Kinetic energy to thermal energy

[nickyfernandezzz]

Hey guys, I need help with the following question.
A lead ball of specific heat capacity 130J/KgK travels with a speed of 100m/s before it hits a wooden cube fixed to floor and gets stuck inside.What will be the rise in temperature of this lead ball by the time it comes to

I attempted the question as follows,
Kinetic energy Thermal energy of
Of the lead ball = the lead ball after hitting
Before hitting the cube on the cube.
1/2×m×v^2 = mc(t1-t2)
1/2 × m×100^2 = m× 130×(t1-t2)
t1-t2 = 38 Celsius.

But apparently the answer I was given is 3 Celsius.
Can someone please tell me where I went wrong and how to answer the question correctly?

Thanks.

 

[TomHart]

Kinetic energy Thermal energy of
Of the lead ball = the lead ball after hitting
Before hitting the cube on the cube.

Huh?

Is there any additional information in the problem that you may have omitted? e.g. Did any of the energy go towards heating up the wood? If all of the kinetic energy of the lead ball was used to heat up the lead, then I can't find any problem with your calculation.

Edit: I think I figured out the key to reading your statement (which I quoted above). It was written in 2 columns with the equal sign in between. So it reads:
"Kinetic energy of the lead ball before hitting the cube = Thermal energy of the lead ball after hitting on the cube."

 

[oz93666]

I think your answer of 38 is correct .

 

[nickyfernandezzz]

Huh?

Is there any additional information in the problem that you may have omitted? e.g. Did any of the energy go towards heating up the wood? If all of the kinetic energy of the lead ball was used to heat up the lead, then I can't find any problem with your calculation.

Edit: I think I figured out the key to reading your statement (which I quoted above). It was written in 2 columns with the equal sign in between. So it reads:
"Kinetic energy of the lead ball before hitting the cube = Thermal energy of the lead ball after hitting on the cube."

It doesn't say whether any energy goes to heating up of the wood. But if it does, how do I do the calculation?

 

[oz93666]

It doesn't say whether any energy goes to heating up of the wood. But if it does, how do I do the calculation?

You can't ...books/sources often have errors

You say ..."But apparently the answer I was given is 3 Celsius. " ... apparently ? was the answer given 3 or 3.8... and who gave it to you?

 

[nickyfernandezzz]

You can't ...books/sources often have errors

You say ..."But apparently the answer I was given is 3 Celsius. " apparently ? was the answer given 3 or 3.8... and who gave it to you?

My physics lecturer gave the answer as 3 Celsius .

 

[oz93666]

My physics lecturer gave the answer as 3 Celsius .

Tell him he's wrong

 

[nickyfernandezzz]

Tell him he's wrong

Yeah,I guess that's what I'll do. Anyway thanks for helping.

 

[TomHart]

It doesn't say whether any energy goes to heating up of the wood. But if it does, how do I do the calculation?

Well, if it doesn't say, then I think it makes sense to assume all of the kinetic energy goes to heat up the lead. If it did tell you that some of the energy was used to heat up the wood, the problem would have to indicate (somehow) how much energy went to the wood and how much to the lead.
And like oz93666 indicated, it is not uncommon that incorrect answers are given.
Regarding telling your lecturer that he is wrong, it may be wise to find a kinder way to address the apparent discrepancy. :)

 

[nickyfernandezzz]

Well, if it doesn't say, then I think it makes sense to assume all of the kinetic energy goes to heat up the lead. If it did tell you that some of the energy was used to heat up the wood, the problem would have to indicate (somehow) how much energy went to the wood and how much to the lead.
And like oz93666 indicated, it is not uncommon that incorrect answers are given.
Regarding telling your lecturer that he is wrong, it may be wise to find a kinder way to address the apparent discrepancy. :)

Yes, that seems wise.Thank you.