Calculate the heat released when 25.0 grams of water freezes at 0 degrees celsius.
a) Look up the molar heat (enthalpy) of fusion of water.
b) Convert 25 grams of H2O to moles by dividing by the formula mass of H2O.
c) Multiply (a) times (b)
GK's method will work but let me save you a step.
The latent heat of fusion of water (the heat released when water turns to ice at 0 C) is 79.8 calories per gram. Many textbooks round it off to 80 cal/g. Multiply 79.8 by 25.0 grams for the answer.
To calculate the heat released when water freezes, you need to use the equation:
q = m * ΔH
where:
q is the heat released or gained (in joules)
m is the mass of the water (in grams)
ΔH is the heat of fusion or enthalpy change (in joules/gram)
The heat of fusion for water is 334 J/g.
Step 1: Convert the mass of water from grams to kilograms.
25 grams = 0.025 kilograms
Step 2: Plug the values into the equation.
q = 0.025 kg * 334 J/g
Step 3: Calculate the heat released.
q = 8.35 J
Therefore, when 25.0 grams of water freezes at 0 degrees Celsius, it releases 8.35 joules of heat.