PHY 122 Ch 16 – 18 Exam Name Exam1 #1
A 20.0 kg crate of fruit (c = 0.9 cal / (gr⋅K)) slides 10.0m down a ramp inclined at an angle of 30.0° below
the horizontal. The crate was at rest at the top of the incline and has a speed of 8.00 m/s at the
bottom. If energy dissipated by friction is absorbed by the crate of fruit and the fruit reaches a
uniform final temperature, what is its temperature change ∆T?
3/3
8/8
3/3
6/6 6/6
TETop
= TEBottom
m g htop + ½ mvtop2 – Edissipated = mghbot + ½ mvbot2
20(10)sin30(10) +
0 – Edissipated = 0 + ½ 20 82
Edissipated = 1000 – 640 = 360 Joules = 1500 cal
4/4
Q
= m
c ∆T
1500 = 20000(.9) ∆T
∆T = 0.083 C°
Exam1 #2
The deepest place in all the oceans is the Marianas Trench, where the depth is 10.9 km. Let density of
salt water be 1.1 g/cc. If a copper (BCu = 14×1010 N/m2) ball 20.0 cm in diameter is taken to the bottom
of the trench, by how much does its volume decrease?
14/14
10/10
P= ρ
g
h
P = 1100 (10)(10,900)
P = 1.2 x 108 Pa
6/6
Bulk Modulus = F/A / ∆V /
V
10
8
14×10
= 1.2 x 10 / ∆V/ (4/3)π(.1)3
∆V/ (4/3)π(.1)3 = 1.2 x 108 / 14×1010
∆V = 3.59 cc
Exam2 #1
An open topped steel (ρsteel = 8.4 gr/cc) cube whose volume outer volume is 64 cc and inner volume is is
27 cc is completely filled with mercury (ρHg = 13.6 gr/cc). When the cube is impacted with multiple lead
bullets (vbullet = 400 m/s, mbullet = 2 grams), a volume of .65 mm3 of mercury overflows the cube. How many
bullets impacted the steel cube. Assume thermally insulated. αsteel = 10.0 × 10−6/K, αHg = 18.0 × 10−6/K,
csteel = 0.100 cal / gr K, cHg = 0.033 cal / gr K
4/4
4/4
2/2
5/5
4/4
5/5
Q=
msteel csteel ∆T +
mHg cHg ∆T
∆V =
∆Vsteel –
∆VHg
Q = (64-27)8.4 (.1) (1) + 27(13.6) (.033) (1)
∆V = (27*3*10*∆T – 27*3*18*∆T) x 10-6
Q = 43.2 cal = 181 Joules
α
Note: β = 3 , and we only need inner volume
∆V = (27*3*10*∆T – 27*3*18*∆T) x 10-6
.65 x 10-9 = (648 *∆T) x 10-6
∆T = 1 C°
KEbullet = ½ m
v2
KEbullet = ½ .002 4002
5/5
KEbullet = 160 Joules, Two bullets
Exam2 #2
(a) Calculate the work W done by the gas during process 2→6
(b) Calculate the work W done by the gas during process 1→2→6→5→1
(c) Calculate the work W done by the gas during process 1→4→6
(a) Zero Joules
5/5
(b)
3/3
2/2
1 Æ 2 = (3Vo – Vo) 3Po = 6 VoPo
4/4
1/1
6 Æ 5 = (Vo – 3Vo) Po = -2 VoPo
(c) 1→ 4→ 6
represents ¼ of a circle. Take
total Area and subtract ¼ of area of a circle.
3/3
2/2 5/5
5/5
W = 4 VoPo - ¼ Acircle
W = 4 VoPo - ¼ π ( r )
* (r)
W = 4 VoPo - ¼ π(3Po-Po) * (3Vo-Vo)
1/1
W = 6VoPo + 0 + -2VoPo + 0
W = 4 VoPo
W = (4 – π) VoPo
This problem was the same on both exams
During the power stroke in a four-stroke automobile engine, the piston is
forced down as the mixture of combustion products and air undergoes an
adiabatic expansion. Assume that (1) the gauge pressure right before the
expansion is 19.0 atm, (2) the volumes of the mixture right before and
after the expansion are 50.0 and 400 cm3, respectively (3) the mixture
behaves like an ideal gas with specific heat ratio, γ = 1.40. Find the work
generated during the expansion.
(a) What is the final pressure?
γ
γ
P V
Pf =
= Pf V
P
3/3
f
γ
( V / Vf )
7/7
1.4
Pf = (19+1) (50/400)
Pf = 1.09 atm = 1.10 x 105 Pa
1/1
(b) What is the transferred heat, Q? zero
(c) How many degrees of freedom in
monoatomic particules (moving in 3D space)?
(a) 3
1/1
(b) 5
(c) 7
(d) What is the translational energy for a
single degree of freedom of N particles
(moving in 3D space)?
(a) a N ½mvx2
(b) N ½ kBT
1/1
(c) ½ nRT
(d) All of the above
(e) Given, Etrans = n CV T, solve for CV in terms of R
for the fuel mixture
Etrans
= n CV T
df(½ nRT) = n CV T
5/5
5/5
5(½ nRT) = n CV T
CV = ½5R
γ=
1.40 = CP / CV = df+2 ½R / df ½R = 7 (½R) / 5 ½R
(f) Given, Work = ΔEint = nCvΔT, solve for the work
generated in the expansion.
Work = n Cv ΔT
Work = n ½5R ΔT
7/7
Work = 5/2 n R ΔT
Work = 5/2 ΔPV
10/10
Work = 5/2 (
Pf Vf
–
P
V)
5
-5
5
Work = 5/2 ( 1.10x10 40x10 – 20.3x10 5x10-5)
Work = -143 Joules