Why peltier elements have smaller heatsink on cold side

A Peltier element consists of two thin ceramic plates with small "cubes" of different materials, which are separated by a small airgap:

Broken peltier element by Frank Andre (Own work) [Public domain], via Wikimedia Commons

Heat is transferred through the cubes from one plate to the other, but not laterally from one cube to its neighbor due to the air gap. Also the ceramic would need to have an exceptional thermal conductivity to spread heat from a small spot over the entire surface, which it does not have. (I guess even copper would not be sufficient, because the plates are so thin)

Also, each cube has the same "cooling power", so the power will not increase where your hot component touches the peltier element.

So, I'd advise you to use a heat spreader, like a (thicker) plate of copper between the peltier element and the hot component.

Please also keep in mind this:

A peltier element pumps a constant amount of heat energy per second (or has a constant "pumping power") from the cool to the warm side:

$$P_{cool}=const.$$

This is one parameter given in each datasheet.

On the other side, head is also transferred back by ordinary thermal conductivity, which depends on the temperature difference:

$$P_{thermal}(T)=C\cdot\Delta T$$

As result, the actual, total (usable) pumping power also depends on temperature difference:

$$P(\Delta T)=P_{cool}-C\cdot\Delta T$$

At a certain temperature difference, both processes cancel out each other, so the actual, total pumping power is zero. This temperature difference is also given in each datasheet.

As example: A 100W peltier element with a maximum temperature difference of 60°C can remove only 50W, when the temperature difference is 30°C.

(There are more effects, but this is the most simple thing one should keep in mind)

Another note: A peltier element can get pretty cold and could cause drops of dewed water near you component. This is another reason why you should use a heat spreader

The simple answer is: maybe. There are a few things you've skipped over. First is the plate thermally insulated? If the plate is covered in thermal insulation, you can certainly cool it to 10 C, and probably get down to about -20 C if the insulation is very good.

100 30-watt TECs will draw 3 kW. A general rule of thumb with Peltiers is that you can get 10% of that in cooling. So. Let's run things backwards. 10 C is about 15 C below room temperature. If you were to run 300 watts of heating into your plate, could you get a 15 C rise in temperature? If so, you have a pretty good shot at cooling it to your specifications. On the other hand, if it's exposed to air, it may be hard to get the effect you need.

It's not reasonable to think that you can pack your TECs perfectly. As KalleMP has pointed out, you need to leave some room to run wires, but let's say you used a 9x9 array of coolers, with about 5 mm between units. Then you would have 81 coolers, with a power dissipation of about 2400 watts, and about 240 watts cooling. Let's assume your controlled plate is perfectly insulated. Let's also assume your plate is 6 mm thick, rather than 3, so as to let the metal spread the temperature evenly. Then the volume of the plate is .015 cu meters, and the weight is 40.5 kg. The specific heat of aluminum is .9 J/deg-kg. To drop the temperature of the plate 15 C will take an energy of $$E = 0.9 \times 15 \times 40.5 = 547 \text{J} $$ and at 240 watts it will only take about 2.3 seconds to drop the temperature you need. The less the insulation, the more power you'll need, and keep in mind that Peltiers do not produce much temperature differential when run at high current.

Your construction approach is appropriate, but as has been mentioned, in principle you might need a pretty good cooling system to handle 2.4 to 3 kW of waste heat.