# Thermal properties of matter caluculation

1. Nov 6, 2007

### exotic rooftile

A squash ball of mass 46g is struck against a wall so it hits with a speed of 40m/s, and rebounds with a speed of 25m/s.
Calculate the temperature rise (s.h.c. of rubber is 1600J/kg/K)

This is fine. I use the equation:

heat energy = mass x shc x temperature change

(40 - 25) = 0.046 x 1600 x temperature change

15/73.6 = temperature change

temperature change = 0.2K

Then it asks why is it unecessary to know the mass?

And I cannot for the life of me think why. Is there another equation I'm supposed to know? Have I made a mistake? Am I overlooking something incredibly obvious?

Any help/hints would be very much appreciated.

Thanks,
Rachael
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Nov 6, 2007

### Staff: Mentor

The part of the solution (40-25) is not the heat energy. It's best to write units with the corresponding values.

The 40 m/s - 25 m/s is simply the change in velocity (which is also the change in specific momentum). The change in energy is the change in kinetic energy and KE = 1/2 mv2. But looking at the righthand side one multiplies the mass * specific heat.

If we deal with the specific kinetic energy and specific heat, we can eliminate mass from the equation. Thus

$\Delta$v2/2 = shc*$\Delta$T