Question

Name Chem 1120 Homework #2: Liquids, Solids, and Intermolecular Forces Ethyl Alcohol 70 J/g C 854 J/g 78。 2.43 J/g C 115。1 105 J/g 0.97 J/g C 4. The heating curve shown above is forethanol. How much heat is required to warm a 50.0 g sample of at-90°C to 88 C?

Answer 1

Given temperatures are Initial temp-90 °C Final temp 88 °C Specific heat for solid 0.97J/g* °OC Specific heat for liquid 2.43 J/g* °C Specific heat for gas 1.70 J/g* °C Melting point of ethyl alcohol - - 115 °C Bpoling point of ethyl alcohol -78 °C At 90 °C, ethyl alcohol is in liquid state. 1. Conversion of liquid at 90 °C to liquid at 78 °C q1 m*C*AT 50.0g* 2.43 J/g* °C (78 (-90) °C 20412 J q2-m*AHy 50 g* 854J/g 42700 J q,-m*C*AT-50 g* 1.70 J/g* °C * (88-(78)"C-850 J 2. Conversion of liquid 78 °C to vapor at 78 °C 3. Conversion of vapour at 78°C to vapour at 88 °C Total heat energy required q1q2q3 - 20412+42700850 63962 J 63.962 kJ