Question

What is the minimum number of kiloJoules needed to change 40.0 grams of water at 100C to steam at the same temperature and pressure

Answer 1

Answer:

90.4Kg

Explanation:

The energy required= latent heat of vaporization

Latent heat of vaporization of water= 2260 KJ/Kg

Mass of water = 0.04 Kg

Heat required= 0.04Kg× 2260 KJ/Kg

Heat required= 90.4Kg

Answer 2

Answer: The minimum number of kilo Joules needed is 90.4KJ.

Explanation:

The energy required= latent heat of vaporization.

Latent heat of vaporization of water= 2260 KJ/Kg

Mass of water = 0.04 Kg

Heat required= 0.04Kg× 2260 KJ/Kg

Heat required= 90.4KJ.

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