A 20.0 -g lead bullet leaves a rifle at a temperature of $47.0^{\circ} \mathrm{C}\( and travels at a velocity of \)5.00 \times 10^{2} \mathrm{m} / \mathrm{s}\( until it hits a large block of ice at \)0^{\circ} \mathrm{C}$ and comes to rest within it. How much ice will melt?

Answer: To find the amount of ice that melts, follow these steps: 1. Calculate the initial heat energy of the bullet: \(Q_{bullet} = (0.020 \text{kg})\times(128 \frac{\text{J}}{\text{kg} \cdot \text{K}}) \times (47 \text{K})\) 2. Calculate the initial kinetic energy of the bullet: \(KE_{bullet} = 0.5\times(0.020 \text{kg})\times(5.00\times10^2\frac{\text{m}}{\text{s}})^2\) 3. Calculate the total energy transferred to the ice: \(E_{total} = Q_{bullet} + KE_{bullet}\) 4. Determine the amount of ice that melts: \(m_{ice} = \frac{E_{total}}{L_f}\), where \(L_f = 3.33\times10^5 \frac{\text{J}}{\text{kg}}\) By substituting the values calculated in steps 1-3 into the formula in step 4, you will find the mass of ice that melts.