DeltaU = Q + W
Q = heat absorbed = 0.615kJ = +0.615kJ (+ve when heat is absorbed) = 615J
W = 0.247calories = 0.247*4.184J = 1.03J = -1.03J (negative when work is done on the system)
Delta U = 615J + (-1.03J) = 613.9Joules = 0.614kJ
Change in Energy = 0.614kJ
CATES UI Work to the system. What is the change in energy 3. What is the...
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1. a) How much work is done by/on the system when is taken from
point a to point b following the
path indicated in the figure?
b) How much heat is absorbed/released by the system when is
taken from point a to point b
following the path indicated in the figure? Express your answer in
kJ
c) What is the change in internal energy, ?U? Express your
answer in kJ.
2. a) How much work is done on the system...
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