Answer: It would require 83,617 J of heat to change 32.5 grams of liquid water into steam, from room temperature (25°C) to 115°C.
Step-by-step explanation:
To calculate the heat required to change 32.5 grams of liquid water into steam , you need to consider two steps: first, heating the liquid water to its boiling point, and second, converting it into steam.
Step 1: To heat 32.5 grams of liquid water from 25°C to 100°C , you can use the formula:
q1 = m * C * ΔT
where q1 is the heat absorbed by the water, m is the mass of water, C is the specific heat of water, and ΔT is the change in temperature. The specific heat of water is 4.184 J/g°C.
q1 = (32.5 g) * (4.184 J/g°C) * (75°C) = 10,189 J
So , 10,189 J of heat is required to heat 32.5 grams of liquid water from 25°C to 100°C.
Step 2 : To convert the liquid water into steam, you can use the formula:
q2 = m * ΔHvap
where q2 is the heat required for vaporization, m is the mass of water, and ΔHvap is the heat of vaporization of water. The heat of vaporization of water is 40.67 kJ/mol, or 2,257 J/g.
To find the mass of water that is converted into steam, you can use the formula:
m = n * M
where n is the number of moles of water vaporized and M is the molar mass of water. The molar mass of water is 18.015 g/mol.
n = (32.5 g) / (18.015 g/mol) = 1.804 mol
m = (1.804 mol) * (18.015 g/mol) = 32.5 g
q2 = (32.5 g) * (2,257 J/g) = 73,428 J
So, 73,428 J of heat is required to convert 32.5 grams of liquid water into steam.
The total heat required is:
q = q1 + q2 = 10,189 J + 73,428 J = 83,617 J