Question

has anyone done the imploding can experiment where you calculate efficiency!

Answer 1

@surry99 Can you help us?

Answer 2

sure....set up the problem

Answer 3

the efficiency n = w/Q where W is work and Q is heat input.
derive an equation for n, and in the simplest form all quantities should be known.
we know the volume of the can, the atmospheric pressure, latent heat, and mass of steam

Answer 4

\[n=(P \Delta V)/L(m)\] or is there more to it?

Answer 5

ok it looks like you...
1) filled a can with water
2) heated it until it became steam
3) let it out of the can
4) the can was then crushed by the atmospheric pressure
correct?

Answer 6

no... after your step two... we grabed it with tongs, quickly turned it upside down and submerged it into ice water in which in imploded. Then filled it with water to measure the new volume

Answer 7

Hold on i'll post you a photo that might help you better understand

Answer 8

oh yes that makes sense...then your formula is correct,
Work = Patm(deltaV)
Q = mL

Answer 9

We were questioning it because we are getting numbers that would give us like 50% efficiency. Our lab instructor told us an engine of a car has like a 20% efficiency so we thought the can should be less. But the can shouldn;t be less because there are less outside sources working against it, correct?

Answer 10

do you know the units that we need to plug into that equation?

Answer 11

No your results sound reasonable and car engines are not very efficient for a variety of reasons.
Good question about units...does an efficiency have units?

Answer 12

We can't find any, when we look at example problems in our text book they have commas next to them... thats why I suggested maybe putting the answer in percentages.

Answer 13

Your efficiency should have no units. The units for work on the top should be the same as the units for heat on the bottom.
What units did you use for:
P
V
M
L

Answer 14

efficiency is a ratio between heat going in and heat going out. An equation i found in my book says that efficiency = 1 - Qinput/Qoutput.......which is why i am questioning our initial equation. if our initial equation says the numerator is W = PdeltaV, and the book equation says 1 - Qin / Qout

Answer 15

That is correct ...it is for a Carnot engine....but did you in anyway determine Qin and Q out.
My guess is you know:
P in atmospheres
V was milliliters
m was in grams
L was in Joules /gram

You need to let me know if these are correct.

Answer 16

I think the book said Latent was kj... i'm trying to find the page...not 100% on that

Answer 17

those are correct, which i then further converted into
P in N/msquared
V in m cubed
Latent in J/kg
Mass in kg
therefor i think it all crosses out to be in Joules/Joules....but that number is VERY small. I did a few calculations and got less than 10% efficient

Answer 18

ok lets summarize:
1) My mistake for indicating 50% was reasonable...I read it fast and thought it said 5%.
2) Physicsgirls units indeed cancel out as they should...excellent work
3) n= 1 - qin/qout is derived from n=w/q....same equation
4) less than 10% looks better

Answer 19

Yeah sorry about the 50% she went on and redid the problem and forget to mention it to me :-P

Answer 20

so then, how do we find Qin? we only have information for PdeltaV..... Qin would be L*m....but do we know that information?

Answer 21

PdeltaV is correct for the work...remember the equations are the same and you measured P and deltaV

Answer 22

i guess a better question is:
how can n = w/q be equal to n = 1- Qin/Qout
when W = Qin - Qout

Answer 23

does that mean PdeltaV = Qin - Qout?

Answer 24

hang on...

Answer 25

yes

Answer 26

Gotta run...you two ok?

Answer 27

yes real quick, if our effency equation comes up with 10 percent, and the carnot equation comes up with 26% is that a reasonable comparison

Answer 28

yes that is...the Carnot efficiency is the best you can possible do if you try to convert heat from a high temperature source to do work because no matter what you will have to release some heat at a lower temperature. That is it, it is not possible to convert all the heat to do work...some heat must always be rejected.

Answer 29

thank you so much!

Answer 30

You are both very welcome. Goodnight to you both.