# Final exam (Thermodynamics 2014)

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

$c(Metal)*m(metal)*(Ti1-T_f) + c(copper)*m(calorimeter)*(Ti1-T_f) =$

$= m(water)*C(water)(T_f – Ti2)$

$C(copper) =385 J/(kg*K)$

$C_{water} = 4811 J/(kg*K)$

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

$19.65*c(metal) =571.725 +23814.45$

$c(metal) =1241.02 J/(kg*K)$

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

$d W= P*dV4$ where $d V$ 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

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

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

$G = m*g = rho*V*g$ for the dry object

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

$rho/(rho-rho(water)) =G/G(submerged)$

$rho/(rho-1000) =400/150$

$rho = (rho-1000)*2.67$

$rho-2.67*rho = -2670$

$1.67*rho =2670$

$rho =1598.8 kg/m^3$

Correct answer is C) 1600 kg/m^3

4. Using the standard form of a wave $x = A cos(omega t)$, 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 $omega$, 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
$Delta(U) = Work$

$N*C_v*(T_f-T_i) = 3750$

where $N$ is the number of gas moles, $C_v = (3/2)*R =12.5 J/(mol*K)$

$4.5*12.5*(T_f-560) =3750$

$T_f-560 =66.67$

$T_f =626.67 K$
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?

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

angular acceleration $epsilon = omega/t = 44.8/2.5 =17.92 rad/s^2$

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, $volume = mass/density = 5.50 kg/19300 kg/m^3 =$

$= 2.85*10^{-4} m^3 = 285 cm^3$.

b) BF = weight of water displaced, so this equals the volume of gold times the density of water:

$2.85*10{-4} m^3 x 1000 kg/m^3 = 0.285 N$

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