Final Temp of Clay Homework: Find Change in Temp & Final Temp

[xpatelsxownage]

Homework Statement

On a pleasant fall day(temperature of 21.0 C) a lump of clay (with mass of .885kg) is thrown against the wall with speed of 38.0 m/s. the clay deforms as it sticks to the wall, noiselessly. assuming no heat escapes into the air, what will the final temp of the clay be.(clay starts at same temp as the air) Cclay=2555j/kgK

Homework Equations

Q=cm(change in Temp)T
Q=change in KE
KE=1/2mv(squared)
1/2mv(squared)=cm times the (change in temp)

The Attempt at a Solution

1/2(.885kg)(38.0 m/s)squared=(2555j/kgK)(.885kg)*Change in temp of clay.
(.4425kg)(1444)=(2261.175)(change in temp of clay)
638.97=(2261.175)(change in temp of clay)
.283 C=change in temp of clay

final temp=.283+21.0
final temp=21.28


i did it three times but the answer doesn't seem right to me. if it loses no energy shouldn't all of it be transformed into heat.
some when please tell me what I am doing wrong.

 

[genneth]

I don't immediately see the problem. It appears correct.

 

[xpatelsxownage]

thank you, i was trying to think of a different way. but this was the only way i could think of and i wanted to make sure this was right because there is a follow up question to this. again thanks

 

[rijo664]

hey suraj i got a different answer to that problem. i did what u did and i got 28 degree celsius but i did not get 21.28 as my final answer.

 

[xpatelsxownage]

how did you get 28 C

 

[dynamicsolo]

hey suraj i got a different answer to that problem. i did what u did and i got 28 degree celsius but i did not get 21.28 as my final answer.

The kinetic energy transformed into heat is 639 J. With the given specific heat and the fact that the lump is about 1 kg in mass, it's hard to see how the temperature change would be *larger* than 1 degree C...

 

[rijo664]

well this thing is due tuesday. i know that. and i got 28 by doing exactly what u did and then i added the 28 with 21 to get my final temperature the 28 is the change in temp.

 

[xpatelsxownage]

did you do some mathematical mistake because 28 rise in temp is huge. i thought this was due monday

 

[rijo664]

he said u got till tuesday. First i found the kinetic energy which came out to be 638.97 and then i found the change in temp by using the kinetic enerygy which is heat. because that is the reason why they gave the velocity. and i got 28 degrees.

 

[dynamicsolo]

well this thing is due tuesday. i know that. and i got 28 by doing exactly what u did and then i added the 28 with 21 to get my final temperature the 28 is the change in temp.

Be sure to check your units or calculator input: the change of 0.28 C comes from

639 J / [(2555 J kg/K)(0.885 kg)] .

 

[rijo664]

yeah that's exactly how i did it to get the change in temp. because u can't do the change in temp without the heat.

 

[xpatelsxownage]

he said u got till tuesday. First i found the kinetic energy which came out to be 638.97 and then i found the change in temp by using the kinetic enerygy which is heat. because that is the reason why they gave the velocity. and i got 28 degrees.

you use 1/2mv squared=cm delta t

 

[rijo664]

yup. that's what i did and when is this thing due do u know for sure i believe he said it's due tommorow.

 

[xpatelsxownage]

yup. that's what i did and when is this thing due do u know for sure i believe he said it's due tommorow.

cm delta t should come out to 2261.175j/k delta t

 

[rijo664]

k my fault make sense yup i forgot about the decimal ur answer is right sry. how did u do the second one and is this thing do tommorow.

 

[xpatelsxownage]

u use the heat in=heat out formula. ill show you in 10 min after i get home. it is due tomorrow.