Question

Ethyl alcohol has a boiling point of 78.0 degrees Celsius, a freezing point of -114 degrees Celsius, a latent heat of vaporization of 879 kJ/K, a latent heat of fusion of 109 kJ/K, and a specific heat of 2.43 kJ/kg.K. How much energy must be removed from 0.510 kg of ethyl alcohol that is initially a gas at 78.0 degrees Celsius so that it becomes a solid at -114 degrees Celsius?

Answer 1

here,

the mass of ethyl alcohol , m = 0.51 kg

initial temperature , Ti = 78 degree C

the final temperature , Tf = - 114 degree C

the heat energy required , Q = m * ( Lv + Ce * ( Ti - Tf) + Lf)

Q = 0.51 * ( 879 + 2.43 * ( 78 - (-114) + 109 ) KJ

Q = 742 KJ