OK, back of the envelope time. Suppose in summer the net inflow of heat is 800 W/m² of the sun-facing aspect of the conservatory (allowing somewhat less than 1 kW/m² for heat loss, particularly through assumed ventilation) and we want to soak up, say, 4 hour's worth of heat (i.e., effectively the warmest part of the afternoon though, of course, actual absorption would spread over more of the day).
Total energy to store is 800 × 4 × 3600 = 11'520 kJ/m².
Suppose the temperature of the water rises from 20 to 25°C (trying to keep the air temperature to, say, 30°C instead of 40°C) the energy absorbed by the water (with a heat capacity of 4.2 kJ·kg⁻¹·K⁻¹) is 5 * 4.2 = 21 kJ·kg⁻¹.
Dividing the second number into the first we need 548 kg of water for each square metre of the conservatory facing towards the sun.
In other words, with these assumptions the tank would need to have a capacity equivalent to one 548 mm thick across the north facing side of the conservatory. That's big but not totally implausible. Still, it shows that something of only a few hundred litres is not likely to help much.
Another consideration is the rate at which heat can be moved in and out of the tank. This might be of interest:
http://njhurst.com/blog/01165734373and look for "wootank" in the middle of this post:
http://njhurst.com/blog/01163886516