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 Nov 24th 2017, 07:02 AM #1 Junior Member     Join Date: Nov 2017 Location: The Jurassic Forests Posts: 25 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 07:06 AM.
 Nov 24th 2017, 07:43 AM #2 Senior Member     Join Date: Aug 2008 Posts: 113 $cm \Delta T = 0.025 \cdot mgh$ $\Delta T = \dfrac{0.025 \cdot gh}{c}$ try again ...
Nov 24th 2017, 08:17 AM   #3
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 Originally Posted by skeeter $cm \Delta T = 0.025 \cdot mgh$ $\Delta T = \dfrac{0.025 \cdot gh}{c}$ try again ...
Thank you!!!

 Dec 2nd 2017, 03:26 PM #4 Senior Member   Join Date: Jun 2010 Location: NC Posts: 402 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 TyrannosaurusWreck likes this.
Dec 2nd 2017, 06:14 PM   #5
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 Originally Posted by TyrannosaurusWreck 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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