Question

Question 5 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 a specific heat of 2.43 kJ/kg-K. How much energy must be removed from 0.539 kg of ethyl alcohol that is initially a gas at 78.0 °C so that it becomes a solid at -114 °C? Number Units the tolerance is +/-2% This answer has no units SAVE FOR LATER SUBMIT ANSWER ° (degrees) m/s^2 N/m kg.m/s or Nos N/m^2 or Pa kg/m^3 m/s^3 times

Answer 1

m = mass of ethyl alchohol = 0.539 kg

L_v = heat of vaporization = 879 kJ/kg

c = specific heat = 2.43 J/(kg-K)

L_f = latent heat of fusion = 109 kJ/kg

$\Delta$T = change in temperature = 78 - (- 114) = 192

Q = Total Heat

Q = Heat of vaporization + Heat due to change in temperature + heat of fusion

Q = m L_v + m c $\Delta$T + m L_f

Q = (0.539) (879) + (0.539) (2.43) (192) + (0.539) (109)

Q = 784.01 kJ