# The heat of vaporization of ethyl alcohol is about 200 cal/g. If 2.4 kg of this fluid were...

## Question:

The heat of vaporization of ethyl alcohol is about {eq}200\ \dfrac {cal}g {/eq}. If {eq}2.4\ kg {/eq} of this fluid were allowed to vaporize in a refrigerator, find what mass of ice would be formed from {eq}0 ^\circ C {/eq} water?

## Energy Associated With Phase Transition

Matter can exists in different phases like solid phase, liquid phase and gaseous phase. Transition from one phase to another or back involves either release or absorption of energy. The transition takes place at definite temperature known as transition temperature. The lowest energy phase is the solid phase. At some higher energy liquid phase and at still higher energy gaseous phase exists. So in these transitions from solid tom liquid, liquid to gaseous state energy has to be supplied and in the reverse transitions energy will be released.

Given data

• Heat of vaporization of ethyl alcohol {eq}L_v = 200 \ cal /g \\ L_v = 200 \times 1000 \times 4.184 = 8.368 \times 10^5 \ J/ kg {/eq}
• Mass of ethyl alcohol vaporized {eq}M = 2.4 \ kg {/eq}
• Latent heat of fusion of water {eq}L = 3.34 \times 10^5 \ J/ kg {/eq}

Let m be the mass of ice formed.

Heat energy absorbed by the alcohol in vaporization {eq}Q_1 = M L_v {/eq}

This amount of heat energy has be taken from water at zero degree Celsius by converting it into water.

Then the amount of heat energy released from water {eq}Q_2 = m L {/eq}

On equating the two expressions for heat energy we get {eq}M L_v = m L {/eq}

Therefore mass of ice formed {eq}m = \dfrac { M L_v } { L } \\ m = \dfrac { 2.4 \times 8.368 \times 10^5 } { 3.34 \times 10^5 } \\ m = 6.013 \ kg {/eq} 