What quantity of heat is required to vaporize 32.5 g of liquid water at 100°C?
heat= mass*Hvaporization
To calculate the quantity of heat required to vaporize a substance, you need to use the formula:
Q = m * ΔHvap
Where:
Q is the quantity of heat (in Joules)
m is the mass of the substance (in grams)
ΔHvap is the molar enthalpy of vaporization (in J/g)
To find the value of ΔHvap for water, you can consult a reference source or use the provided information. In this case, the molar enthalpy of vaporization of water is approximately 40.7 kJ/mol or 40.7 J/g.
Now, let's calculate the quantity of heat required to vaporize 32.5 grams of liquid water at 100°C:
Q = 32.5 g * 40.7 J/g
Q = 1321.75 J
Therefore, the quantity of heat required to vaporize 32.5 g of liquid water at 100°C is approximately 1321.75 Joules.