Question

Ethyl alcohol has a boiling point of 78.0°C, a freezing point of −114°C, a heat of vaporization of 879 kJ/kg, a heat of fusion of 109 kJ/kg, and specific heat of 2.43 kJ/kg·K. How much energy must be removed from 0.605 kg of ethyl alcohol that is initially a gas at 78.0°C so that it becomes a solid at −114°C?
________kJ
Please show WORK and ANSWERS! Thanks!

Answer 1

heat of fusion, it can not be kJ/kgK, must be kJ/kg, cause the process occurs at a fixed temp.

Now, let's operate the problem:

if we want it to change from gas to solid, it must overcomes 3 processes, which mean that the energy must be

given to 3 processes:

1. the latent heat at vaporization temp: Q = L1 * m
2. when it is at liquid state, the heat required to change the temp from 78 to -144 is Q = c * m * (t2-t1)

3. the latent heat at freezing point Q= L2*m

--> total energy = L1 * m + c * m * (t2-t1) + L2*m

= 879* 0.606 + 2.43 * 0.606* (78+144) + 109 * 0.6 06

=.9256kj