Question

A certain quantity of steam has a temperature of 100.0 oC. To convert this steam into ice at 0.0 oC, energy in the form of heat must be removed from the steam. If this amount of energy were used to accelerate the ice from rest, what would be the linear speed of the ice? For comparison, bullet speeds of about 700 m/s are common.

Answer 1

Answer:

2452.79432 m/s

Explanation:

m = Mass of ice

$L_s$ = Latent heat of steam

$s_w$ = Specific heat of water

$L_i$ = Latent heat of ice

v = Velocity of ice

$\Delta T$ = Change in temperature

Amount of heat required for steam

$Q_1=mL_s\\\Rightarrow Q_1=m(2.256\times 10^6)$

Heat released from water at 100 °C

$Q_2=ms_w\Delta T\\\Rightarrow Q_2=m4186\times (100-0)\\\Rightarrow Q_2=m0.4186\times 10^6$

Heat released from water at 0 °C

$Q_3=mL_i\\\Rightarrow Q_3=m(333.5\times 10^3)\\\Rightarrow Q_3=m(0.3335\times 10^6)$

Total heat released is

$Q=Q_1+Q_2+Q_3\\\Rightarrow Q=m(2.256\times 10^6)+m0.4186\times 10^6+m(0.3335\times 10^6)\\\Rightarrow Q=3008100m$

The kinetic energy of the bullet will balance the heat

$K=Q\\\Rightarrow \frac{1}{2}mv^2=3008100m\\\Rightarrow v=\sqrt{2\times 3008100}\\\Rightarrow v=2452.79432\ m/s$

The velocity of the ice would be 2452.79432 m/s