Impacts typically occur at a range of 20 to 70 km/s.
Let's look at the 20 km/s one.
1 kg of mass at 20 km/s = 1kg * 20,000 m/s * 20,000m/s /2 = 200 million joules.
To bring ice from absolute zero to melting point requires 1000 * 273 * 0.5 = 136 thousand joules. Yes, melting it will take energy and so will vapourization, but this is VERY small potatoes compared to the kinetic energy. (Water ice is the largest part of comet ice. Ammonia has a somewhat higher specific heat, methane, far lower.)
So as a source of coolth, a comet impact doesn't rate.
So let's instead, invite Archanangel Gabriel to come down and gently lay the comet down to melt and chill the area.
Large comet 20 km diameter. Assume solid ice for the sake of simplicity. 20 km sphere is 4/3 * pi * 10^3 = about 4200 cubic km of ice.
Wow. Lots of ice cubes for my martini!
Comparison: Canada has an area of 10 million km2. If you had 1/10 of a meter of ice (4 inches...) you would have 1/10,000 of a km * 10 million km = 1000 km3. So your comet has about the four times the chilling effect as spring in Canada. (Ok, 10 cm is arbitrary.)
Or the same as 40 cm of ice on Canada.
Now a square meter with 40 cm of ice would be about 400 kg. (Please, do NOT muddle the waters with the different density of water and ice. This is a BOTE calc.)
1 kg of ice takes about 500 kJ to melt, assuming it's starting from close to absolute zero. 400 kg takes 200,000 kJ
Solar constant is about 1 kW/m2 at noon. (Yes this is variable depending on latitude, time of year, cloud cover, and things that otherwise were thought only to affect the flavour of pizza) So 200,000 seconds of noonday sun would melt the ice. About 55 hours.
Which would be about 200 hours due to night, and low angles. 8 days. Ish.
However Canada isn't the whole world. Only about 1/50 of it. so 8/50 or about 4 hours sunlight over the earth to melt the comet.
Like throwing an ice cube from your drink into a campfire.
You're left with nuclear winter possibilities. Others have addressed that.