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Ethyl alcohol has a boiling point of 78.0 C, a freezing
Chapter , Problem 39(choose chapter or problem)
Ethyl alcohol has a boiling point of \(78.0^{\circ} \mathrm{C}\), a freezing point of \(-114^{\circ} \mathrm{C}\), a heat of vaporization of 879 kJ/kg, a heat of fusion of 109 kJ/kg, and a specific heat of \(2.43 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\). How much energy must be removed from 0.510 kg of ethyl alcohol that is initially a gas at \(78.0^{\circ} \mathrm{C}\) so that it becomes a solid at \(-114^{\circ} \mathrm{C}\)?
Questions & Answers
QUESTION:
Ethyl alcohol has a boiling point of \(78.0^{\circ} \mathrm{C}\), a freezing point of \(-114^{\circ} \mathrm{C}\), a heat of vaporization of 879 kJ/kg, a heat of fusion of 109 kJ/kg, and a specific heat of \(2.43 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\). How much energy must be removed from 0.510 kg of ethyl alcohol that is initially a gas at \(78.0^{\circ} \mathrm{C}\) so that it becomes a solid at \(-114^{\circ} \mathrm{C}\)?
ANSWER:Step 1 of 5
Given data:
Boiling point of Ethyl Alcohol: \(T_{1}=78^{\circ} \mathrm{C}=273+78=351 \mathrm{~K}\)
Freezing point of Ethyl Alcohol: \(T_{2}=-114^{\circ} \mathrm{C}+273=159 \mathrm{~K}\)
Latent heat of vapourization: \(L_{v}=879 \mathrm{~kJ} / \mathrm{kg}\)
Latent heat of fusion: \(L_{f}=109 \mathrm{~kJ} / \mathrm{kg}\)
Specific heat capacity: \(C=2.43 \mathrm{~kJ} / \mathrm{kg}-K\)
Mass = \(m=0.51 \mathrm{~kg}\)