Question

A 3.00g lead bullet at 30.0°C is fired at a speed of 240 m/s into a large block of ice at 0°C, in which it becomes embedded. What quantity of ice melts?

Answer 1

Answer:

$m=2.94*10^{-4}kg\\ 0r\\m=0.294g$

Explanation:

The total energy transferred from bullet to the ice is

$Q=K.E_{i}+m_{bullet}C_{lead}|0^{o}C-30.0^{o}C |\\ Q=(1/2)m_{bullet}v_{i}^{2}+m_{bullet}C_{lead}\\ Q=(1/2)(3.00*10^{-3}kg )[(240m/s)^{2}+128*30]\\ Q=97.9J$

The mass of ice that melts when this quantity of thermal energy absorbed is:

$m=\frac{Q}{L_{water} } \\m=\frac{97.9}{3.33*10^{5} } }\\m=2.94*10^{-4}kg\\ 0r\\m=0.294g$