Chem+160+Handout+Problems+Answers - HANDOUT PROBLEMS...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
HANDOUT PROBLEMS SOLUTIONS H1CH12. Start problems of this kind with a statement of conservation of energy in words: heat lost by the hot coffee = heat needed to warm the ice to 0°C + heat needed to melt the ice + heat gained by the melted ice. Convert this statement into symbols: –q coffee = q ice + q fusion + q water –q coffee = ms T = m coffee (4.184 J/g°C)(T final – 98.4)°C q ice = m ice s ice (0 –(–2.2))°C q water = m water s water (T final – 0)°C m ice = m water s ice s water q fusion = 7685.39(=7690) J Putting all of this together and solving for T final gives T final = 83.2°C. H2CH12. (a) Δ vap = 22.2 kJ/mol (b) –185.4 – 22.2 = –207.6 kJ/mol H3CH12. (a) The vapor pressure depends on the temperature, which has not changed. Therefore, the partial pressure of water in the compressed gas will be 19.8 torr: some liquid must condense. (b) 0.0388 g (calculated from n = PV/RT). H4CH12.
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

This document was uploaded on 10/30/2011 for the course CHEM 160:160 at Rutgers.

Ask a homework question - tutors are online