System Win = -4J Qn = 40 J 15 J out in Qout = - 25 J Wout 10 J AU = Q - W= 15 J - 6 J = +9J (a) W = - 159 J System 159 J 150 J in out Qout = - 150 J AU = Q - W = - 150 J -(- 159 J) = +9J (b) Figure 15.4 Two different processes produce the same change in a system. (a) A total of 15.00 J of heat transfer occurs into the system, while work takes out a total of 6.00 J. The change in intemal energy is AU = Q - W = 9.00 J . (b) Heat transfer removes 150.00 J from the system while

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Chapter1: Units, Trigonometry. And Vectors
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(a) Suppose there is heat transfer of 40.00 J to a system, while the system does 10.00 J of work. Later, there is heat transfer of 25.00 J out of the system while 4.00 J of work is done on the system. What is the net change in internal energy of the system?
(b) What is the change in internal energy of a system when a total of 150.00 J of heat transfer occurs out of (from) the system and 159.00 J of work is done on the system?

System
Win = -4J
Qn
= 40 J
15 J
out
in
Qout = - 25 J
Wout
10 J
AU = Q - W= 15 J - 6 J = +9J
(a)
W = - 159 J
System
159 J
150 J
in
out
Qout = - 150 J
AU = Q - W = - 150 J -(- 159 J) = +9J
(b)
Figure 15.4 Two different processes produce the same change in a system. (a) A total of 15.00 J of heat transfer occurs into the system, while work
takes out a total of 6.00 J. The change in intemal energy is AU = Q - W = 9.00 J . (b) Heat transfer removes 150.00 J from the system while
Transcribed Image Text:System Win = -4J Qn = 40 J 15 J out in Qout = - 25 J Wout 10 J AU = Q - W= 15 J - 6 J = +9J (a) W = - 159 J System 159 J 150 J in out Qout = - 150 J AU = Q - W = - 150 J -(- 159 J) = +9J (b) Figure 15.4 Two different processes produce the same change in a system. (a) A total of 15.00 J of heat transfer occurs into the system, while work takes out a total of 6.00 J. The change in intemal energy is AU = Q - W = 9.00 J . (b) Heat transfer removes 150.00 J from the system while
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