Which amount of heat is released when 75.0 g of water at 0° C is frozen to ice?

Q = m·∆H(fusion) = 75.0 gms (80 calories/grm) = 6000 calories = (6000 calories)(4.184 joules/calorie) = 25,104 joules

To find the amount of heat released when water freezes to ice, we need to use the equation:

Q = m * ΔHf

where Q is the amount of heat released, m is the mass of the substance undergoing the phase change, and ΔHf is the heat of fusion, which is the amount of heat required to convert one unit of mass from a liquid to a solid at a constant temperature.

The heat of fusion for water is 334 J/g.

Given that the mass of water is 75.0 g and the temperature change is from 0°C to 0°C (the freezing point of water), we can substitute these values into the equation:

Q = 75.0 g * 334 J/g

Now we can calculate the amount of heat released:

Q = 25050 J

Therefore, 25,050 J (joules) of heat is released when 75.0 g of water at 0°C is frozen to ice.