Answer:
ΔT=3.781 Kelvin
Explanation:
Given data
m=5.00g 0r 0.005kg
v=44.0 m/s
c (specific heat)=128 j/kg
Kinetic Energy=[tex]\frac{1}{2}mv^{2}[/tex]
KE=[tex]\frac{1}{2}(0.005)(44)^{2}[/tex]
KE=4.84 joules
1/2 Kinetic energy goes to heat=1/2×4.84
=2.42 J
Q=mcΔT
ΔT=[tex]\frac{Q}{mc}[/tex]
ΔT=[tex]\frac{2.42}{0.005*128}[/tex]
ΔT=3.781 K
