Re: 嗯...
: 1.Show that if a system undergoes an irreversible isothermal process
from state a to b without doing any work,
then the total entropy production is given by;
S tot = Wba /T
: Where Wb is the work that must be done in a reversible process
that restores the system to its initial state.
:
2.Calculate the total entropy change when a 2.00 Kg mass at 300 K
falls freely a distance of 25.0 m into a heat bath at 300 K.
:
3.Compute the change in entropy of water when 55.5 mol of water is
converted from ice at –10.0 oC and 1 atm
to water vapor at 110 oC and 1 atm.
: For H2O(s): Hfus = 6.01 KJ/mol Cp = 37.66 K-1mol-1 T fus = 273.15 K
: For H2O(l): Hvap = 44.0 KJ/mol Cp = 75.32 K-1mol-1 T vap = 373.15 K
: For H2O(g): Cp = 30.54 +0.0103 T K-1mol-1
:
4.Pure water that is free of dust can be supercooled,
that is, cooled below the freezing point.
Under carefully controlled condition water can be supercooled to–45 oC.
Consider an adiabaticlly isolated 100 g sample of supercooled water at
: t –10.0 oC athat spontaneously begins to freeze.
Ice is more ordered than liquid water at the same temperature.
How do you reconcile the above spontaneous process with the second law?
Calculate the total entropy change for the process.
See above problem for data.
:
5.Problem 4.5 of the textbook.
:
6.Problem 4.23
:
: 修整過了.....
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