Answer:
a. 0.01 C
b. dissipated to outside environment
Explanation:
Let the specific heat of copper be 0.3846 kJ/kg-K or 384.6 J/kg-C
(a)The original kinetic energy of the block is:
[tex] E_k = \frac{mv^2}{2} = \frac{1.5*3^2}{2} = \frac{1.5*9}{2} = 6.75 J[/tex]
As 85% of this kinetic energy is converted to block internal heat energy, with specific heat we can calculate the rise in temperature:
[tex]E_h = 0.85E_k = 0.85*6.75 = 5.7375 J[/tex]
[tex]mc_c\Delta T = 5.7375[/tex]
[tex]1.5*384.6\Delta T = 5.7375[/tex]
[tex]\Delta T = \frac{5.7375}{1.5*384.6} \approx 0.01^oC[/tex]
(b) the remaining 15% energy would probably be dissipated to outside environment as heat energy
