A beaker contains 166 mL of ethyl alcohol at 25 ^°C. What is the minimum amount of energy that must be removed to produce solid ethyl alcohol?

 

Given are:

Density of ethyl alcohol = 789 kg/m³

V = 166 mL or 166 * 10^-6 m³

T =  25 °C

m = density*volume

= 789 kg/m³ * 166 * 10^-6 m³ = 0.130974kg

Freezing point = T_f = -114.1°C

Latent heat = L = 10800 J/kg

Specific heat of ethyl alcohol is 2.460 J/g°C or 2460 J/kg°C

The minimum amount of energy that must be removed to produce solid ethyl alcohol is given by:

e = m*s (T - T_f) + m*L

= 0.130974(2460) * (25-(-114.1)) + 0.130974( 10800)

= 46231.988364 or 46232 J