Solving the Mystery of Clay Temperature Change

[rijo664]

k the question states

On a pleasant fall day (temperature of 21.0 degree celsius) a lump of clay (with mass of
.855 kg) is thrown against the wall with a 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 be the final temperature of the clay? (Assume the clay starts at the same temperature as the air; Specific heat of clay is 2555 J/kgK.

I did:
Temperature original= 21 degree F
Mass= .885c kg
V= 38.0 m/s
Specific Heat of clay= 2555 J/kgK
Temperature final= ?
Formula= Q=cm delta T
Delta T= Q/cm

that what i have done so far. i know what c and m is but i don't know how to find the Q because i don't have the Change in Temperature. Can u help.

This is what i have so far.
Delta T= Q/(2555J/kgK)(.885kg)

This is what i got so far.

In order to find Q i need the change in temperature. which is what i am stuck on. and i also don't get what i am suppose to do with that velocity.

 

[olgranpappy]

the initial and final kinetic energies of the clay are not equal. that energy had to go somewhere. the wording of the problem implies that you are to consider that energy which was lost is turned into heat.

 

[Doc Al]

The change in temperature is what you're asked to find. But you should be able to figure out what Q is. Hint: Why did they give you the speed of the clay?