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• Mia L. Ayuningtyas

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5. Systematic use of solar energy can yield a large saving in the cost of winter space heating for a typical house in the north central United States. If the house has good insulation, you may model it as losing energy by heat steadily at the rate 6000 W on a day in April when the average exterior temperature is 4 °C, and when the conventional heating system is not used at all. The passive solar energy collector can consist simply of very large windows in a room facing south. Sunlight shining in the daytime is absorbed by the floor, interior walls, and objects in the room, raising their temperature to 38 °C. As the sun goes down, insulating draperies or shutters are closed over the windows. During the period between 5:00 P.M. and 7:00 A.M. the temperature of the house will drop, and a sufficiently large “ thermal mass” is required to keep it from dropping to far. The thermal mass can be a large quantity of stone ( with specific heat 850 J/kg·°C) in the floor and the interior walls exposed to sunlight. What mass of stone is required if the temperature is not to drop below 18°C overnight? Solution The stone energy reservoir has a large area in contact with air and is always at nearly the same temperature as the air. Its overnight loss of energy is

ρ=

Q mc ∆ T = ∆t ∆t

s (−6000 J /s)(14 h)(3600 ) ρ∆t h 3.02 x 108 J ∙ kg ∙° C 4 m= = = =1.78 x 10 kg c ∆ T (850 J /kg ∙° C)(18 ° C−38 ° C) 850 J (20° C)

14. Two thermally insulated vessels are connected by a narrow tube fitted with a valve that is initially closed. One vessel, of volume 16.8 L, contains oxygen at a temperature of 300 K and a pressure of 1.75 atm. The other vessel, of volume 22.4 L, contains oxygen at a temperature of 450 K and a pressure of 2.25 atm. When the valve is opened, the gases in the two vessels mix, and the temperature and pressure become uniform throughout. (a) What is the final temperature? (b) What is the final pressure? Solution Vessel one contains oxygen according to PV = nRT : 5 −3 3 PV 1.75 ( 1.013 x 10 Pa ) 16.8 x 10 m nc = = =1.194 mol RT 8.314 Nm/mol ∙ K (300 K )

Vessel two contains much oxygen :

2.25 ( 1.013 x 105 ) 22.4 x 10−3 nk = mol=1.365 mol 8.314 ( 450 ) (a)

The gas comes to an equilibrium temperature according to

( mc ∆ T )cold =−( mc ∆ T )hot nc Mc ( T f −300 K ) +n k Mc ( T f −450 K )=0 The moral mass M and specific heat divide out:

1.194 T f −358.2 K +1.365 T f −614.1 K=0 T f= (b)

972.3 K =380 K 2.559

The pressure of the whole sample is its final state is

P=

nRT 2.559 mol 8.314 J 380 K 5 = =2.06 x 10 Pa=2.04 atm −3 3 V molK ( 22.3+16.8 ) x 10 m

21. In an insulated vessel, 250 g of ice at 0°C is added to 600 g of water at 18.0°C. (a) What is the final temperature of the system? (b) How much ice remains when the system reaches equilibrium? Solution (a) Since the heat required to melt 250g of ice at 0°C exceeds the heat required to cool 600g of water from 18°C to 0°C, The final temperature of the system(Water+ice) must be 0°C (b) Let m represent the mass of ice that melts before the system reaches equilibrium at 0°C

Qcold =Q hot m L f =−m w c w (0 ° C−T i)

(

m 3.33 x 105

J =−(0.6 kg)(4186 J /kg ∙ ° C)( 0° C−18 °C) kg

)

m = 136g the ice remaining = 250g – 136g = 114g

23. A sample of ideal gas is expanded to twice its original volume of 1.00 m 3 in a quasi-static process for which

P=∝ V 2 , with

∝ = 500 atm/m6, as shown in

Figure P20.23. How much work is done on the expanding gas?

Solution f

W if =−∫ PdV i

The work done on the gas is the negative of the area under the curve

P=∝ V

2

between Vi f

W if =−∫ PdV = i

and Vf

−1 ∝(V 3f −V 3i ) 3

Vf = 2Vi = 2(1.00m3) = 2.00 m3

1

Wif = - 3

[(5.00atm/m6)(1.013 x 105 Pa/atm)][(2.00m3)3 + (1.00 m3)3] =

-1.18 MJ

26. An ideal gas is enclosed in a cylinder that has a movable piston on top. The piston has a mass m and an area A and is free to slide up and down, keeping the pressure of the gas constant. How much work is done on the gas as the temperature of n mol of the gas is raised from T 1 to T2? Solution f

f

W =−∫ PdV =−P∫ dV =−P ∆ V =−nR ∆ T =−nR (T 2−T 1 ) i

i

29. A thermodynamic system undergoes a process in which its internal energy decrease by 500 J. At the same time, 220 J of work is done on the system. Find the energy transferred to or from it by heat. Solution

∫ ¿=Q+W ∆ E¿

∫ ¿−W =−500 J−200 J =−720 J Q=∆ E ¿ The negative sign indicates that positive energy is transferred from the system by heat.

39. A 2.00-mol sample of helium gas initially at 300 K and 0.400 atm is compressed isothermally to 1.20 atm. Nothing that the helium behaves as an ideal gas, find (a) the final volume of the gas, (b) the work done on the gas, and (c) the energy transferred by heat. Solution (a)

Pi V i =P f V f =nRT =2 mol ( 8.314 J / K ∙mol ) (300 K ) =4.99 x 103 J

V i=

Vf=

nRT 4.99 x 10 3 J = Pi 0.4 atm

nRT 4.99 x 103 J 1 3 = = V i=0.041 m Pf 1.2 atm 3

(b)

W =−∫ PdV =−nRT ln

Vf 1 =−( 4.99 x 103 ) ln =5.48 kJ Vi 3

(c)

∫ ¿=0=Q+ W ∆ E¿ Q = -5.48

kJ

43. A bar of gold is in thermal contact with a bar of silver of the same length and area (Fig. P20.43). One end of the compound bar is maintained at 80.0°C while the opposite end is at 30.0°C. When the energy transfer reaches steady state, what is the temperature at the junction?

Solution In the steady state condition

k Au A Au

ρ Au=ρ Ag

( ∆∆ Tx )

Au

=k Ag A Ag

( ∆∆ Tx )

Ag

A Au= A Ag ∆ x Au=∆ x Ag ∆ T Au=80−T ∆ T Ag=T −30 T is the temperature of the junction

k Au (80−T )=k Ag (T −30) T = 51.2 °C

51. The intensity of solar radiation reaching the top of the Earth’s atmosphere is 1340 W/m2. The temperature of the Earth is affected by the so-called greenhouse effect of the atmosphere. That effect makes our planet’s emissivity for visible light higher than its emissivity for infrared light. For comparison, consider a spherical object with no atmosphere, at the same distance from the Sun as the

Earth. Assume that its emissivity it the same for all kinds of electromagnetic waves and that its temperature is uniform over its surface. Identify the projected area over which it absorbs sunlight and the surface area over which it radiates. Compute its equilibrium temperature. Chilly, isn’t it? Your calculation applies to (a) the average temperature of the moon, (b) astronauts in mortal danger aboard the crippled Apollo 13 Spacecraft, and (c) global catastrophe on the Earth if widespread fires should cause a layer of soot to accumulate throughout the upper atmosphere, so that most of the radiation from the Sun were absorbed there rather than at the surface below the atmosphere. Solution The sphere of radius R absorbs sunlight over the area of its day hemisphere, projected as a flat circle perpendicular to the light : radiates in all directions, over area

4 π R2 . Then, in steady state.

ρ¿ =ρout

(

e 1340

[

T=

W π R2 =eσ (4 π R 2)T 4 2 m

)

1340 W /m 2 W 4 (5.67 x 10−8 2 4 ∙) m K

1/ 4

]

=277 K =4 ° C

2

πR .

It

70. The inside of a hollow cylinder is maintained at a temperature T a while the outside is at a lower temperature, Tb(Fig.P20.70). The wall of the cylinder has a thermal conductivity k. Ignoring end effects, show that the rate of energy conduction from the inner to the outer surface in the radial direction is

[

T −T b dQ =2 πLk a dt ln b/a

]

(Suggestion: The

temperature gradient is energy current passes cylinder of are

dT/dr. Note that a radial through a concentric

2 πrL . )

Solution For a cylindrical shell of radius r, height L, and thickness dr, the equation for thermal conduction

dQ dT =−k A dt dx

becomes

dQ dT =−k (2 πrL) dt dr

Under equilibrium conductions,

dT =

−dQ 1 dT dt 2 πkL r

(

)

and

dQ dt

is constant

T b−T a=

−dQ 1 b ln dt 2 πkL a

(

)

but

T a> T b

dQ 2 πkL(T a −T b) = dt ln b/a

71. The passenger section of a jet airliner is in the shape of a cylindrical tube with a length of 35.0 m and an inner radius of 2.50 m. Its walls are lined with an insulating material 6.00 cm in thickness and having a thermal conductivity of 4.00 x 10-5 cal/s·cm·°C. A heater must maintain the interior temperature at 25.0 °C while the outside temperature is -35.0 °C. What power must be supplied to the heater? (Use the result of Problem 70.) Solution From problem 70 the rate of energy flow through the wall is

dQ 2 πkL(T a −T b) = dt ln b/a

s ∙ cm∙ ° C 4.00 x 10−5 cal/ ¿ ¿ (3500 cm)(60.0 ° C) 2π¿ dQ =¿ dt dQ =2.23 x 103 cal/s=9.32 k W dt This is the rate of energy loss from the plane by heat, and consequently is the rate at which energy must be supplied in order to maintain a constant temperature.