HOw much energy is required to heat 36.0g H2O from a liquid at 65C to a gas at 115C? The following physical data may be uselful.
Hvap=40.7kJ/mol
C liquid=4.18
C gas=2.01
C solid= 2.09
T melting=0 C
T boiling=100 C
The answers about 87.7 kJ
However I'm not sure how you get this,
so H=H heating liquid+H vap+H heating gas. I'm not getting the same answer
q for liquid is mass x specific heat x delta T. or for gas is mass x specific hat gas x delta T.
q for phas change is mass x delta Hvap or delta H fusion
THEN add all.
To calculate the energy required to heat 36.0g of water (H2O) from a liquid at 65°C to a gas at 115°C, we need to consider several steps: heating the liquid, vaporization, and heating the gas.
1. Heating the liquid:
The specific heat capacity of liquid water is given as C liquid = 4.18 J/g°C. We need to calculate the energy required to heat the liquid water from 65°C to 100°C, which is the boiling point.
Energy for heating liquid =
mass × specific heat capacity × ΔT
= 36.0g × 4.18 J/g°C × (100°C - 65°C)
= 36.0g × 4.18 J/g°C × 35°C
= 5257.8 J
2. Vaporization:
To convert liquid water at its boiling point (100°C) to water vapor at the same temperature, we need to consider the heat of vaporization (Hvap), which is given as 40.7 kJ/mol.
First, we need to calculate the number of moles of water:
moles = mass / molar mass
= 36.0g / 18.015 g/mol (molar mass of water)
= 1.997 moles
Energy for vaporization = moles × Hvap
= 1.997 moles × 40.7 kJ/mol
= 81.31 kJ (Note: kJ to J conversion is needed to use consistent units)
3. Heating the gas:
Now, we need to calculate the energy required to heat the water vapor at its boiling point (100°C) to 115°C.
Energy for heating the gas =
mass × specific heat capacity × ΔT
= 36.0g × 2.01 J/g°C × (115°C - 100°C)
= 36.0g × 2.01 J/g°C × 15°C
= 1085.4 J
Now, let's sum up the energies for each step:
Total energy required = Energy for heating liquid + Energy for vaporization + Energy for heating gas
= 5257.8 J + 81310 J + 1085.4 J
= 87653.2 J
Converting Joules to kilojoules:
Total energy required = 87653.2 J / 1000
≈ 87.7 kJ (rounded to one decimal place)
So, the answer is approximately 87.7 kJ.
To find the total energy required to heat and vaporize the water, you need to break it down into three steps: heating the liquid water, vaporizing the liquid water, and heating the resulting steam to the final temperature.
Step 1: Heating the liquid water
First, you need to calculate the energy required to heat the water from 65°C to its boiling point at 100°C. The specific heat capacity of liquid water (C liquid) is given as 4.18 J/g°C.
ΔT1 = (100 - 65) = 35°C (change in temperature)
q1 = m * C liquid * ΔT1
= 36.0 g * 4.18 J/g°C * 35°C
≈ 5245.4 J
Step 2: Vaporizing the liquid water
Next, you need to calculate the energy required to completely vaporize the liquid water. The enthalpy of vaporization (Hvap) is given as 40.7 kJ/mol. First, you need to convert the mass of water into moles.
1 mole of H2O = 18.02 g
Moles of water = (36.0 g) / (18.02 g/mol)
≈ 1.998 mol
q2 = Hvap * moles of water
= 40.7 kJ/mol * 1.998 mol
≈ 81.34 kJ
Step 3: Heating the steam
Finally, you need to calculate the energy required to heat the resulting steam from the boiling point (100°C) to the final temperature (115°C). The specific heat capacity of steam (C gas) is given as 2.01 J/g°C.
ΔT3 = (115 - 100) = 15°C (change in temperature)
q3 = m * C gas * ΔT3
= 36.0 g * 2.01 J/g°C * 15°C
≈ 1085.4 J
Now, calculate the total energy required by summing up the energy from each step:
Total energy = q1 + q2 + q3
= 5245.4 J + 81.34 kJ + 1085.4 J
≈ 87.7 kJ
Therefore, the total energy required to heat and vaporize 36.0 g of water from 65°C to 115°C is approximately 87.7 kJ.