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Unformatted text preview: Quiz 7. Monday, March 24‘“, 2008 PHY 252 Name. SOLu—nozu 1. Two blocks of copper, each exactly of 1 kg mass, are placed in a well
insulated container so that no heat can flow in or out of the system (:12. Bl°°k l BIOCI‘ 2 it is a closed system). Block 1 is initially at temperature Tl =60.0 °C,
block 2 is initially at temperature T2 = 20.0 °C. Cu Cu
1.0 kg 1.0 kg
(a) Calculate the equilibrium temperature Tf. (1 point) 600 DC 200 0C
From lsi: law Qt°t=o = Q‘+Qz QI+ at. = MICi (T; T;) + MIC1CTgT;) ‘‘‘ O :2) T3, = “‘31; IMICITI .__ éCTliTg) = 41(60‘c.+2.o°c) 2' 40°C — 'A‘
(b) Calculate the change in e‘ii‘ti’cfﬁy of the system, ASJr once equilibrium is reached.
(CCll =390J/kgoK) (1 point)
Téd Tédﬂ T 1"
ASed» =' A5. + A5,. = J .3: + 3 __.=== mgj‘il: +M1ctj‘ﬂ’
"r. T 1.‘ T T 'T‘ T T
‘ I.
Astot = nigh1% +“:Ca1HIi 7:. “cauLAI—E
‘ Ta, TIT},
1.
as“ = (1:3)(aaomna.n)u[_&_m_ .. lé'J'IK *
(z‘i‘SHCSBI) (c) Now consider the system before equilibrium is reached. After some time a certain amount of
heat Q is transferred from block 1 to block 2. The temperature of block 1 has fallen by an amount
AT , and the temperature of block 2 has risen by an equal amount AT (since they have the same
mass and specific heat, and heat is conserved by the 1st law), but neither has yet reached Tf. Derive down an expression for the entropy change of the system AS in terms of AT , m, CONT“ and T2. (1 point)
Frau lsi law Ql =  at ‘= MCAT
1151“ “Rur
AS ::. AS.+ASL == ‘73! + 9L9. :_ WICcu +1“ EAT)
T‘ T T T T. 1I
I. ((1) Now solve for the value of AT that maximizes AS . (Hint, solve d(AT) = 0 ), and comment brieﬂy on the physical insight revealed by your result. (2 points)
Differentiate AS: dcas) :: urn—GEE. + lid—‘1‘. = a forum: orm‘u‘
cl CAT) Fr! "AT TadHAT '
% TI+AT +T1+AT =0 ==> AT = 1"“ : 50°C“2°°‘ ... 206°
'2 2 “There is atumiuﬂ Poiulr for AS when
coltickle T5 .
To electJ: ‘15 N.” is a. Maximum or CLICAS) .__ ._ WIng “(6“ h. atcm‘ ("n—ATP (mm: Eiuilibr‘fum CorrespauaLs to 'the. maximum eu‘Lropd change that? is
canalsten‘t witL.__'l:ke "its! low . =‘T.AT = 6010 = 40°C Mimmum 6! :Hereu'tliat'e ad q in < O alwaﬂL .‘. 45 .15 mxﬁﬂﬂhd‘ﬁ Quiz 7. Monday, March 24‘“, 2008 PHY 252 2. An ideal heat engine has a theoretical efﬁciency of 25.0 % when the high temperature reservoir is at
TH = 127 °C . (a) What is the temperature of the low temperature reservoir, TL ? (1 point)
e , .__= l __ “Tc.
Ideal. 3F:
0.25. ___ l.._ r¢_ a 1.. :5 @119: trooth—oar)
2?3N2?K ‘HDOK Tc = 300 k .....__.. +
In one cycle, a real engine extracts 250. J of heat from the hot reservoir while doing 40. J of work.
(b) What is the actual efﬁciency of the engine? (1 point)
._. _. T ._..
ecu114.4 ‘ 19/— — “'0' = 0.14 :2 16% — =14
Qu 2 s 0.3
(c) How much heat per cycle is exhausted to the cold reservoir? (1 point)
[Q4 : [QH[_[w] = 230.2:— A—oq‘ = 210:: i4
3. 1.00 g of ice at 0.0 °C melts in a large lake of water whose temperature is very slightly above 0.0 °C.
Estimate the entropy change of the ice? (LF = 333 J/ g) (2 points)
AS"; '2 77 5"— ~= ..M_L_§ 56: a.“ pretend
e T T .1.
Ast'ce = C "005) “_(_33__3 We) :. 1.22 T/K 9(
2? 3 K
What is the entropy change of the ice—lake system? (1 bonus point) 'qu. tux—.th macaw from rum at 0°C, .‘. Aside :: — :LF 2:.  1.7.2. T/K
an Aswan = 45,“ +AsMe := L22 r.22 = OU/k* 1e. JUL» imam; Promass L4 rem—AM. ...
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 Spring '08
 Treacy
 Thermodynamics, Heat, entropy change, equilibrium temperature Tf.

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