Answer: The increase in temperature of the nail after the three blows is 8.0636 Kelvins. The correct option is (d).
Explanation:
Kinetic energy of the hammer ,K.E.=
[tex]\frac{1}{2}mv^2=\frac{1}{2}1.00 kg\times (8.50 m/s)^2=36.125 J[/tex]
Half of the kinetic energy of the hammer is transformed into heat in the nail.
Energy transferred to the nail in one blow =
[tex]\frac{1}{2}K.E.=\frac{1}{2}\times 36.125 J=18.0625 J[/tex]
Total energy transferred after 3 blows,Q =[tex]3\times 18.0625 J=54.1875 J[/tex]
Mass of the nail = 15 g = 0.015 kg
Change in temperature =[tex]\Delta T[/tex]
Specif heat of the steel = c = 448 J/kg K
[tex]Q=mc\Delta T[/tex]
[tex]54.1875 J=0.015 kg\times 448 J/kg K\times \Delta T[/tex]
[tex]\Delta T=8.0636 K\approx 8.1 K[/tex]
The increase in temperature of the nail after the three blows is 8.1 Kelvins.Hence, correct option is (d).
