Why is the intensity of Hawking radiation dependent on the size of the black hole it comes from? I am asking for a non mathematical answer. I think I have a pretty good understanding of physics without the maths, which unfortunately eludes my understanding. It may be the answer would be obvious if I understood the maths, but I don't. As I see it, the event horizon has the same property for any size of black hole; it is the point from which light cannot escape. Given that this property is the same regardless of the size of the hole, it seems reasonable to suppose that the intensity of the Hawking radiation would be constant per unit area of the horizon. I know I am wrong, but I don't know why. A non mathematical explanation would be appreciated.
 A: Unfortunatately maths is necessary for quantum effects and Hawking radiation is a quantum effect . His original calculations gave

that the black hole  should glow like a blackbody of temperature
                6 × 10-8/M kelvins,


You can see that the calculations gave the temperature  inversely proportional to the mass  of the black hole, and the size of the black hole depends on the mass ( see table).
A: It is hard to give a good answer without using math, but here is a try. The Hawking- radiation of a black hole is thermal radiation (blackbody radiation), so its intensity only depends on the temperature of the black hole. The Hawking-temperature of a black hole is inversely proportional to its mass. This makes a massive black hole radiate at a much lower intensity than a small one. A more massive black hole also has a larger area to radiate from, but since the intensity of radiation is so strongly dependent on the temperature, the total radiated power from a black hole is inversely proportional to the mass squared. 
A: As I mentioned in a comment to John Rennie's answer, a Schwarschild black hole has only one non-zero curvature invariant -- a quantity that can be defined in a coordinate-independent way. The usual form of this invariant is $$\Psi_2 = -\frac{r_s}{2r^3}$$
where $r_s$ is the radius of the black hole, $r_s = 2Gm/c^2$. (The notation $\Psi_2$ is standard). This is the only curvature invariant in the sense that any other is necessarily a function of $\Psi_2$. The Kretschmann scalar mentioned by John is indeed proportional to the square of $\Psi_2$. 
As you can see $\Psi_2$ has the units of $\text{m}^{-2}$. Since this is the only coordinate-independent quantity describing the black hole, we should expect that the typical wavelength is proportional to the value of $\lambda = 1/\sqrt{-\Psi_2}$ on the event horizon, which is $$\lambda = {\sqrt{2} r_s} = \frac{\sqrt{8} Gm}{c^2}$$
simply because up to a constant, this is the only physical quantity with units of length.
In quantum mechanics energy is inversely proportional to wavelength, $E = 2\pi\hbar c/\lambda$, and for a black-body spectrum the temperature is proportional to the typical energy, $T = aE/k_B$, where $a$ is some number that one can look up. Putting together all the steps, $$T \propto 1/m.$$
You have to do a more careful analysis to find the proportionality constant (or perhaps more importantly: argue that the black hole radiates at all...) but I think you can understand that $T \propto 1/m$ from the uniqueness of $\Psi_2$ and this dimensional analysis.
A: Any non-mathematical answer is obviously going to be an oversimplification, but as long as you're happy with that here is my oversimplification.
The temperature of a big black hole is lower than a small black hole because the curvature at the event horizon decreases with radius. It's the curvature, i.e. the bending, that determines the temperature - more curvature = higher temperature. The curvature is a fourth rank tensor, but a convenient measure of it is the Kretschmann scalar:
$$ R_{abcd}R^{abcd} = \frac{12r_s^2}{r^6} $$
where $r_s$ is the radius of the event horizon and $r$ is the distance from the centre of the black hole. At the event horizon we have $r = r_s$ and therefore the Kretschmann scalar curvature is:
$$ R_{abcd}R^{abcd} = \frac{12}{r_s^4} $$
So the Kretschmann scalar falls as the fourth power of the black hole size. That's basically why bigger black holes are colder.
A: The wavelength of radiation escaping from a black hole is comparable to the radius of the black hole. So larger black holes have longer wavelength Hawking radiation, and thus lower energy.
Why is this true? I have a vague intuition that if the wavelength of some particle is smaller than the black hole, the black hole will just absorb it because it is acting classically with respect to the black hole, but if the wavelength is larger than the black hole, the black hole really can't absorb it because it doesn't "fit" inside. However I don't see how this vague intuition fits together with the math. 
