Well, in answer to the caption, lessee...
Assuming for now that there's no air resistance or anything like that... she just starts spinning at the start of the night, slowing the Earth, and then stops spinning at the end, speeding the Earth back up again. This is probably not even remotely like what would really happen, over such a long time air resistance would become substantial, but this system is a lot simpler... I don't want to think about what would happen with the atmosphere...
The moment of inertia of Earth
I = 8.034Ã—10^37 kg m^2.
Now, suppose we wanted to delay dawn by a total of 1 second, over the course of a 12 hour period (say, the entire night). The period of the Earth during that time would increase from 24 hours to 24 hours 2 seconds, or a difference in angular velocity of:
Δω = 2π/86400 - 2π/86402
= 2π(86402 - 86400)/(86402 * 86400)
= 4π/(86402 * 86400)
= 1.6833Ã—10^-9 rad/s
This gives the difference in rotational inertia as
ΔL = IΔω
= 1.352Ã—10^29 Js
The moment of inertia of a human body is much harder to find... the best I could find is a model
which gives the moment of inertia around a vertical axis as around:
I = 1.5 kg m^2
Now, with the rotational inertia calculation above:
ω = L / I
= 9.016Ã—10^28 rad/s
or 1.434Ã—10^28 rotations per second.
And that's assuming she's standing on the north pole... if she's standing elsewhere on the planet, you get confusing stuff in three dimensions... but I'm pretty sure she'd have to spin even faster just to get that one extra second.
Going the other way, assuming she spins about once per second:
ω = 2π rad/s
L = 4.189 Js
Δω = 5.214Ã—10^-38
which comes to a difference in period of about 1.112Ã—10^-48 seconds.
Now, you'll notice that the speed in the first part would mean most of her body is travelling significantly faster than the speed of light, and that the time in the second is less than the Planck time. But since I couldn't be bothered getting into relativity or quantum theory, that'll do for me.
Sorry for the downer.