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Volgvan
3 years ago
12

Determine the heat energy required to vaporize 13.9 grams of liquid water at 100° C. O 2,006 cal O 47.8 cal O 7,506 cal O 24.9 c

al
Physics
1 answer:
Harlamova29_29 [7]3 years ago
6 0

Answer:

7506 cal

Explanation:

Given: Mass of water = m = 13.9 grams.

Heat energy required to determine the conversion of 13.9 grams of water to steam or vapor is m L,

where L = Latent heat of vaporization of water=540 cal/g.

Q = mL = (13.9)(540) = 7,506 cal.

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