Question

Heat is extracted from a certain quantity of steam at

Answer 1

Answer:$v=2452.91 m/s$

Explanation:

Given

initially steam is at $100^{\circ}C$ and converted to $0^{\circ} C$ ice

Let m be the mass of steam

latent heat of fusion and vaporization for water is

$L_f=3.33\times 10^5 J/kg$

$L_v=2.26\times 10^6 J/kg$

Heat required to convert steam in to water at $100^{\circ}C$

$Q_1=m\times L_v=m\cdot 2.26\times 10^6 J$

Heat required to lower water temperature to $0^{\circ}C$

$Q_2=m\times c\times \Delta T$

$Q_2=m\times 4.184\times (100)$

$Q_2=4.184m\times 10^5 J$

Heat required to convert $0^{\circ}C$ water to ice at $0^{\circ}C$ is

$Q_3=m\times L_f$

$Q_3=m\times 3.33\times 10^5=3.33m\times 10^5 J$

$Q=Q_1+Q_2+Q_3$

$Q=(2.26+0.4184+0.33)m\times 10^6 J$

$Q=3.0084m\times 10^6 J$

So this energy is equal to kinetic energy of  bullet of mass m moving with velocity v

$Q=\frac{1}{2}mv^2$

$3.0084m\times 10^6=\frac{1}{2}mv^2$

$v^2=3.0084\times 2\times 10^6$

$v=2.452\times 10^3 m/s$

$v=2452.91 m/s$