Thermal properties of matter caluculationby exotic rooftile Tags: caluculation, matter, properties, thermal 

#1
Nov607, 03:30 PM

P: 1

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
Nov607, 04:43 PM

Admin
P: 21,628

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 mv^{2}. 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]v^{2}/2 = shc*[itex]\Delta[/itex]T 


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