# Homework Help: Expanding Steel, Can't Follow Equation

1. Mar 25, 2004

### holly

Gosh, I can't stand the new look.

Q. Place a 1 kg block of iron at 40 degrees C into a kg of water at 20 degrees C and the final temp will be: More than 30 degrees C, Less than 30 degrees C, At or Around 30 degrees C.

I felt it would be At or Around 30 degrees C, because I tried to do an equation where I have 1(40 - T) = 1(T - 20), giving me 60 = 2T, giving me 30. But I don't really know if that's right. I don't know where I got the equation from...

Q. There's a span of steel 1.3km long, how long would it get if its temp was raised 15 degrees C. Using the formula Delta L = L(sub nought) alpha Delta T, which was then substituted to read (1300m) (11 x 10^-6/degrees C)(15 degrees C) = 0.21 meters.

I can't follow that. I can't follow the 11 x 1/10000000 part. Can't make it come out right with a number like that hanging around.

Any help appreciated. T minus 12 hours to the Big Test.

2. Mar 25, 2004

I don't like te new look, either. It's too cluttered.

1. If I'm not mistaken (I've never much studied this kind of stuff...), heat will continue to transfer so long as there is a temperature difference between the steel and the water. So the steel will keep cooling and the water will keep heating until they equalize. That doesn't much help the temperature it'll end at, though. That depends on the amount of energy in the system.

Q = cmT

Use that to calculate the energy in the system water + steel, and then we know that they will end at the same temperature, so,

$$Q = c_\textrm{water}(1\textrm{kg})T_f + c_\textrm{steel}(1\textrm{kg})T_f$$
Edit: Q = (c_water)(1kg)(T_f) + (c_steel)(1kg)(T_f)

factor out the $T_f$ (Edit: T_f), Q is known, the two c's can be looked up, and the 1kg is known. So 1 unknown and 1 equation!

2. Er.. What's the question again?

Edit: LaTeX is acting weird.

Last edited: Mar 25, 2004
3. Mar 25, 2004

### ShawnD

I agree

The temperature should be less than 30 degrees because water has a much higher heat capacity. Iron is only about 460 J/(kg*C), but water is 4186 J/(kg*C).

The equation is m1c1(Tf - Ti)1 + m2c2(Tf - Ti)2 = 0. That equation will look a lot better when tex is working again. It's equal to 0 because the net change in energy is 0.

$$m_1c_1 \Delta T_1 + m_2c_2 \Delta T_2 = 0$$

Remember to add the original length. 0.21 is only the change.

Last edited: Mar 25, 2004
4. Mar 25, 2004

### holly

Well, thank you for the answers. I don't follow them, but thank you.

I kept trying to do the one about the steel expanding, and I was supposed to be using 10^-6, which was a millionth, not a 10-millionth, so I got it to come out right finally. Thanks for reminding me to add on the length.

I do NOT at all follow the equations about the piece of metal being put in the water. Usually, I can't follow the equations even when properly displayed, but surely these ones are not displaying all the way? I'm allergic to latex, anyway.

*sigh* Maybe the old look will suddenly return and the new look will go away? Remember what happened to New Coke!

oops, had to go back & put in the minus sign...

Last edited: Mar 25, 2004
5. Mar 25, 2004

### holly

Okay, I am now noticing, I had the words Radio Wave by my ferret, and now I just have Registered User. That's not as good.

cookiemonster still has "I like cookies," but I got zip. That's not fair. No wonder I can't do physics...too upset, having my little title taken away like that, probably because I was obnoxious elsewhere on the boards...boo hoo [b(]

And ShawnD has Narbo on his! And mine was removed! NOT FAIR! NOT FAIR!

6. Mar 25, 2004

I think you get to pick your own title...

And ShawnD's explanation is the short version of mine. I don't know why LaTeX isn't working, though, you'd have to ask Greg. =\

7. Mar 25, 2004

### ShawnD

I'll try to explain that formula.

(m)(c)(delta T) represents the change in thermal energy for a substance. When substances of different temperatures are mixed together, the energy of each substance changes but the net energy in the system does not change.

Thermal energy formulas are a lot like momentum formulas. Seeing how the energy formula is like the momentum formula may put it in terms you are familiar with. You may remember this formula (i means initial, f means final):
(m1)(Vi1) + (m2)(Vi2) = (m1)(Vf1) + (m2)(Vf2)

It shows that momentum (like thermal energy) is conserved after a collision (mixing substances together). Now supposed you put everything on the left side, then it would look like this:
(m1)(Vi1) - (m1)(Vf1) + (m2)(Vi2) - (m2)(Vf2) = 0

Now factor out masses:
(m1)(Vi1 - Vf1) + (m2)(Vi2 - Vf2) = 0

Rewrite it again
(m1)(delta V1) + (m2)(delta V2) = 0

That looks a lot like
(m1)(c1)(delta T1) + (m2)(c2)(delta T2) = 0

Does it make any more sense now?

You can also add phase changes into the equation. Add or subtract the phase change based on whether the substance changing phase gained or lost energy. When ice melts, it gains energy; therefore, you would add that term. When steam condenses, it loses energy; therefore, you would subtract that term. Another great thing is that since the formula always equals 0, you can put the formula into equation solver if you have a Texas Instraments graphing calculator, or put it as a graph and have the calculator find the zeros.

Last edited: Mar 25, 2004
8. Mar 26, 2004

### holly

Okay Brilliant Ones, thank you again.

It's almost Test Time. I'm forgetting things as fast as I can remember them.

I think I'll be able to do the "mixing different temp items" problems now, because I DO remember the momentum equation! I like both equations. cookiemonster's reminds me of the word Q = "cat" (if you fudge on the m) so it's easy to remember. And the other one I do remember in a hazy way! Us sonography students are not supposed to use calculators unless we absolutely have to...our program leader really doesn't want us "dependent" on them. "If there's an emergency, you won't have a calculator handy! You must be able to think for yourselves! Your patients are depending on you!" Yes, and I guess we'll just happen to have a big old ultrasound machine in our back pocket, ha ha.

Thanks again.