Is My Calculation of the Final Temperature of Clay Correct?

[rijo664]

This was the question

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

These are the variables.
Here is what my answer was.

First i used the Kinetic Energy Formula that is
K=.5(m)(v squared)
I plugged in what i know that is the mass and the velocity
and i got 638.97 that is my kinetic energy.
In order to find the delta T i plugged what i got when i did the kinetic energy formula to the Specific heat formula that is
Delta T=638.97/the specific heat of clay and the mass of clay. and then i subtracted what i got with the initial temperature so am i right so far or wrong just need to know.

 

[learningphysics]

Is the mass 0.855 kg or 0.885kg? I used 0.855 and got a different kinetic energy...

Your method is correct. But remember that temperature increases... kinetic energy converts to heat. So you add delta T to the initial temperature.