|Nov6-07, 03:30 PM||#1|
Thermal properties of matter caluculation
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.
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
|Nov6-07, 04:43 PM||#2|
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
[itex]\Delta[/itex]v2/2 = shc*[itex]\Delta[/itex]T
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