How to Calculate Potential Energy and Latent Heat in Physics?

[Ashleykins]

I have two homework questions that I am desperate for some help on. I've posted them on our class discussion page, and I'm on my last attempt. We get three attempts on our homework. The first question deals with SPE and the second deals with latent heat.

For Question 1:
1. Homework Statement
#1
You have a copper-beryllium alloy spring of length 0.020 m, spring constant 224.0 N/m. spring-states.png

You pull the block out to xmax = 0.033 m and hold it there. Then you release it. As it moves back toward equilibrium, it passes by position x2 = 0.021 m. Calculate the change in potential energy from xmax to x2.

Type in your answer to the nearest 0.001 Joule of potential energy. Use a - minus sign if the ΔPE is negative. E.g., if your answer is -1.2088 Joules, then type in -1.209.

Homework Equations

SPE=1/2kx^2

The Attempt at a Solution

So I've tried a few different things. I know that k=224.0 For x I subtracted xmax-x2, giving me x=0.012 So the equation would be SPE=(1/2)(224)(0.012)^2 = 0.016 This was marked wrong
My next attempt was a solution I found on our class discussion page which stated the following:
SPE=1/2k(xmax) and then SPE=1/2(k)(x2) then subtract the solution from x2-xmax. This also came out wrong. Please help!

Question 2: Latent Heat. So we've done this equation before 100 times using water and it's been fine. I am not understanding what is going wrong now...

Homework Statement

While shopping at Yankee Candle, you notice that the price of their candles ($0.06/g) is increasing at a higher rate than your work income.
You decide to make your own candles and buy 454 g of scented paraffin wax from Amazon.com ($0.02/g). To form the candle in your mold, you have to heat the solid wax from room temperature (297 K) to 330 K. Once you pour the liquid wax into the mold, it must cool back down to room temperature (297 K) before removing it from the mold.


Substance: Paraffin Wax:
Melting Point: 327K
Specific Heat (SOLID): 0.57 cal/g K
Specific Heat (LIQUID): 0.58 cal/g K
Latent heat of fusion: 44 cal/g



CALCULATE: total calories of heat is used for this process before pouring it out?

Type in the numeric part of your answer to the nearest calorie. E.g., if your answer comes in at 88.5 calories, then type 89 in the answer box.

2. So I know that Q=mc(delta t)

The Attempt at a Solution

. [/B]
So the first time, I thought maybe it was just a mathematical error so I tried it again. Still wrong. I have no idea what it could be here. This is how I calculated it:
for Q1: (454g)(0.57)(327-297) --> (454)(0.57)(30) =7763.40
Q2: (454g)(44) =18160
Q3: (454)(0.58)(330-327) --> (454)(0.58)(3) =789.96
add them up. 26713.36
Round to nearest whole number -->26713

This HW is due at 9am tomorrow, I've spent an entire weekend on it and I still have a test for another class to study for tonight. I'm like this close to just giving up. PLEASE help me.

 

[CWatters]

My next attempt was a solution I found on our class discussion page which stated the following:
SPE=1/2k(xmax) and then SPE=1/2(k)(x2) then subtract the solution from x2-xmax. This also came out wrong. Please help!

What do you mean by "subtract the solution from x2-xmax"? Subtract what from x2-xmax?

You might well be doing it right (calculating the change in energy) but that line doesn't make a lot of sense. You didn't show your working or even the answer you entered so we can't see where or if you made a mistake.

PS The energy stored in a spring depends on the change in length not the stretched length so don't forget to subtract the relaxed length.

 

[CWatters]

Q2: (454g)(44) =18160

Might be other errors as well but check that.

 

[Ashleykins]

Okay. so my calculator apparently hates me... I did it again and Q2 came out to 19976

 

[Ashleykins]

What do you mean by "subtract the solution from x2-xmax"? Subtract what from x2-xmax?

You might well be doing it right (calculating the change in energy) but that line doesn't make a lot of sense. You didn't show your working or even the answer you entered so we can't see where or if you made a mistake.

Oops Sorry, I'll edit. I was in a bit of a rush, I've been working on this homework assignment for no lie like 3 days and I have a test tomorrow for another class that I haven't even started studying for. I honestly can't even think straight right now...I had it all written out but forgot to type it!
So what I did, subtract the answer from the x2 calculation from the answer from the xmax calculation. The answer from the x2 calculation was 2.535. The answer from xnew was 0.131. 2.535-0.131=2.404

 

[CWatters]

So what I did, subtract the answer from the x2 calculation from the answer from the xmax calculation.

Change in energy is the final energy minus the initial energy.

More in a moment.

 

[gneill]

Hi Ashleykins,
Welcome to Physics Forums.

Please post one question per thread. With multiple questions being handled in one thread it can cause confusion amongst the resulting discussions.

For your spring question, the equation for potential energy should be applied to one position of the spring at a time. An argument that is the difference of two positions will not work. So calculate them separately, then take the difference in the energies. Make sure you get the order correct when you take the difference as it affects the sign of the result.

 

[CWatters]

I can't see your drawing spring-states.png but note that the energy stored in a spring depends on the change in length relative to the relaxed length which might be

x2-xrelaxed = 0.021 - 0.020

 

[CWatters]

Unfortunately I've got to go as it's 1am here.

 

[Ashleykins]

Hey guys. I finally got it. I honestly don't know what I was doing wrong or what "clicked." But something did. Maybe I'm just tired. Thanks anyways. Happy Halloween! I will remember to only do one equation per thread in the future! Sorry about that!