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