SOLUTIONS MANUAL
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Solutions Chapter 1 Page 1
D E Winterbone Current edition: 13/02/2015
, Chapter 1 Solutions
P1.1 This example shows how the First Law can be applied to individual processes and how these
can make up a cycle.
A cycle of events is shown in Fig A.10. It is made up of four processes, and the heat and work
associated with those processes is as given.
Fig A.10: Cycle of events made up of four processes.
Process 12: Q = +10J W = -18J
Process 23: Q = +100J W = 0J
Process 34: Q = -20J W = +70J
Process 41: Q = -10J W = +28J
Calculate the values of Un - U1, the net work, the net heat transfer and the heat supplied for the cycle.
Solution
This problem can be solved by applying the First Law to each of the processes in turn, when the change
in internal energy is dU Q W .
Process 12: dU 12 U2 U 1 Q 12 W 12 10 18 28J
Process 23: dU 23 U3 U2 Q23 W 23 100 0 100J
Process 34: dU 34 U4 U3 Q34 W 34 20 70 90J
Process 41: dU 41 U1 U4 Q41 W 41 10 28 38J
The change in internal energy relative to the internal energy at point 1, U1, is given by
dU1n Un U1 Un Un1 Un1 Un2 ....... U2 U1
dUn1,n dUn2,n1 ....... dU12
Hence:
dU 12 U 2 U 1 28J
dU 13 dU 12 dU 23 28 100 128J
dU 14 dU 12 dU 23 dU 34 28 100 (90) 38J
dU 11 dU 12 dU 23 dU 34 dU 41 28 100 (90) (38) 0J
The result dU11 = 0 confirms that the four processes constitute a cycle, because the net change of state
is zero.
The net work done in the cycle is
W W
cycle
12 W23 W34 W41
18 0 70 28 80J
The positive sign indicates that the system does work on the surroundings.
The net heat supplied in the cycle is
Solutions Chapter 1 Page 1
D E Winterbone Current edition: 13/02/2015