Respuesta :
Answer:[tex]v=2452.91 m/s[/tex]
Explanation:
Given
initially steam is at [tex]100^{\circ}C[/tex] and converted to [tex]0^{\circ} C[/tex] ice
Let m be the mass of steam
latent heat of fusion and vaporization for water is
[tex]L_f=3.33\times 10^5 J/kg[/tex]
[tex]L_v=2.26\times 10^6 J/kg[/tex]
Heat required to convert steam in to water at [tex]100^{\circ}C[/tex]
[tex]Q_1=m\times L_v=m\cdot 2.26\times 10^6 J[/tex]
Heat required to lower water temperature to [tex]0^{\circ}C[/tex]
[tex]Q_2=m\times c\times \Delta T[/tex]
[tex]Q_2=m\times 4.184\times (100)[/tex]
[tex]Q_2=4.184m\times 10^5 J[/tex]
Heat required to convert [tex]0^{\circ}C[/tex] water to ice at [tex]0^{\circ}C[/tex] is
[tex]Q_3=m\times L_f[/tex]
[tex]Q_3=m\times 3.33\times 10^5=3.33m\times 10^5 J[/tex]
[tex]Q=Q_1+Q_2+Q_3[/tex]
[tex]Q=(2.26+0.4184+0.33)m\times 10^6 J[/tex]
[tex]Q=3.0084m\times 10^6 J[/tex]
So this energy is equal to kinetic energy of bullet of mass m moving with velocity v
[tex]Q=\frac{1}{2}mv^2[/tex]
[tex]3.0084m\times 10^6=\frac{1}{2}mv^2[/tex]
[tex]v^2=3.0084\times 2\times 10^6[/tex]
[tex]v=2.452\times 10^3 m/s[/tex]
[tex]v=2452.91 m/s[/tex]
