A $$100$$ g of iron nail is hit by a $$1.5$$ kg hammer striking at a velocity of $$60$$ ms$$^{-1}$$. What will be the rise in the temperature of the nail if one fourth of energy of the hammer goes into heating the nail? [Specific heat capacity of iron $$= 0.42$$ J g$$^{-1}$$ °C$$^{-1}$$]

The mass of the nail is $$100$$ g, the mass of the hammer is $$1.5$$ kg, the velocity of the hammer is $$60$$ m/s, and the specific heat of iron is $$0.42$$ J g$$^{-1}$$ $$°$$C$$^{-1}$$. The kinetic energy of the hammer is calculated as $$KE = \frac{1}{2}mv^2 = \frac{1}{2} \times 1.5 \times (60)^2 = \frac{1}{2} \times 1.5 \times 3600 = 2700 \text{ J}$$. One fourth of this energy is transferred to the nail, so $$Q = \frac{1}{4} \times 2700 = 675 \text{ J}$$.

Applying $$Q = ms\Delta T$$, where $$m$$ is the mass of the nail, $$s$$ is the specific heat capacity, and $$\Delta T$$ is the temperature rise, gives $$675 = 100 \times 0.42 \times \Delta T$$ and hence $$675 = 42 \times \Delta T$$. Solving for $$\Delta T$$ yields $$\Delta T = \frac{675}{42} = 16.07°\text{C}$$. Therefore, the rise in temperature of the nail is $$16.07°$$C. The correct answer is Option A.