CCDs are exceptional image sensors in a scientific camera. They're very sensitive to light, and offer extremely predictable and repeatable results at any given exposure and temperature. However, CCDs are also sensitive to heat, not just to the visible light that you're trying to record. Cooling a CCD reduces the buildup of this "thermal current" in images. Here's a graph showing how mean thermal current builds at various temperatures and exposure lengths on a QSI 583 with a Kodak KAF8300 Sensor. This data was collected by subtracting a master bias frame at each target temperature from each of 10 individual dark frames and then calculating the mean thermal current for each data point.
QSI 583 (KAF8300) Mean Thermal Current (ADUs, Gain = 0.5 e/ADU)
Temp
 60 sec
 300 sec
 600 sec

0C

3.12 
13.91 
25.98

10C 
1.77 
6.04 
11.10

15C 
1.31 
4.17

6.95

20C 
1.03 
3.43

5.57

The values in the table and chart above are expressed in ADUs. In the table below the thermal current is converted to electrons using the QSI 583 High Gain setting of 0.5 e/ADU.
QSI 583 (KAF8300) Mean Thermal Current (e)
Temp
 60 sec
 300 sec
 600 sec

0C

1.56 
6.96 
12.99

10C 
0.88 
3.02

5.55

15C 
0.65 
2.09

3.48

20C 
0.52 
1.72

2.78

Note that even with a 10 minute exposure (600 sec) at 10C, the mean contribution of thermal current is just 5.55 electrons. Keep in mind that dark current is unwanted signal, not noise. The dark current can be subtracted but the noise from the dark current combines with other sources of noise to determine the total noise in the image. The noise in any signal is the square root of the signal, or in this case, SQRT(5.5) = 2.35 e RMS. Read noise of the QSI 583 is ~8e RMS.
The major sources of noise in an image are Shot noise (from the light you're trying to record), Thermal Current noise and Read noise. Noise combines as the square root of the sum of the squares.
A 10 minute dark frame at 10C has the following major sources of noise: Read Noise = 8e RMS Thermal Noise = 2.35e RMS Total noise in a 10min dark at 10C is SQRT (8^2 + 2.35^2) = 8.3e RMS Note that the larger source of noise dominates the result.
Reducing the temperature to 20C would yield these results: Thermal Noise = SQRT(2.78) = 1.67e RMS
Total noise in a 10min dark at 20C is SQRT (8^2 + 1.67^2) = 8.2e RMS
Dropping the temperature from 10C to 20C results in a drop in total noise of just 0.1e RMS in a 10minute exposure.
Now consider a 10minute light frame at 10C. Assume the brightest stars are close to saturation. The brightest areas of nebulosity are at 3200 ADUs (1600e) and the dimmest areas are near the sky background of 450 ADUs (225e).
Total noise in areas of bright nebulosity in this example 10minute light frame at 10C: Read noise = 8e RMS Thermal noise = 2.35e RMS Shot noise = SQRT(1600e) = 40e RMS *The square root of the signal level Total noise = SQRT (40^2 + 8^2 + 2.35^2) = 40.9 e RMS Note again that the largest source of noise, the shot noise in this case, dominates the total noise in the image.
And for the faintest areas of nebulosity near the sky background: Read noise = 8e RMS
Thermal noise = 2.35e RMS
Shot noise = SQRT(225e) = 15e RMS *The square root of the signal level
Total noise = SQRT (15^2 + 8^2 + 2.35^2) = 17.2 e RMS
Even at these faint signal levels, the shot noise still dominates the total noise in the image, reducing the noise from thermal current to insignificance.
See "Understanding CCD Noise" for a more complete discussion of the sources of noise in CCD Images.
For more background on Shot Noise and the quantum nature of light, see this Wikipedia article>>
