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 0.601 kg of ethyl alcohol that is initially a gas at 78.0°C so that it becomes a solid at −114°C?

Answer 1

874.2 KJ

Step-by-step explanation:

Ethyl alcohol boiling point = 78oC, freezing point = -114oC heat of vaporization = 879 kJ/kg heat of fusion = 109kJ/Kg specific heat capacity (c) of ethyl alcohol = 2,43KJ/Kg.k

calculating each heat drawn singly

Quantity loss due vaporization = m × Lv where m is mass in kg and Lv is the heat of vaporization

Quantity of heat loss due to fusion = m × Lf ( heat of fusion)

Quantity of heat loss due to cooling from 78oc to -114 = m × c × ΔT 9 (change in Temperature)

Quantity of heat removed due to vaporization = 0.601 ×879 = 528.279 KJ

Quantity of heat lost due to cooling from 78oc to -114oc = mcΔT= 0.601 × 2.43 × ( 78 - (-114)) = 0.601 ×2.43 × 192 = 280.4 kJ

Quantity of heat lost due to fusion during freezing = 0.601 × 109 = 65.509 KJ

Total heat removed = sum of the each quantity of heat removed from the gaseous ethyl alcohol = 528.279 + 280.4 + 65.509 = 874.2 KJ