# Finding specific latent heat of vaporization.

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#1
Question: A sample of alcohol is heated with a 40 W heater until it boils. As it boils, its mass decreases at a rate of 2.25 g per minute. Assuming that 80% of the energy supplied by the heater is transferred to the alcohol, estimate the specific latent heat of vaporization of the alcohol. Give your answer in J kg-1.

I used the formula E=mL and then changed it into this form: where P = power supplied by heater.

Substituting the values t=60s, m=2.25 x10-3kg and P=40W, gives me L = 0.85 x 106 J kg-1 as the answer, whereas the answer in the book is 1.4 x 104 J kg-1. Where am I doing a mistake? 0
9 years ago
#2
I haven't checked the answers but yours is 60 times the book's.
Have you converted correctly from minutes to seconds? (If you have. Did the book?)
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#3
(Original post by Stonebridge)
I haven't checked the answers but yours is 60 times the book's.
Have you converted correctly from minutes to seconds? (If you have. Did the book?)
The questions states the rate of decrease of mass in grams per minute, so yes, I've converted 1 minute into 60 seconds. As a joule contains a second as part of its base units, so I guess the book hasn't converted it into seconds? 0
9 years ago
#4
I agree with your answer. Looks like the book is wrong. (Again)
0
9 years ago
#5
Are you taking CIE?
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#6
(Original post by Stonebridge)
I agree with your answer. Looks like the book is wrong. (Again)
Thanks. Sorry, but I didn't understand what you meant by 'again'?
(Original post by maths time is here)
Are you taking CIE?
Yeah, how do you know? Have you got this coursebook?
0
9 years ago
#7
(Original post by Zishi)
Thanks. Sorry, but I didn't understand what you meant by 'again'?

Yeah, how do you know? Have you got this coursebook?
I don't think actual calculations of specific latent heat of vapourisation is in the CIE curiculum, yeah I do
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#8
(Original post by maths time is here)
I don't think actual calculations of specific latent heat of vapourisation is in the CIE curiculum, yeah I do
Well, it is.
0
9 years ago
#9
(Original post by Zishi)
Thanks. Sorry, but I didn't understand what you meant by 'again'?

There have been many cases on here (that I've had to check) where the book was wrong. I didn't mean necessarily your book.
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#10
(Original post by Stonebridge)
There have been many cases on here (that I've had to check) where the book was wrong. I didn't mean necessarily your book.
Yeah, this annoys me.

I know it's off topic, can you please see the attachment? I think that underline expression is wrong - it shouldn't be mg cos(theta). . .
0
9 years ago
#11
(Original post by Zishi)
Yeah, this annoys me.

I know it's off topic, can you please see the attachment? I think that underline expression is wrong - it shouldn't be mg cos(theta). . .
which book is this from?
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#12
(Original post by maths time is here)
which book is this from?
A-Level physics by Roger Muncaster.
0
9 years ago
#13
(Original post by Zishi)
Yeah, this annoys me.

I know it's off topic, can you please see the attachment? I think that underline expression is wrong - it shouldn't be mg cos(theta). . .
It's correct. (Muncaster's books are excellent)
mg cos theta is the component of the weight along the line PO (towards the centre)
From this you subtract R to get the resultant (centripetal) force towards the centre.
0
9 years ago
#14
http://www.engineeringtoolbox.com/fl...eat-d_147.html

According to that link the LHoV of ethanol is 846 kJ/kg, which can be rounded up the answer you have.
I'm gonna go with the book being wrong (as mentioned above by Stonebridge. I'd trust his word on this sort of thing)
0
#15
(Original post by Stonebridge)
It's correct. (Muncaster's books are excellent)
mg cos theta is the component of the weight along the line PO (towards the centre)
From this you subtract R to get the resultant (centripetal) force towards the centre.
Wow, I was such a fool to think of it as the other way round.
Thanks again.

(Original post by Manitude)
http://www.engineeringtoolbox.com/fl...eat-d_147.html

According to that link the LHoV of ethanol is 846 kJ/kg, which can be rounded up the answer you have.
I'm gonna go with the book being wrong (as mentioned above by Stonebridge. I'd trust his word on this sort of thing)
Well, I'd trust his word on every sort of thing. 0
9 years ago
#16
gah of course it's in the syllabus lol, did you finish the end of chapter questions and the 'exam-style questions' at the end of that chapter??
0
4 years ago
#17
15 Rate of loss of mass = 2.25 g per minute= 0.002 2560 = 3.75 × 10–5 kg s–1Rate of supply of energy to alcohol = 40 × 80%= 32 WSo specific latent heat of vaporisation= rate of supply of energyrate of loss of mass = 323.75 × 10–5= 853 000 J kg–1 ≈ 850 kJ kg–1
in new book
0
4 years ago
#18
(Original post by Zishi)
Question: A sample of alcohol is heated with a 40 W heater until it boils. As it boils, its mass decreases at a rate of 2.25 g per minute. Assuming that 80% of the energy supplied by the heater is transferred to the alcohol, estimate the specific latent heat of vaporization of the alcohol. Give your answer in J kg-1.

I used the formula E=mL and then changed it into this form: where P = power supplied by heater.

Substituting the values t=60s, m=2.25 x10-3kg and P=40W, gives me L = 0.85 x 106 J kg-1 as the answer, whereas the answer in the book is 1.4 x 104 J kg-1. Where am I doing a mistake? 15 Rate of loss of mass = 2.25 g per minute= 0.002 2560 = 3.75 × 10–5 kg s–1Rate of supply of energy to alcohol = 40 × 80%= 32 WSo specific latent heat of vaporisation= rate of supply of energyrate of loss of mass = 323.75 × 10–5= 853 000 J kg–1 ≈ 850 kJ kg–
0
4 years ago
#19
15 Rate of loss of mass = 2.25 g per minute
= 0.002 25
60 = 3.75 × 10–5 kg s–1
Rate of supply of energy to alcohol = 40 × 80%
= 32 W
So specific latent heat of vaporisation
= rate of supply of energy
rate of loss of mass = 32
3.75 × 10–5
= 853 000 J kg–1 ≈ 850 kJ kg–1
its in new book
0
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