CBE30361 Lecture 7: Thermo lecture 8

• School: ND
• Department: Chemical & Biomolecular Engr.
• Course: CBE30361
• Professor: Ruilan Guo

Document Summary

Energy (cid:1847)(cid:1755)(cid:1755) (cid:1872)(cid:1867)(cid:1872)(cid:1853)(cid:1864) (cid:1861)(cid:1866)(cid:1872)(cid:1857)(cid:1870)(cid:1866)(cid:1853)(cid:1864) (cid:1857)(cid:1866)(cid:1857)(cid:1870)(cid:1859)(cid:1877) (cid:1847)(cid:1755)(cid:1755)= (cid:1865)(cid:1847) (cid:1875)(cid:1860)(cid:1857)(cid:1870)(cid:1857) (cid:1847) = (cid:1871)(cid:1868)(cid:1857)(cid:1855)(cid:1861)(cid:1858)(cid:1861)(cid:1855) (cid:1861)(cid:1866)(cid:1872)(cid:1857)(cid:1870)(cid:1866)(cid:1853)(cid:1864) (cid:1857)(cid:1866)(cid:1857)(cid:1870)(cid:1859)(cid:1877) In general, we consider problems where kinetic energy + potential << internal. I. e. , we can neglect it (they are either negligible or they don"t change). T, v (temperature and specific volume) --> u is fixed. Just need two variables to specify it. Put in ice bath, take to 50 (cid:1829), measure the amount of ice melted (determine transfer of energy) Burn methane until temp. is 120*c, measure the amount of methane that was burned (determine transfer of energy) Even if you do a bunch of weird things to the system: The amount of ice you need to melt to get the liquid at state 7 to 50*c is the same as that same liquid at state 1. Therefore, the net gain/loss of energy is 0, and the internal energy is fixed at this point. Neglect kinetic and potential (cid:1856)(cid:1847) = (cid:2034)(cid:1843) +(cid:2034)(cid:1849)