52,475.39 J thermal energy is needed to completely vaporize 36.04 g of water at 100°C.
Thermal energy is the energy contained within a system which is responsible for the temperature.
Given data:
36.04 grams of water.
Since the molecular weight of water is about 18 g/mol.
[tex]Moles = \frac{36.04 grams}{18 g/mol}[/tex]
Moles = 2
Q = mcΔT
Q = 36.04 x 4.184 J/g-K x 348 J/g-K x 348
Q = 52,475.39 J
Hence, 52,475.39 J thermal energy is needed to completely vaporize 36.04 g of water at 100°C.
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