1. A 0.300 kg piece of metal is heated to 88°C and placed in a copper calorimeter with mass 0.150 kg which contains 0.500 L of water that are both initially at 12.6°C. Over a short period of time the mixture comes to an equilibrium temperature of 22.5°C, what is the specific heat of the metal?

A) c. 1500 J/(kg ∙ °C)

B) d. 3400 J/(kg ∙ °C)

C) b. 1075 J/(kg ∙ °C)

D) a. 1250 J/(kg ∙ °C)

The heat given by the metal and calorimeter = The
heat taken by water

c(Metal)*m(metal)*(Ti1-Tf) +
c(copper)*m(calorimeter)*(Ti1-Tf) = m(water)*C(water)(Tf - Ti2)

C(copper) =385 J/kg/K

Cwater = 4811 J/kg/K

C(Metal)*0.3*(88-22.5) = 385*0.15*(22.5-12.6) +0.5*4811*(22.5-12.6)

19.65*c(metal)
=571.725 +23814.45

c(metal) =1241.02 J/kg/K

__Correct answer is D) 1250 J/kg/K__

2. A gas is
taken through the cycle illustrated here. During one cycle, how much work is
done by an engine operating on this cycle?

A) 4PV

B) 3PV

C) 2PV

D) PV

Infinitesimal work is dW= P*dV, dV is the infinitesimal change of
volume

At constant volume (vertical transformations in the figure) there is no work done.

At the same pressure (horizontal transformations in figure) the work is

W =2P*(4V-V) - P*(4V-V) = P*3V = 3PV

At constant volume (vertical transformations in the figure) there is no work done.

At the same pressure (horizontal transformations in figure) the work is

W =2P*(4V-V) - P*(4V-V) = P*3V = 3PV

__Correct answer is B) 3PV__

3. An
object has a weight of 400 N when it is dry. When it is completely submerged in
water is has a weight of 150 N. What is the density of the material is the
density of water is 1000 kg/m3?

A) c. 1200 kg/m3

B) b. 4500 kg/m3

C) d. 1600 kg/m3

D) a. 2300 kg/m3

G = m*g = ro*V*g for the dry object

G(submerged) = [ro-ro(water)]*V*g

ro/(ro-ro(water)) =G/G(submerged)

ro/(ro-1000) =400/150

ro = (ro-1000)*2.67

ro-2.67*ro = -2670

1.67*ro =2670

ro =1598.8 kg/m^3

__Correct answer is C) 1600 kg/m^3__

4. Using
the standard form of a wave x = A cos(wt), for this wave: 2.5 cos(6.0 t) please
determine the following with the correct units:

a. The amplitude.

b. The frequency.

c. The maximum and minimum speed

d. The maximum acceleration

a. A is the amplitude so it is just 2.5 m.

b. the value next to the t is w, so f = omega/2p = 6.0
/(2p) = 0.955 Hz

c. The maximum speed is omega*A so this is (6.0
rad/s)(2.5 m) = 15.0 m/s. The minimum speed is when it stops to “turn around”
so it is 0.

d. The maximum acceleration is -omega*2A = -(6.0
rad/sec)2(2.5 m) = -90 m/s2.

5. 4.50
moles of an ideal gas is at 560K. Is the gas undergoes an adiabatic compression
and 3750 J of work is done on the gas, what is the final temperature of the
gas?

A) b. 572 K

B) a. 627 K

C) c. 2590 K

D) d. 977 K

Adiabatic compression it means there is no heat
exchanged.

variation of internal energy = work done

delta(U) = Work

N*Cv*(Tf-Ti) = 3750 , N is the number f gas moles, Cv = (3/2)*R =12.5 J/mol/K

4.5*12.5*(Tf-560) =3750

Tf-560 =66.67

Tf =626.67 K

__Correct answer is B) 627 K__

6. A CD
with a diameter of 12.5 cm goes from rest to a tangential velocity of 5.6 m/s.
If it does this in 2.5 seconds, what is its angular acceleration?

A) d. 16 rad/s2

B) a. 13 rad/s2

C) c. 24 rad/s2

D) b. 18 rad/s2

Angular velocity omega = V/R = 5.6/0.125 = 44.8
rad/sec

angular acceleration epsilon = omega/t = 44.8/2.5
=17.92 rad/s/s

__Correct answer is D) 18 rad/s^2__

7. A piece
of gold with a mass of 5.50 kg and density of 19300 kg/m3 is suspended from a
string and then totally immersed in a beaker of water. Using density of water
is 1000 kg/m3

a) Determine the volume of the piece of gold.

b) Determine buoyant force on the gold when it is
submersed.

a) rho = mass/volume so now, volume = mass/density
= 5.50 kg/19300 kg/m3 = 2.85 x 10-4 m3 = 285 cm3.

b) BF = weight of water displaced, so this equals
the volume of gold times the density of water: 2.85 x 10-4 m3 x 1000 kg/m3 =
0.285 N

8. If a car horn honking that is 450 m is heard 1.3 seconds
after it goes off, what is the temperature in degrees Celsius?

A) a. 25 °C

B) c. 56 °C

C) b. 34 °C

D) d. 12 °C

speed of sound V = Distance/Time = 450/1.3 =346.15
m/s

dependence of speed of sound with temperature

V(T) = 331.3 + 0.606*T , T in
Celsius degree

346.15 =331.3 +0.606*T

T =24.5 Celsius degrees

__Correct answer is A) 25 Celsius degrees__