Physics 211 Test 4 - · PDF filef = 3:33 105J=kg, L ... (15 points) If...

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Physics 211 Test 4 NAME: SECTION: Physics 211 Test 4 June 20, 2011 useful constants: c w = 4186J/kgK , L f =3.33 × 10 5 J/kg, L v = 22.6 × 10 5 J/kg, c ice = 2100J/kgK , k B =1.38 × 10 -23 J/K , σ =5.67 × 10 -8 W/(m 2 K 4 ), ρ steel =7.8 × 10 3 kg/m 3 , α steel = 11 ×10 -6 C -1 , c steel = 460J/kgK , k steel = 46W/m C , α al = 24 ×10 -6 C -1 , c al = 920J/kgK , k al = 240W/m C , α glass =9.0 × 10 -6 C -1 , c glass = 840J/kgK , k glass = 0.84W/m C , STP : T =0 C and P =1atm = 101.3kPa. Multiple choice questions 2 points each. 1. Both the pressure and volume of a given sample of an ideal gas double. This means its temperature in Kelvins must (a) reduce to one-half of is original value. (b) reduce to one-fourth of its original value. (c) remain unchanged. (d) double. (e) quadruple. 2. It is a well known fact that water has a higher specific heat than steel. Now consider equal masses of water and iron that are initially in thermal equilibrium. The same amount of heat, 30. calories, is added to each. Which statement is true? (a) They are no longer in thermal equilibrium; the water is warmer. (b) It is impossible to say without knowing the exact mass involved and the exact specific heat capacities. (c) They remain in thermal equilibrium. (d) They are no longer in thermal equilibrium; the iron is warmer. 3. A gas is expanded to twice its original volume with no change in its temperature. This process is (a) isobaric. (b) adiabatic. (c) isentropic. (d) isothermal. (e) isometric. C W Fay IV: Summer2011 1

Transcript of Physics 211 Test 4 - · PDF filef = 3:33 105J=kg, L ... (15 points) If...

Page 1: Physics 211 Test 4 -   · PDF filef = 3:33 105J=kg, L ... (15 points) If 0:040kg0:060kg0:080kgof ice at 0 Cis added to 0:300kgof water at ... Physics 211 Test 4 NAME: SECTION: 12

Physics 211 Test 4 NAME: SECTION:

Physics 211 Test 4June 20, 2011

useful constants: cw = 4186J/kgK, Lf = 3.33 × 105J/kg, Lv = 22.6 × 105J/kg,cice = 2100J/kgK, kB = 1.38 × 10−23J/K, σ = 5.67 × 10−8W/(m2K4), ρsteel = 7.8 ×103kg/m3, αsteel = 11×10−6C−1, csteel = 460J/kgK, ksteel = 46W/m◦C, αal = 24×10−6C−1,cal = 920J/kgK, kal = 240W/m◦C, αglass = 9.0 × 10−6C−1, cglass = 840J/kgK, kglass =0.84W/m◦C, STP : T = 0◦C and P = 1atm = 101.3kPa.Multiple choice questions 2 points each.

1. Both the pressure and volume of a given sample of an ideal gas double. This meansits temperature in Kelvins must

(a) reduce to one-half of is original value.

(b) reduce to one-fourth of its original value.

(c) remain unchanged.

(d) double.

(e) quadruple.

2. It is a well known fact that water has a higher specific heat than steel. Now considerequal masses of water and iron that are initially in thermal equilibrium. The sameamount of heat, 30. calories, is added to each. Which statement is true?

(a) They are no longer in thermal equilibrium; the water is warmer.

(b) It is impossible to say without knowing the exact mass involved and the exactspecific heat capacities.

(c) They remain in thermal equilibrium.

(d) They are no longer in thermal equilibrium; the iron is warmer.

3. A gas is expanded to twice its original volume with no change in its temperature. Thisprocess is

(a) isobaric.

(b) adiabatic.

(c) isentropic.

(d) isothermal.

(e) isometric.

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Physics 211 Test 4 NAME: SECTION:

4. Phase change occurs

(a) as the temperature increases.

(b) as the temperature decreases.

(c) as the temperature remains the same.

(d) all of the above.

5. A circular piece is cut from an aluminum sheet at room temperature. The sheet andthe circular piece are separately raised to 100◦C.

(a) The circular piece will be smaller than the hole.

(b) The circular piece will be larger than the hole.

(c) The circular piece will exactly fit the hole.

6. A coal-fired plant generates 500MW 550MW 520MW of electric power. The plantuses 4.8×106kg of coal each day. The heat of combustion of coal is 3.3×107J/kg. Thesteam that drives the turbines is at a temperature of 300◦C and the exhaust water isat 37◦C.

(a) (10 points) What is the overall efficiency of the plant for generating power?

(b) (10 points) What is the Carnot efficiency?

given:Pout = 600MW m = 4.8× 106kg C = 3.3× 107J/kgTH = 300◦C = 573K TC = 37◦C = 310K

a)

ε =Pout

Pin

=Pout

mC/(24 ∗ 3600)= 27%(30%)(28%) (1)

b)

εc = 1− TcTH

= 46% (2)

7. (5 points) State in your own words the second law of thermodynamics?

8. (15 points) If 0.040kg 0.060kg 0.080kg of ice at 0◦C is added to 0.300kg of water at25◦C in a 0.100kg aluminum calorimeter cup, What is the final temperature of thewater? TF =

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given:mi = 0.050kg Ti = 0◦C Tf =? mw = 0.300kg(0.400kg)(0.500kg)Tw = 25◦C mc = 0.100kg

miLf +micw∆T1 +mccc∆T2 +mwcw∆T2 = 0 (3)

miLf +micw(Tf − Ti) + (mccc +mwcw)(Tf − Tw) = 0 (4)

miLf +micwTf −micwTi + (mccc +mwcw)(Tf )− (mccc +mwcw)(Tw) = 0 (5)

(micw +mccc +mwcw)Tf = micwTi + (mccc +mwcw)(Tw)−miLf (6)

Tf =(micwTi + (mccc +mwcw)(Tw)−miLf )

(micw +mccc +mwcw)= 13◦C(8.6◦C)(4.2◦C) (7)

9. (15 points) A certain object with a surface temperature of 150◦C 200◦C 250◦C is ra-diating heat at a rate of 200J/s. To double the object’s rate of radiation energy, whatshould be its surface temperature in Kelvin? T =

given:P1 = 200W T1 = 100◦C(120◦C)(150◦C) = 373K(393K)(423K)T2 =?

P = σAe(T 4) (8)

P2 = 2P1 (9)

eσA(T 42 ) = 2eσA(T 4

1 ) (10)

T 42 = 2T 4

1 (11)

T2 = 4√

2T1 = 443K(467K)(503K) (12)

10. (5 points) Glass has more than 3 times larger thermal conductivity than air. Yet glass-wool (glass-fibers and air) is preferable to insulate wall spaces rather than leaving”empty” air spaces. Why is glass-wool better even though glass has a larger conduc-tivity than air?Solution: Convection is very important in the transfer of heat through air.

11. At room temperature (20.0◦C) the area of a hole in a sheet of material is 5.505cm2.When heated to 830◦C the area increases by 30.0mm2 40.0mm2 20.0mm2 .

(a) (10 points) What is the thermal coefficient of expansion for the material? α =

(b) (5 points) If the temperature had decreased to −120◦C, what would be the newarea of the hole? A =

given: A0 = 5.505cm2 ∆A = 29.0mm2 = 0.290cm2 Tf2 = −20◦C Ti = 20◦CTf1 = 830◦C(800◦C)(770◦C)a)

∆A = A02α∆T (13)

α =∆A

2A0∆T= 3.25× 10−5C−1(4.49× 10−5C−1)(2.24× 10−5C−1) (14)

b)

A = A0(1 + 2α∆T ) = 5.557cm2(5.574cm2)(5.540cm2) (15)

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12. (15 points) A container of ideal gas at STP undergoes an isothermal expansion and itsentropy changes by 3.00J/K 4.00J/K 5.00J/K . How much work does it do? W =SOLUTIONGiven: T = 0◦C = 273K ∆S = 3.66J/K(4.66J/K)(5.66J/K) W =?

∆S =Q

T(16)

∆U =3

2nR∆T = 0 (17)

Q = ∆U +W (18)

W = Q = ∆ST = 819J(1090J)(1370J) (19)

13. (10 points)Take home problem The temperature of 2.0mol of ideal gas is increasedfrom 150◦C to 250◦C by two different processes. In Process A, 2500J of heat is addedto the gas; in process B, 3000J of heat is added. (a) In which case is more work done(Explain) (b) Calculate the change in internal energy and work done for each process.

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