Answer: The thermal energy needed to completely vaporize 36.04 g of water at 100 degrees Celsius is approximately 1471.35 joules.
Step-by-step explanation:
To calculate the thermal energy required to complete vaporize 36.04 g of water at 100 degrees Celsius, we need to use the formula:
Q = m * ΔH_vap
where:
Q is the thermal energy (in joules) needed for vaporization,
m is the mass of the substance (in grams), and
ΔH_vap is the molar enthalpy of vaporization (in J/g) of the substance.
For water, the molar enthalpy of vaporization (ΔH_vap) is approximately 40.79 J/g at 100 degrees Celsius.
Now, plug in the values:
m = 36.04 g
ΔH_vap = 40.79 J/g
Q = 36.04 g * 40.79 J/g
Q ≈ 1471.3476 J
So, the thermal energy needed to completely vaporize 36.04 g of water at 100 degrees Celsius is approximately 1471.35 joules.