Answer:
2452.79432 m/s
Explanation:
m = Mass of ice
[tex]L_s[/tex] = Latent heat of steam
[tex]s_w[/tex] = Specific heat of water
[tex]L_i[/tex] = Latent heat of ice
v = Velocity of ice
[tex]\Delta T[/tex] = Change in temperature
Amount of heat required for steam
[tex]Q_1=mL_s\\\Rightarrow Q_1=m(2.256\times 10^6)[/tex]
Heat released from water at 100 °C
[tex]Q_2=ms_w\Delta T\\\Rightarrow Q_2=m4186\times (100-0)\\\Rightarrow Q_2=m0.4186\times 10^6[/tex]
Heat released from water at 0 °C
[tex]Q_3=mL_i\\\Rightarrow Q_3=m(333.5\times 10^3)\\\Rightarrow Q_3=m(0.3335\times 10^6)[/tex]
Total heat released is
[tex]Q=Q_1+Q_2+Q_3\\\Rightarrow Q=m(2.256\times 10^6)+m0.4186\times 10^6+m(0.3335\times 10^6)\\\Rightarrow Q=3008100m[/tex]
The kinetic energy of the bullet will balance the heat
[tex]K=Q\\\Rightarrow \frac{1}{2}mv^2=3008100m\\\Rightarrow v=\sqrt{2\times 3008100}\\\Rightarrow v=2452.79432\ m/s[/tex]
The velocity of the ice would be 2452.79432 m/s
