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fomenos
22 days ago
10

A 5.00 g sample of water vapor, initially at 155°C is cooled at atmospheric pressure, producing ice at –55°C. Calculate the amou

nt of heat energy lost by the water sample in this process, in kJ. Use the following data: specific heat capacity of ice is 2.09 J/g×K; specific heat capacity of liquid water is 4.18 J/g×K; specific heat capacity of water vapor is 1.84 J/g×K; heat of fusion of ice is 336 J/g; heat of vaporization of water is 2260 J/g.
Physics
1 answer:
Ostrovityanka [3.2K]22 days ago
3 0

Answer:

Total energy lost is 16.1 KJ

Explanation:

The heat of vaporization for water is  1.84 J/gK, indicating a loss of energy

The heat loss when cooling from 155 to 100 = 1.84 x 5 x (155 + 273 - 100 + 273) = 506 J

Latent heat of steam = 2260 J/g x 5 g = 11300 J

The heat lost upon cooling from 100 to 0 = 4.18 x 5 x (100 - 0) = 2090 J

The fusion heat is calculated as 336 J/g x 5 g = 1680 J

Heat to cool ice from 0 to -55 =  2.09 J/gK x 5 g x (0 -(-55)) = 574.75 J

Summing all heats gives H of the reaction

506 J + 11300 J + 2090 J + 1680 J + 574.75 J = 16150.75 J = 16.1 KJ

Total energy lost is 16.1 KJ

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According to Newton's second law of motion, the overall force on the proton can be expressed as follows.

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Applying the energy balance equation,

Energy In = Energy Out

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7 0
1 month ago
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