# Finding specific latent heat of vaporization.

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**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

^{-3}kg and P=40W, gives me L = 0.85 x 10

^{6}J kg

^{-1}as the answer, whereas the answer in the book is 1.4 x 10

^{4}J kg

^{-1}. Where am I doing a mistake?

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#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?)

Have you converted correctly from minutes to seconds? (If you have. Did the book?)

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(Original post by

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?)

**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?)

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(Original post by

I agree with your answer. Looks like the book is wrong. (Again)

**Stonebridge**)I agree with your answer. Looks like the book is wrong. (Again)

(Original post by

Are you taking CIE?

**maths time is here**)Are you taking CIE?

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#7

(Original post by

Thanks. Sorry, but I didn't understand what you meant by 'again'?

Yeah, how do you know? Have you got this coursebook?

**Zishi**)Thanks. Sorry, but I didn't understand what you meant by 'again'?

Yeah, how do you know? Have you got this coursebook?

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(Original post by

I don't think actual calculations of specific latent heat of vapourisation is in the CIE curiculum, yeah I do

**maths time is here**)I don't think actual calculations of specific latent heat of vapourisation is in the CIE curiculum, yeah I do

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#9

**your**book.

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(Original post by

There have been many cases on here (that I've had to check) where the book was wrong. I didn't mean necessarily

**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.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). . .

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#11

(Original post by

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). . .

**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). . .

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(Original post by

which book is this from?

**maths time is here**)which book is this from?

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#13

**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). . .

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.

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#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)

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)

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(Original post by

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.

**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.

Thanks again.

(Original post by

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)

**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)

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#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??

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#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

in new book

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#18

(Original post by

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

**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

^{-3}kg and P=40W, gives me L = 0.85 x 10^{6}J kg^{-1}as the answer, whereas the answer in the book is 1.4 x 10^{4}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–

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#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

= 0.002 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

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