classifies first order deformations of the affine cone over $X$. If the depth at the vertex $v$ of the affine cone satisfies   $depth_v B \geq 3$ there is a long exact sequence in cohomology (coming from the relative tangent) 
$$ \ldots H^1(X, \mathcal{O}_X) \to (T^1_B)_0 \to H^1(T_X) \to H^2(X, \mathcal{O}_X) \to \ldots$$
The depth condition implies that $ H^1(X, \mathcal{O}_X)=0$, but $H^2(X, \mathcal{O}_X)$ can be non-zero.
In this case $(T^1_B)_0$ is just a subspace of $H^1(T_X)$ (actually corresponding to the conical deformations of the affine cone).
An example to bear in mind is a K3 surface (for example as quartic in $\mathbb P^3$). Then $T^1(B)_0$ is 19 dimensional, whereas $H^1(T_X) \cong \mathbb C^{20}$.
In the case of a quintic threefold $W$ however, $H^1(W, \mathcal O_W)=H^2(W, \mathcal O_W)=0$, and this agrees with your computation.
As a reference, you may look at the original Schlessinger's paper (Schlessinger, Michael. "On Rigid Singularities." Rice Institute Pamphlet - Rice University Studies, 59, no. 1 (1973)), or at https://arxiv.org/pdf/1512.00835.pdf for more general results on the topic.
