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Old Apr 15th 2017, 05:16 AM   #1
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Join Date: Apr 2017
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Application of First law of Thermodynamics

Hi people,

I was stuck on this problem and needed some assistance. I have provided the question and answer below.

Q. The specific latent heat of vaporisation of steam is 2.26 MJ/kg. When 50 cm^3 of water is boiled at standard atmospheric pressure of 1.01*10^5 Pa, 83*10^3 cm^3 of steam are formed.

Density of water = 1000 kg/m^3

1. Calculate the mass of water boiled.

Answer: 50*10^-3 KG.

Attempted solution:

So I know that the molecules have to do work against the atmospheric pressure to create the increased volume. Therefore, the work done is equal to the pressure multiplied the difference in final volume and the initial volume.

W = PdV = (1.01*10^-5)(0.083-5*10^-5) = 8377.95 J

Then I thought of using q = ml, but that wouldn't make sense because the calculated quantity above is not the energy required to change phase from water into steam?

I'm a bit confused and would appreciate any sort of help! Thank you in advance.
skengdo410 is offline   Reply With Quote
Old Apr 15th 2017, 01:32 PM   #2
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Join Date: Apr 2015
Location: Somerset, England
Posts: 508
Where do you get such rubbish questions?

Firstly water vapourises, steam condenses.

The latent heat is the same in either direction.

So 50cc of water at vapourises absorbing the latent heat of vapourisation.

When water vapourises to dry steam it expands approximately 1600 times ie (Google)

So the approx new vol is 50 x 1600 = 80 x 10^3 cc

We are told that the new volume is 83 x 10^3 cc

So all the water boils.

We are told the density of the water before vapourisation is 1000 kg/cum

Thus the mass of boiled water is 50 x 10^-6 x 1000 = 50 x 10^-3 kg

What was the point of this question or was there more that you did not say?
studiot is offline   Reply With Quote

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