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Joule machine

~christina~
Gold Member
712
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[SOLVED] joule machine..

1. Homework Statement
In joule's apparatus below, the mass of each block is 1.50kg, and the insulated tank is filled with 200g of water. What is the increase in the emperature of he water after the bocks fall through a distance of 3.00m?

http://img101.imageshack.us/img101/5377/34746360en6.th.jpg [Broken]


2. Homework Equations
[tex]\Delta U_{total}= W_{total} [/tex]


3. The Attempt at a Solution

Hm...not quite sure. I would think that I need to find the calories but I'm not sure how to find that either.

I know that the work done would = potential energy from when the blocks drop the 3m but to find the change in temperature would I use this equation ([tex]Q= C\Delta T[/tex] and equate the the Q to the [tex] \Delta U[/tex] )?



I know:
h= 3.00m
m= 1.50kg
g= 9.8m/s

mass of water= 2.00g=> 0.002kg
help please..
 
Last edited by a moderator:

Answers and Replies

23
0
Try conservation of energy

The change in gravitational potential energy equals the heat gained by the water.
 
~christina~
Gold Member
712
0
Try conservation of energy

The change in gravitational potential energy equals the heat gained by the water.
so would it be:

[tex]mgh= mc\Delta T[/tex] ?
 
23
0
Yes. The first m should be the mass of both blocks since they both fall through that distance.
 
~christina~
Gold Member
712
0
Yes. The first m should be the mass of both blocks since they both fall through that distance.
so it's 2mgh
 
23
0
~christina~
Gold Member
712
0
if you want m to be 1.5 kg, then yes
okay I found the temp change to be 10.54 deg C
so that sounds alright to me.

Thanks for your help mike :smile:
 
23
0
Hmm, are you sure that your units are consistent on both sides?

The specific heat of water is approximately 4.18 J/(g*C) which is equal to 4,180 J/(kg*C).

The temperature change that I get is about 0.105 degrees C.
 
~christina~
Gold Member
712
0
Hmm, are you sure that your units are consistent on both sides?

The specific heat of water is approximately 4.18 J/(g*C) which is equal to 4,180 J/(kg*C).

The temperature change that I get is about 0.105 degrees C.
uh oh..I thought that it was 2g of water. That is what was wrong with it.

I get the same thing after I finish correcting.

Thank you for catching that :redface:
 

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