Physics 101 Quiz #8 SolutionNovember 20, 2002 Name

20 Minutes. Box your answers.

density specific heat thermal coefficient ofcapacity conductivity thermal expansion

(kg/m3) (J/kg◦C) (J/m s ◦C) (1/◦C)water 1000 4186 0.60 69× 10−6

copper 8890 387 390. 17× 10−6

aluminum 2700 900 240. 23× 10−6

1. A 4 kg aluminum rod has a temperature of 100◦C. It is placed into an insulatedbucket containing 50 kg of water at 0◦C.

(a) [4 points] What is the equilibrium temperature of the water andaluminum?

Qgained by water = Qlost by aluminum

mH20 cH20 (T − 0◦C) = mAl cAl (100◦C− T )

(50 kg)(

4186 Jkg◦C

)(T − 0◦C) = (4 kg)

(900 J

kg◦C

)(T − 0◦C)

2.09× 105 (T − 0◦C) = 3600 (100◦C− T )2.09× 105 T = 3.6× 105◦C− 3600T2.13× 105 T = 3.6× 105◦C

T = 1.7◦C

(If you don’t get the answer to part a, use T = 50◦C in part b.)

(b) [2 points] The original length of the aluminum rod was 0.50 m. By howmuch did its length change?

∆L = αL0∆T = (23× 10−6)(0.5 m)(100.0◦C− 1.7◦C) = 0.0011 m = 1.1 mm

Continued on the other side....

2. In an industrial lab, a vat of liquid at 20◦C must be heated. This is accomplishedby running a hot water pipe of length 0.50 m and radius 4.0 cm (0.04 m) throughthe vat. The pipe is made of 2.0 mm (0.002 m) thick copper. The water flowingthrough the pipe has an average temperature of 60◦C.

(a) [2 points] How much heat flows through the wall of the pipe in ten seconds?

The pipe is a cylinder of length L = 0.5 m and radius r = 0.04 m, so the pipewall has area A = 2πLr = 0.126 m2. It has thickness l = 0.002 m, so

Q =kA∆Tt

l=

(390 J

ms◦C

)(0.126 m2)(60◦C− 20◦C)(10 s)

0.002 m= 9.8× 106 J

(b) [2 points] The water in the pipe flows at a velocity of 20.0 m/s. How muchdoes the temperature of the water change as it flows through the pipe?(Hints: (1) The temperature change is small, so your answer to part a isvalid even though T is changing; (2) What mass of water flows throughthe pipe in 10 seconds? How can you use this number?)

(Note: this solution uses different definitions of A and ∆T than part a. Flowrate is written as Qf to distinguish it from heat, which is written as Q.)The cross-sectional area of the pipe is A = πr2 = π (0.04 m)2 = 0.00503 m2.The volume flow rate is Qf = vA = (20 m/s)(0.00503 m2) = 0.101 m3/s. Int = 10 s, V = Qf t = (0.101 m3)(10 s) = 1.01 m3 of water flows through thepipe. This has mass m = ρV = (1000 kg/m3)(1.01 m3) = 1010 kg. The waterloses the energy that passes through the pipe wall, so Q = −9.8× 106 J, and

Q = mc∆T ⇒ ∆T =Q

mc=

−9.80× 106 J

(1010 kg)(

4186 Jkg◦C

) = −2.3◦C

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Physics 101 Quiz #8 SolutionNovember 19, 2004 Name

20 Minutes. Box your answers.

Constants needed in the quiz:Radius of the earth: 6.38 · 106 mDensity of snow: 200 kg

m3

Density of water: 1000 kgm3

Specific heat capacity of water: cH2O = 4186 Jkg·!C

Latent heat of fusion for water/ice: Lf = 33.5 · 104 Jkg

Stefan-Boltzmann constant: ! = 5.67 · 10!8 JK4·m2·s

Problem 1 Soon after the earth was formed, heat released by the decay of radioactiveelements raised the average internal temperature from 300 to 3000 K which roughlyis the value today.

(a) [2 points] Assume that the coe!cient of volume expansion is " = 3 · 10!5 K!1,how much did the radius of the earth increase when it heated. (Hint: 3 · # = ")

!R = ! · R · !T =13· " · R · !T

=13· 3 · 10!5 1

K· 6.38 · 106 m · 2700 K

!R = 172 km

(b)[2 points] The heat of the inner parts of the earthis conducted to the surface and can be measured. InNorth America in a surface area of A = 1 m2 the heatflow coming from underneath is Q/t = 54 · 10!3 J/s(see figure on the right). If the surface temperature is15"C and the thermal conductivity of the rocks in theupper part of the earth is k = 2.5 J

s·m·!C , what is thetemperature at a depth of d = 35 km?

Ad

earth�surfaceheat�flow�Q/tfrom�the�centerof�the�earth

Q = k · A

d· !T · t

=! !T =Q

t· d

k · A = 54 · 10!3 Js· 35000 m2.5 J

sm!C · 1 m2= 756"C

The absolut temperature at 35 km is then T = 756"C + 15"C " 770"C.

Continued on the other side....

(c)[2 points] If an area of 1 m2 at the earth surface emits Q/t = 54 · 10!3 J/s inradiation, what would the temperature at the earth surface be, if there would be noother energy sources? Assume that the earth is a perfect black body emitter (e = 1).

Q = e · # · T 4 · A · t, e = 1

T = 4

!Q

t· 1# · A = 4

"54 · 10!3

Js· 15.67 · 10!8 J

K4·m2·s · 1 m2" 31 K

(d)[2 points] The amount of heat that reaches the earth from the sun is much largerthan this. A surface area of 1 m2 receives 1400 J per second from the sun. If all theenergy would be absorbed by a 0.5 m layer of snow, how long would it take to meltthe snow?

mass of the snow: msnow = $snow · V = 200 kgm3 · 0.5 m3 = 100 kg

heat needed to melt the snow: Q = msnow · Lf = 100 kg · 33.5 · 104 Jkg = 33.5 · 106 J

time needed to melt the snow: Q1400J/s = 33.5·106 J

1400 J/s = 24000 s = 6.6 h

(e)[2 points] After the snow melted, the water of 0"C stays on the surface and willabsorb 10.1 · 105 J during the rest of the day. What is the final temperature of thewater?

Q = cH2O · mH2O · !T

!T =Q

cH2O · mH2O=

10.1 · 105J4186 J

kg·!C · 100 kg= 2.4"C

Rewrite and sign the honor pledge: “I pledge my honor that I have notviolated the Honor Code during this examination.”

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QUIZ 8 Physics 101, Fall 2005

SOLUTIONS

Problem 1: Ideal Monatomic Gas Engine

One mole of an ideal, monoatomic gas iscontained in a cylinder closed on the topby a movable piston, and is taken throughthe cycle shown on the figure (A → B →C → D → A). The temperature at pointA is TA = 293K. The volume at the pointA of the cycle is VA = 0.023m3.

A

B C

D

VA

3VA

PA

2PA

a. (1 pt) Determine the total internal energy UA at point A. Box your answer.

UA =32nRTA =

32(1 mol)(8.31 Jmol−1 K−1)(293K) = 3650 J

b. (2 pts) Determine the net work performed by the gas on the surroundings (W ) in one

complete cycle (A→ B → C → D → A). Box your answer.

PA =nRTA

VA=

(1 mol)(8.31 Jmol−1 K−1)(293K)0.023 m3

= 105 Pa

WABCDA = (2PA − PA)(3VA − VA) = 2PAVA = (2× 105 Pa)(0.023 m3) = 4600 J

c. (2 pts) Determine the net heat exchanged with the surroundings (Q) in one completecycle (A→ B → C → D → A). Hint: pay attention to the sign! Q is positive if heat enters

the system, negative if heat exits the system. Box your answer.

∆UABCDA = QABCDA −WABCDA = 0

QABCDA = WABCDA = 4600 J

QUIZ 8 Physics 101

d. (2 pts) Determine the temperatures at points B, C, and D (TB, TC , and TD respectively).

Box your answers.

PAVA

TA=

PBVB

TB→ TB = TA

PBVB

PAVA= TA

2PAVA

PAVA= 2TA = 2(293 K) = 586 K

In a similar way, for points C and D:

TC = TAPCVC

PAVA= TA

2PA3VA

PAVA= 6TA = 6(293 K) = 1758 K

TD = TAPDVD

PAVA= TA

PA3VA

PAVA= 3TA = 3(293 K) = 879 K

Problem 2: Liquid Helium Storage(3 pts) Liquid helium is stored at its boiling point temperature of 4.2K in a sphericalcontainer with radius = 0.3m. The container is a perfect blackbody radiator (i.e.: emis-sivity e = 1). The container is surrounded by a spherical shield cooled at liquid nitro-gen temperature (77K). The space between the container and the shield is under vacuum.The latent heat of vaporization of Helium is LV = 2.1 × 104 J/kg. What mass of heliumboils away through the venting valve in one hour? Hint: the Stefan-Boltzmann constant isσ = 5.67× 10−8 J s−1 m−2 K−4). Box your answer.

mLV = Q = eσ4πr2(T 4N − T 4

He)t

m =eσ4πr2(T 4

N − T 4He)t

LV=

(1)(5.67× 10−8 J s−1 m−2 K−4)(4π)(0.3 m)2[(77K)4 − (4.2 K)4](3600 s)2.1× 104 J kg−1

m = 0.39 kg