Finding latent heat of vaporization from immersion heater data

[Student-]

A student suspects that the power rating on an immersion heater is 50 W but he is not sure.He sets up an apparatus as shown. After the liquid has been brought to its boiling point, he finds that 3.5g of vapor escaped each minute through the outlet tube.The boiling point of the liquid is 80 degrees celsius.
Calculate the heat required to convert 3.5 g of liquid at its boiling point to 3.5g of vapor at the same temperature.

My attempt :
I know that to find the energy required for a phase change would be q=mass x latent heat of vaporization Q=mxl_v

But i don't know what it is the latent heat of vaporization of the material.

I'm thinking something like :
50 W = 50 joules per second so in one minute 50x60 3000 joules would be expended...but the answer that my book has is 2940J ! and I don't know where to move from here. Any help at all would be awesome ! Thanksss a lot :D

 

[sammy4u]

Okay, let us use the 50 W first.
We know that:

P = ΔE / Δt

where P is power, E is energy and t is time.
so if we plug in the one minute given and 50 W we get the amount of energy.

50 = ΔE / 3600
ΔE = 180000

Now we know that it is already at the boiling point so we don't need to worry about the energy being used to heat up the substance. So we can go straight to Q = mL_v. We know that Q is the energy we found (ΔE) which is 180,000. So we plug that in, and mass we know is 3.5 grams.

180,000 = 0.0035 * L_v
L_v = 51428

I hope you can do the rest on your own, since you were only looking for L_v.

 

[gneill]

Okay, let us use the 50 W first.
We know that:

P = ΔE / Δt

where P is power, E is energy and t is time.
so if we plug in the one minute given and 50 W we get the amount of energy.

50 = ΔE / 3600
ΔE = 180000

Watts are Joules per second so you need the number of seconds in a minute, not in an hour.
A minute is only 60 seconds.

The problem seems to be lacking some vital information (or a stated purpose!) What's the meaning of the value found for the heat of vaporization if we don't trust the heat source? Is there something to compare the result to? Is the given problem the complete problem statement?

 

[sammy4u]

Sorry I thought it said an hour. Thank you gneill, but yes using the same steps you can find the Lv and you should be able to find the answer. That is what student- wanted to know.

 

[gneill]

Sorry I thought it said an hour. Thank you gneill, but yes using the same steps you can find the Lv and you should be able to find the answer. That is what student- wanted to know.

Actually, it's not entirely clear what answer the question is looking for. Given that it starts by casting doubt upon the specifications of the heat source, what confidence can one have in any result obtained by using its untrustworthy parameters? I can only see this problem being useful if there is some missing part that would allow one to confirm the result.

 

[sammy4u]

I see what you mean, as there could be something wrong with the heat source and untrustworthy parameters. However I believe that this is probably a high-school introductory physics question that really wouldn't have the students worrying about other things.