How many joules of heat are removed to completely covert 70.0 grams of water to ice at zero Celcius?
q = 80cal/gm x 70gms = 5600cals x 4.184J/cal = 23,430 joules ~ 23,000 joules with 2 sig gigs.
To calculate the amount of heat removed to completely convert 70.0 grams of water to ice at zero degrees Celsius, we need to use the heat formula:
q = m * ΔH
where:
q is the amount of heat,
m is the mass of the substance, and
ΔH is the heat of fusion for the substance.
For water, the heat of fusion (ΔH) is 334 J/g (joules per gram). Therefore, we can substitute these values into the formula:
q = 70.0 g * 334 J/g
q ≈ 23,380 J
So, approximately 23,380 joules of heat are required to completely convert 70.0 grams of water to ice at zero degrees Celsius.