# Need Answer Check - Potential Energy

• ch3570r
In summary, the water at the top of Niagara Falls has a temperature of 10.0 degrees celsius and 505 kg of water falls a distance of 50.0 m. Using the equation PE = mgh and the given assumption that all potential energy goes into increasing internal energy, the temperature of the water at the bottom of the falls will be 62 degrees celsius. To find the required energy to raise the water's temperature by 1 degree, multiply 4186 J/kg by 505 kg, giving a result of 2,113,930 J.

#### ch3570r

1. Water at the top of Niagara Falls has a temperature of 10.0 degrees celsius. Assume that all of the potential energy goes into increasing the internal energy of the water and that it takes 4186 J/kg to increase the water's temperature 1 degree celsius. If 505 kg of water falls a distance of 50.0 m, what will the temperature of the water be at the bottom of the falls?

2. PE = mgh

3. Now, I am not too sure if I did this right, because its somehow seems too simple.
I simply used the equation above and plugged in the numers, (505 x 50 x 98), which gives me 247450. So then, I divided that by 4186, which gives me 59 (degrees celsius). Then I add the 59 to the original 10, giving me 62 degrees celsius. Does that seem right? Maybe I am over simplifying it?

ch3570r said:
I simply used the equation above and plugged in the numers, (505 x 50 x 98), which gives me 247450. So then, I divided that by 4186, which gives me 59 (degrees celsius).
Note that 4186 has units of J per kg.

does that mean that first i divide the 505 kg by 4186, then use (505/4186) x 50 x 9.8? That gives me 59, added to the original 10, 69?

Life would be easier if you tried solving the problem symbolically, waiting until the last step to plug numbers in. (You'll see that some things will cancel.)

But if you are more comfortable working with the numbers, no problem. Answer this: How much energy is required to heat the given mass of water by 1 degree?

well let's see. It takes 4186 J/kg to increase the temperature by 1 degree. So if I take that 4186 and divide it by 505, that gives me about 8.3.

ch3570r said:
well let's see. It takes 4186 J/kg to increase the temperature by 1 degree. So if I take that 4186 and divide it by 505, that gives me about 8.3.
Why are you dividing by the mass here? (It may help you to write out the units.)

I would think that in order to find the Joules per Kg, I would have to divide the given ratio by the given mass.

The problem itself states that 4186 J/kg is needed to raise the temp. 1 degree

ch3570r said:
I would think that in order to find the Joules per Kg, I would have to divide the given ratio by the given mass.

The problem itself states that 4186 J/kg is needed to raise the temp. 1 degree
You are given the Joules per kilogram. You have to find the Joules.

If you divide J/kg by kg, you get (J/kg)/kg. That's not Joules.

Silly example: The cost of apples is \$5 per pound. How much does 3 pounds cost? Would you divide here or multiply?

ok, I would multiply. If I multiply 4186J/kg by 505kg, I get 2113930.

ch3570r said:
ok, I would multiply. If I multiply 4186J/kg by 505kg, I get 2113930.
OK. That's how many Joules of energy is required to raise that water one degree. Compare that with the increase in internal energy of the water to figure out how many degrees it will rise.

does the internal energy equal the potential energy?

From your statement of the problem:

"Assume that all of the potential energy goes into increasing the internal energy of the water"

So I think you are free to make that assumption in your calculation. 