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Old Nov 24th 2017, 08:02 AM   #1
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Specific Heat Problem

Hello again,
I'm trying to figure out this problem-

A 2.4 kg iron ball is dropped from a height of 14 m onto a concrete roadway, and 2.5% of its kinetic energy at the time it reaches the ground is transformed into internal energy in the ball itself. (The rest of the energy is transmitted to the ground, converted into sound energy, and so on.) What is the ball's increase in temperature?

So far I've used the equations PE=mgh and Q=cm△T, with c=449, m=2.4kg, h=14m, g=9.8.
My work: PE=mgh=24(9.8)(14)=3292.8 J.
Q=PE/0.025
PE=0.025cm△T.=
3292.8=0.025(449)(2.4)△T.=
3292.8=26.94△T
△T=12.2K

The answer key says its 7.6e-3 K, so I tried again with 25%(329.28)= 8.23 and tried 329.28=8.23(449)(2.4)△T. Needless to say it's still wrong, and I don't know what to do with it now Any help would be appreciated!

Last edited by TyrannosaurusWreck; Nov 24th 2017 at 08:06 AM.
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Old Nov 24th 2017, 08:43 AM   #2
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$cm \Delta T = 0.025 \cdot mgh$

$\Delta T = \dfrac{0.025 \cdot gh}{c}$

try again ...
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Old Nov 24th 2017, 09:17 AM   #3
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Originally Posted by skeeter View Post
$cm \Delta T = 0.025 \cdot mgh$

$\Delta T = \dfrac{0.025 \cdot gh}{c}$

try again ...
Thank you!!!
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Old Dec 2nd 2017, 04:26 PM   #4
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Hey Wreck...

You might be interested in where the energy equation you used came from.

4.03 Energy Equation Constant Pressure | THERMO Spoken Here!

jp
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Old Dec 2nd 2017, 07:14 PM   #5
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Originally Posted by TyrannosaurusWreck View Post
Hello again,
I'm trying to figure out this problem-

A 2.4 kg iron ball is dropped from a height of 14 m onto a concrete roadway, and 2.5% of its kinetic energy at the time it reaches the ground is transformed into internal energy in the ball itself. (The rest of the energy is transmitted to the ground, converted into sound energy, and so on.) What is the ball's increase in temperature?

So far I've used the equations PE=mgh and Q=cm△T, with c=449, m=2.4kg, h=14m, g=9.8.
My work: PE=mgh=24(9.8)(14)=3292.8 J.
Okay, this is the potential energy when the ball is 14 m above the ground which will have become the kinetic energy when it hits the ground.

Q=PE/0.025
What? You said, above, that 2.5% of the kinetic energy is converted to heat. That is the original potential energy times 0.025, not divided by 0.025! The increase in heat is (3292.8)(0.025)= 82.38.

PE=0.025cm△T.=
3292.8=0.025(449)(2.4)△T.=
3292.8=26.94△T
△T=12.2K

The answer key says its 7.6e-3 K, so I tried again with 25%(329.28)= 8.23 and tried 329.28=8.23(449)(2.4)△T. Needless to say it's still wrong, and I don't know what to do with it now Any help would be appreciated!
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