Solving Joule Machine Problem [SOLVED] joule machine

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The discussion revolves around solving a physics problem involving Joule's apparatus, where two blocks fall and transfer energy to water. Participants clarify that the gravitational potential energy lost by the blocks equals the heat gained by the water, leading to the equation mgh = mcΔT. The correct mass of water is confirmed to be 200g, and the specific heat of water is noted as 4.18 J/(g°C). After some calculations and corrections, one participant arrives at a temperature change of approximately 10.54°C, while another initially miscalculates it as 0.105°C due to confusion over the water mass. The conversation emphasizes the importance of consistent units and careful calculations in solving energy conservation problems.
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[SOLVED] joule machine..

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


Homework Equations


\Delta U_{total}= W_{total}


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 (Q= C\Delta T and equate the the Q to the \Delta U )?



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

mass of water= 2.00g=> 0.002kg
help please..
 
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Try conservation of energy

The change in gravitational potential energy equals the heat gained by the water.
 
mike115 said:
Try conservation of energy

The change in gravitational potential energy equals the heat gained by the water.

so would it be:

mgh= mc\Delta T ?
 
Yes. The first m should be the mass of both blocks since they both fall through that distance.
 
mike115 said:
Yes. The first m should be the mass of both blocks since they both fall through that distance.

so it's 2mgh
 
~christina~ said:
so it's 2mgh

if you want m to be 1.5 kg, then yes
 
mike115 said:
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:
 
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.
 
mike115 said:
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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