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Old Apr 16th 2019, 12:00 AM   #1
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Finding the power used to change phase?

I have figured the heat added to the system so far, but can't figure out where to go from here to get the change in internal energy to work out the work(W)?

https://imgur.com/a/4SiM6fd

It is question number 3 right at the bottom of the photo.

Thank you for any help.
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Old Apr 16th 2019, 12:57 AM   #2
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We are told latent heat of evaporation is 2256 KJ/Kg ....

0.3 Kg evaporates so to do that requires 2236 x 0.3 = 676.8KJ

But they want the answer in Watts , since the time period is 1 HR (3600secs) then the number of joules is the wattage operating for 3600 secs (1 W is 1J/sec)

W x 3600 = 676800 ......W = 188
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Old Apr 16th 2019, 01:18 AM   #3
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Originally Posted by oz93666 View Post
We are told latent heat of evaporation is 2256 KJ/Kg ....

0.3 Kg evaporates so to do that requires 2236 x 0.3 = 676.8KJ

But they want the answer in Watts , since the time period is 1 HR (3600secs) then the number of joules is the wattage operating for 3600 secs (1 W is 1J/sec)

W x 3600 = 676800 ......W = 188
OOHOH MYYY GOOODD. I LITERALLY DID NOT SEE THE TIME. I need to read better....thank you.

Gah that's something they keep telling us and emphasising all the time. "READ the question". You know what I do? "aahh pfftt, I can read. Lol. Who doesn't read the question properly duh psh" Also me: this

Cheers friend!!!
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Old Apr 16th 2019, 02:27 AM   #4
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Talking Unfit drillers

Oz is a very good technician, first rate. He goes to the comet, here is one follower.
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