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Thermodynamics Problem Set: Carnot Cycle and Specific Heat Capacity of Copper, Exams of Thermal Physics

University of Illinois - ChicagoThermal Physics

Solutions to problems 4.4, 4.5, 4.9, 5.2, and 5.4 from adkins' thermodynamics textbook. The problems involve calculating the work done by a carnot engine, the specific heat capacity of copper, and the entropy change of water during heating and cooling. Detailed calculations and formulas.

Typology: Exams

Pre 2010

Uploaded on 07/22/2009

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1
Fall 2008
Phys 461
Solution Set 3
Adkins 4.4
Let TR = 283 K be the temperature of the river, and Tw be the temperature of the tank of water,
initially at 373K. The temperature of the water tank will decrease continuously from 373K to
283K as a Carnot cycle operates between the water tank and the river.
In each infinitesimal cycle of the Carnot engine, the heat absorbed by the engine from the water
tank is dQ1, heat rejected to the river is dQ2, and work done by the engine is dW
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛−=−=
w
R
T
T
dQdQdQdW 1
121
while ww dTCdQ −=
1 where Cw is the heat capacity of the water tank, and the minus sign takes
care of the fact that dQ1 is defined as the heat absorbed by the engine from the water tank (i.e.
dQ1 is positive when dTw is negative)
Thus, ⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛−=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛−−= 11
w
R
ww
w
R
ww T
T
dTC
T
T
dTCdW
The total work extracted by the engine is given by integrating the expression for dW from the
initial temperature of the water tank to the final temperature
()
()
JoulesKKJ
m
m
KgJcmgm
TT
T
T
TcVdTdT
T
T
CW if
i
f
Rw
T
T
T
T
ww
w
R
w
f
i
f
i
10
36
33
333
10585.11/18.4
10
10
90
373
283
ln283//18.4/110
ln
×≅××=
⎥
⎦
⎤
⎢
⎣
⎡+
⎟
⎠
⎞
⎜
⎝
⎛
×××=
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡−−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡−=
−
∫∫
ρ
pf3
pf4
pf5

Partial preview of the text

Download Thermodynamics Problem Set: Carnot Cycle and Specific Heat Capacity of Copper and more Exams Thermal Physics in PDF only on Docsity!

Fall 2008

Phys 461

Solution Set 3

Adkins 4.

Let TR = 283 K be the temperature of the river, and T w be the temperature of the tank of water,

initially at 373K. The temperature of the water tank will decrease continuously from 373K to

283K as a Carnot cycle operates between the water tank and the river.

In each infinitesimal cycle of the Carnot engine, the heat absorbed by the engine from the water

tank is dQ 1 , heat rejected to the river is dQ 2 , and work done by the engine is dW

w

R

T

T

dW dQ 1 dQ 2 dQ 11

while dQ 1 =−Cw dTwwhere Cw is the heat capacity of the water tank, and the minus sign takes

care of the fact that dQ 1 is defined as the heat absorbed by the engine from the water tank (i.e.

dQ 1 is positive when dTw is negative)

Thus,

w

R w w w

R w w T

T

C dT T

T

dW C dT

The total work extracted by the engine is given by integrating the expression for dW from the

initial temperature of the water tank to the final temperature

( )

( )

J K K Joules m

m

m g cm J g K

T T

T

T

dT dT V c T T

T

W C f i

i

f w R

T

T

T

T

w w w

R w

f

i

f

i

10 6 3

3 3

3 3 3

10 1 / 4. 18 / / 283 ln

ln

= × × ≅ ×

= × × ×

−

∫ ∫

Adkins 4.

dW =dQ 1 −dQ 2 =−mw cwdT 1 −mwcwdT 2 =−mwcw( dT 1 +dT 2 )

Work can be extracted until the two bodies of water reach the same final temperature T f = T 1 T 2

Therefore,

( K K K) kJ

gK

J

m

g m

W m c dT dT mwcwTf T Tf T mwcwT T TT

T

T

T

T

w w

f

i

f

i

6 3

3 3

1 2 1 2 1 2 12

= × × × + − × ≅

−

−

Adkins 4.

Specific heat capacity of copper at low temperature is given by

kJ kg K

T

c (^) p 30. 5 / /

3

⎟ ⎠

, where θ = 348 K is the Debye temperature of copper.

The amount of heat required to heat 100 g of copper from 4K to 20K is given by

[ ]

⎟ = × − ≅

= ×

−

K

K

K

K

mc cpdT kg kJ kg K T dT kJ Joules

20

4

8 4 4

20

4

3 3

To cool the copper block back from 4K to 20K, using a refrigerator that can transfer heat from

the copper block to room temperature, is done using a Carnot engine running in reverse, as

shown below:

(a) The entropy change of the reservoir can be written as

R

p

R

R R T

CdT

T

Q

S

∫

∆ = = , where QR , the

heat absorbed by the reservoir, is negative of the heat absorbed by the water.

Thus SR 4. 18 10 J/K 80 K/ 373 K 896. 5 J/K

3 ∆ =− × × =−

⇒∆S (^) universe =∆Sw+∆SR= 103. 5 J/ K

(b)

2 1 2

2

1

1 1 2

2

1

1

R

T

T

p

R

T

T

p

R

R

R

R R R R T

CdT

T

CdT

T

Q

T

Q

S S S

R

R

R

i

∫ ∫

∆ =∆ +∆ = + = , where Ti = 20°C is the initial

temperature of the water, and TR1 = 50°C and TR2 = 100°C are the temperatures of the two

reservoirs, respectively.

Thus, J K

K

K

K

K

S R J K 948. 6 /

3 =− ⎥ ⎦

∆ =− × +

⇒∆S (^) universe = 51. 4 J/ K

(c)

∫ ∫

f

i

T

T

w R

R S J K

T

dQ

T

dQ S 1000 /

1 2

⇒ ∆Suniverse = 0 as expected for a reversible process.

Adkins 5.

kJ K T

J

T

T

gK

J

kg

T T

CdT S S S

i f

f

T

T f

p w heating vaporization

f

i

ln

Latentheat

6

=

×

= ×

∫

using Ti = 288K and Tf = 373K.

To check for irreversibility, we need to look at the entropy change of the universe, and hence the

entropy change of the reservoir.

J K J K kJ K

kg J g K K K J K T T

CdT

S

f f

T

T

p

R

f

i

Latentheat

3 3

6

=− × − × = −

− =− × × − ×

∫

Therefore, ∆ S universe =∆Sw+∆SR= 130 J/K> 0 ⇒irreversible process.