This Problem Was the Same on Both Exams

PHY 122Ch 16 – 18 ExamName

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 ΔT?

3/3 8/83/3
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 / 6/6 6/6 4/4
Q = m c ΔT
1500 = 20000(.9) ΔT
ΔT = 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
P = ρ g h
P = 1100 (10)(10,900)
P = 1.2 x 108 Pa / 10/10 6/6
Bulk Modulus = F/A / ΔV/ V
14×1010 = 1.2 x 108 / ΔV/ (4/3)π(.1)3
ΔV/ (4/3)π(.1)3 = 1.2 x 108 / 14×1010
ΔV = 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
ΔV = ΔVsteel – ΔVHg
ΔV = (27*3*10*ΔT – 27*3*18*ΔT) x 10-6
Note: β = 3α, and we only need inner volume
ΔV = (27*3*10*ΔT – 27*3*18*ΔT) x 10-6
.65 x 10-9 = (648 *ΔT) x 10-6
ΔT = 1 C° / Q = msteel csteel ΔT + mHg cHg ΔT
Q = (64-27)8.4 (.1) (1) + 27(13.6) (.033) (1)
Q = 43.2 cal = 181 Joules
KEbullet = ½ m v2
KEbullet = ½ .002 4002 5/5
KEbullet = 160 Joules, Two bullets 1/1

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  2 = (3Vo – Vo) 3Po = 6 VoPo
4/4 1/1
6  5 = (Vo – 3Vo) Po = -2 VoPo
W = 6VoPo + 0 + -2VoPo + 0
W = 4 VoPo / (c) 1→4→6represents 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)
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?
P Vγ = Pf Vγf /
Pf = P ( V / Vf )γ
3/37/7
Pf = (19+1) (50/400)1.4
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
(c) ½ nRT 1/1
(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 = Eint = nCvT, solve for the work generated in the expansion.
Work = n Cv T
Work = n½5RT 7/7
Work = 5/2 n R T
Work = 5/2 PV 10/10
Work = 5/2 ( Pf Vf – P V )
Work = 5/2 ( 1.10x105 40x10-5 – 20.3x105 5x10-5)
Work = -143 Joules