OK, first, I wanted to determine the volume of the Enterprise. I've been informed that it's 680 by 240 by 87 meters. Because of the sheer scale of a planet, however, I'm going to knock it to kilometers now and save myself a tiny bit of trouble later. So that's .68 by .24 by .087 km. Now, the Enterprise isn't a box, so I can't just multiply them and call that the volume. But I'm going to make the guessimation that if it were put in such a box, it would fill it about 3/4ths the volume. Is that fair?
OK, so it's (and tell me if I fucked up the math anywhere and I'm rounded to the third place):
(.68 x .24 x .087)(.75) =
(.014)(.75) = .011 km³
Right, so the guessimated volume of the Enterprise is .011 km³. Now we have a ratio! So it's .011 km³ per 7 min of charge time.
But what of the Earth? According to the Planetary fact sheet, the Earth has a mean radius of 6371.0 km. We need to find the surface area, then calculate land area. The Earth is 70% covered in water, so I'm just going to assume he Toaster Ovens the land parts where people live.
So:
(4 x pi x (6371)²)(.7) =
(4 x 3.142 x 40,589,641)(.7) =
(510,130,608.088)(.7) = 357,091,425.662 km²
Fabulous. Now lets say that he's got to irradiate a kilometer high, just to cube that. So the Scimitar needs to zap 357,091,425.662 km³. This volume is a good bit larger than the Enterprise, eh?
Anyway:
(357,091,425.662 / .011) = 32,462,856,878.327
That's right, the volume is 32,462,856,878.327 that of the Enterprise. Multiply that times 7 minutes and we get 227,239,998,148.290 minutes. That's 432,048 years and change!

(Disclaimer: this thread is rated not-to-serious. The numbers, to the best of my admittedly rusty ability are real, but I'm just having a bit of fun. Flame and look like a fool.)