Calculate the amount of heat (in kJ) required to convert 87.5 g of water to steam at 100°C. (The molar heat of vaporizaion of water is 40.79 kJ/mol.)

q=mass H2O x delta Hvap.

Calculate the amount of heat required to raise the temperature of a 31 sample of water from 9 to 26

How much heat in Joules is needed to evaporate a cup of water (250g) at 100 degrees Celsius?

To calculate the amount of heat required to convert water to steam, you need to consider two steps:

Step 1: Calculate the amount of heat required to raise the temperature of water from 0°C to 100°C using the specific heat capacity of water.
Step 2: Calculate the amount of heat required to convert the water at 100°C to steam at 100°C using the molar heat of vaporization.

Let's break down the calculations for each step:

Step 1: Calculate the amount of heat required to heat water from 0°C to 100°C.
The specific heat capacity of water is 4.184 J/g°C (joules per gram per degree Celsius). We need to convert the mass of water (87.5 g) to joules.

Q1 = m * c * ∆T
where:
Q1 = heat energy in joules
m = mass in grams
c = specific heat capacity in J/g°C
∆T = change in temperature

Q1 = 87.5 g * 4.184 J/g°C * (100°C - 0°C)
Q1 = 365,350 J

Step 2: Calculate the amount of heat required to convert water at 100°C to steam at 100°C.
The molar heat of vaporization of water is given as 40.79 kJ/mol. To calculate the heat required, we need to convert grams of water to moles using the molar mass of water (18.015 g/mol).

n = m / M
where:
n = number of moles
m = mass in grams
M = molar mass in g/mol

n = 87.5 g / 18.015 g/mol
n = 4.862 moles

Q2 = n * ΔHvap
where:
Q2 = heat energy in joules
n = number of moles
ΔHvap = molar heat of vaporization in J/mol

Q2 = 4.862 moles * 40.79 kJ/mol * 1000 J/kJ
Q2 = 198,830 J

Now, to find the total heat required, you need to add up the heat from both steps:

Total heat required = Q1 + Q2
Total heat required = 365,350 J + 198,830 J
Total heat required = 564,180 J

To convert the heat energy to kilojoules (kJ), divide by 1000:

Total heat required = 564,180 J / 1000
Total heat required = 564.18 kJ

Therefore, the amount of heat required to convert 87.5 g of water to steam at 100°C is 564.18 kJ.