The requirement "widely applied in scientific applications" is probably too high of a bar to get anything of interest, but I can think of several examples in real analysis where cardinalities higher than $2^{{\aleph}_0} = c$ are applied.
The standard argument that there exists a Lebesgue measurable set that isn't a Borel set is an example: The Cantor middle thirds set has $2^c$ many subsets, all of which have Lebesgue measure zero, but there are only $c$ many Borel sets.
A similar argument shows there exist Lebesgue measurable functions that are not Borel measurable: There are $2^c$ many characteristic functions of subsets of the Cantor middle thirds set, but there are only $c$ many Borel measurable functions.
Miroslav Chlebík, in this 1991 Proc. AMS paper, showed there exists $2^c$ many symmetrically continuous functions from the reals to the reals, and thus there exist symmetrically continuous functions that are not Borel measurable. When Chlebík's paper appeared, it had been a long unsolved question whether there even exists a symmetrically continuous function that isn't a Baire one function. See also the math StackExchange question Does $\lim_{h\rightarrow 0}\ [f(x+h)-f(x-h)]=0$ imply that $f$ is continuous?.
Because the union of interior of the unit disk in ${\mathbb R}^2$ with any subset of its boundary is a convex set, there exist $2^c$ many convex sets in ${\mathbb R}^{2}.$ Since there are only $c$ many Borel subsets of ${\mathbb R}^{2},$ it follows that there exist convex sets in ${\mathbb R}^2$ that are not Borel. (Note that this is so not true in ${\mathbb R}.)$