The article says that the water is returned to the lake at about the same temperature that it leaves...though I'm unclear how that's possible.
The most plausible explanation is that the article is wrong, and that the author meant to say that the lake is about the same temperature as it was before. Which is generally true; although the water being returned is slightly warmer, once you get more than a few meters away the difference is negligible. There is a ton of research on this phenomena with respect to the lake source cooling plant in Ithaca, because people obviously had the same concerns about that one before it was built.
Although L. Ontario's surface can hardly be described as "warm"
1 BTU is about 1kJ, so every 1 square foot of cooling (per hour) could theoretically be offset by 25,000J / 4.18J = ~5980g of water being heated 1 degree C. Density of water is 1g/mL, so it takes approximately 6L per hour per square foot if you assume that you don't raise the temperature of the water more than 1 degree C.
They're talking about 50 million square feet with this system, so that's 300 million liters per hour, or about 80 million gallons per hour, or 1.3M gallons per minute. For context, the third water tunnel project in NYC was supposed to deliver 1.3B gallons per day, or about 903,000 GPM assuming constant flow. That's a big pipe.
I tried to find actual citations of their flow rate, but haven't found anything yet. Regardless, I am skeptical of the claim that they aren't returning warmer water to the lake. I'm not worried about it (lake Ontario is gigantic), but I don't believe the article.
(edit: 1BTU is the energy required to heat 1lb of water by 1 degree F at maximum density, so you could work it out directly from the bizarro-world units of HVAC systems.)
The system has the ability to absorb:
H2o_heat_capacity*mass flow rate * delta(T) = cooling_power.
The first term is know and very large. The last term is determined by the number of users and is estimated with forcast usage models. Ask an Env Eng a desired delta(T) and you get the mass flow rate required.
It's a lot of flow.
This is net cooling L. Ontario because Darlington (hence the its cooling loop) doesn't have to run as hard to power all the displaced ACs
Return pipe is municipal wast treatment, which is probably enough to make that water essentially the same temperature as surface level waters once it returns to the lake.