10.3 Solutions
449
(c)
Fig. 10.12 (a) Carnot cycle consisting of two isothermic processes AB and CD; two adiabatic
processes BC and DA. (b) Sterling cycle consisting of two isothermic process AB and CD, two
isochoric processes BC and DA
10.51 (a) n =
PV
RT
=
1.013 × 10 5 × 1 × 10 −3
8.31 × 273
= 0.04465 mol
(i) dQ = nC P dT
S =
T f
T i
dQ
T
= nc p
T f
T i
dT
T
= nc p ln
T f
T i
= 0.04465 × 21 × ln(500/273) = 0.567 J/K
(ii) dQ = nC v dT
S =
T f
T i
dQ
T
= nc v
T f
T i
dT
T
= nc v ln
T f
T i
= 0.04465 × 12.7 × ln(500/273) = 0.343 J/K
(b) dQ 1 = mC v dT
S 1 = m C v
T f
T i
dT
T
= mC v ln
T f
T i
= 1.0 × 4.13 × 10
3
× ln(313/273) = 695 J/K
When 0.5 kg water at 0 ◦ C is mixed with 0.5 kg water at 100 ◦ C, the final
temperature would be 50 ◦ C.
S 2 = 0.5 × 4.13 × 10
3
× ln(313/273)
+ 0.5 × 4.13 × 10
3
× ln(313/373)
= 282.35 − 362.15 = −79.8 J/K