Temperature Change and heat loss

[William Bush]

1. The problem statement:
A 5.00 x 10^2 kg object is attached by a rope through a pulley to a paddle-wheel shaft that is placed in a well-insulated tank holding 25.0 kg of water. The object is allowed to fall, causing the paddle wheel to rotate, churining the water. If the object falls a verticle distance of 1.00 x 10^2 m at constant speed, what is the temperature change of the water? (specific heat = 4186 J/kg and gravity = 9.81 m/s^2).



2. Homework Equations :
Heat Lost = Heat Gained
Mm Cm (Ti-Tf) = Mw Cw (Tf-Ti)



3. The Attempt at a Solution
The instructor has not lectured on this type of problem but has asked us to read the text and give it a shot. I can't seem to find a way forward with this problem. Is this a calorimetry problem? If so, I don't understand why the distance the object fell and the fact that it fell at a constant speed is important. Any help will be appreciated...thanks in advance!

 

[Doc Al]

Hint: Does the energy of the falling object change?

 

[AlephZero]

Mechanical work and heat are two different forms of energy. So, think what "conservation of energy" means in this situation.

Re the "constant speed", since the mass is not acclerating you know the tension on the rope is equal to the weight mg, so you can calculate the work done by the rope on the paddle wheel. If the mass was acclerating the tension in the rope would be different.

 

[William Bush]

Conservation of energy say that energy remains constant, although it may change forms; so I would say that at the top, the object had "potential energy" and at the bottom it had "kinetic energy". If my calculations are correct, the object has 490,500J of potential energy. Does this mean that the object introduces 490,500J of heat energy to the water? If so, wouldn't I have to know what the initial temp of the water to determine how much it would change?

 

[William Bush]

I believe that I can determine the work done by the rope on the paddle with the following equation:
W=FdCos0
So, the work done is:
500(100)Cos 90 degress = 50,000

I'm not sure how this applies to my problem though??

 

[Doc Al]

Conservation of energy say that energy remains constant, although it may change forms; so I would say that at the top, the object had "potential energy" and at the bottom it had "kinetic energy".

Does the object's KE change as it falls?

If my calculations are correct, the object has 490,500J of potential energy. Does this mean that the object introduces 490,500J of heat energy to the water?

I didn't check your calculations, but that's the idea: The object transfers mechanical energy as it falls to the paddle, which in turn transforms that energy into "heat". (Note that AlephZero's advice is equivalent--understand what's he's saying.)

If so, wouldn't I have to know what the initial temp of the water to determine how much it would change?

No. To find the final temperature you'd need to know the initial temperature--but all you are asked to find is the change in temperature.

 

[Doc Al]

I believe that I can determine the work done by the rope on the paddle with the following equation:
W=FdCos0
So, the work done is:
500(100)Cos 90 degress = 50,000

With what force does the rope pull on the paddle? Reread AlephZero's post.

 

[William Bush]

I see that I made an error calculating the work done by the rope...I forget to multiply the mass by the gravity. After recalculating, I see that the work done by the rope is 490,500. I suppose that it is no coincidence that this is the same as the value I got for potential energy. So If the object introduces 490,500 jules of energy to the water, I now have to figure out how this much energy effects that temp of water...any clues!

 

[Doc Al]

specific heat

Here's a clue: http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/spht.html"

 

[William Bush]

Doc Al...you are the man! With your help, I was able to solve the problem. I used the following formula: Q=cm(delta T)
Of course I had to rearrange it to solve for "T" but when I did, I determined that the temperature change was 4.69 degress. I've got more problems to solve but thanks a lot for helping me with this one!