Version 001 – Thermo Quest – tubman – (IBII201516)
This print-out should have 11 questions.
Multiple-choice questions may continue on
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before answering.
Coffee Equilibrium
001 10.0 points
What is the final equilibrium temperature
when 20 g of milk at 8◦ C is added to 107 g of
coffee at 79◦ C? Assume the specific heats of
milk and coffee are the same as that of water,
and neglect the specific heat of the container.
Correct answer: 67.8189◦C.
1
1. The specific heat of A is greater than that
of B.
2. The specific heat of B is greater than that
of A. correct
3. The specific heats of A and B are equal.
Explanation:
Heat is m c ∆T.
The heat that escaped from A is equal to
the heat gained by B. The temperature change
is the same, so a greater mass for A means a
smaller specific heat for A.
Explanation:
Let :
mm
Tm
mc
Tc
= 20 g ,
= 8◦ C ,
= 107 g ,
= 79◦ C .
and
The amount of heat given out by coffee
should be equal to the amount of heat absorbed by milk, so
Water to Boiling Point
003 10.0 points
A mass of 4 kg of water is raised from 28◦ C to
the boiling point 100◦ C.
How much heat is required? The specific
heat of water is 1000 cal/kg ·◦ C.
Correct answer: 2.88 × 105 cal.
Explanation:
Let :
cm mm (Tf − Tm ) = cc mc (Tc − Tf )
mm Tf − mm Tm = mc Tc − mc Tf
(mm + mc ) Tf = mc Tc + mm Tm
mc Tc + mm Tm
mm + mc
(107 g) (79◦ C) + (20 g) (8◦ C)
=
20 g + 107 g
Tf =
= 67.8189◦ C .
Conceptual 11 Q25
002 10.0 points
Two hundred grams of liquid A is at a temperature of 100◦ C. One hundred grams of liquid
B is at a temperature of 0◦ C. When the two
liquids are mixed, the final temperature is
50◦ C.
Which material has a higher specific heat?
m = 4 kg ,
c = 1000 cal/kg ·◦ C ,
Tf = 100◦ C , and
Ti = 28◦ C .
The heat required to raise the temperature
of an object by ∆T is
Q = m c ∆T = m c (Tf − Ti )
= (4 kg) (1000 cal/kg ·◦ C) (72◦ C)
= 2.88 × 105 cal .
Bullet Into Wall
004 10.0 points
A cowboy fires a silver bullet of mass 6 g with
a muzzle speed of 292 m/s into the pine wall
of a saloon.
What is the temperature change of the bullet? Assume that all the internal energy generated by the impact remains with the bullet.
The specific heat of silver is 234 J/kg ·◦ C.
Version 001 – Thermo Quest – tubman – (IBII201516)
2
The heat absorbed by the water is the same
as the heat lost by the brass, so
Correct answer: 182.188 C.
◦
Explanation:
Let :
m = 6 g,
v = 292 m/s , and
c = 234 J/kg ·◦ C .
The kinetic energy of the bullet is
1
1
m v 2 = (0.006 kg) (292 m/s)2
2
2
= 255.792 J .
Ek =
Nothing in the environment is hotter than
the bullet, so the bullet gains no thermal
energy. Its temperature increases because the
255.792 J of kinetic energy became 255.792 J
of extra internal energy. Thus
Ek = Q = m c ∆T
255.792 J
Q
=
∆T =
mc
(0.006 kg) (234 J/kg ·◦ C)
∆Qw = ∆Qb
mw cw (T − Tw ) = mb cb (Tb − T )
(mw cw + mb cb ) T = mb cb Tb + mw cw Tw
mb cb Tb + mw cw Tw
T =
.
mw cw + mb cb
Since
mw cw + mb cb = (0.265 kg) (4180 J/kg ·◦ C)
+ (0.489 kg) (376 J/kg ·◦ C)
= 1291.56 J/◦ C ,
the final temperature will be
mb cb Tb + mw cw Tw
mw cw + mb cb
(0.489 kg) (376 J/kg ·◦ C) (114◦ C)
=
1291.56 J/◦ C
(0.265 kg) (4180 J/kg ·◦ C) (32◦ C)
+
1291.56 J/◦ C
T =
= 43.6733◦ C .
= 182.188 C .
◦
Drop Brass into Water
005 10.0 points
A 489 g sample of brass at 114◦ C is placed in
a calorimeter cup that contains 265 g of water
at 32◦ C.
If you disregard the absorption of heat
by the cup, what will be the final temperature? Assume the specific heat of brass is
376 J/kg ·◦ C, and the specific heat of water
is 4180 J/kg ·◦ C.
Correct answer: 43.6733◦C.
mb
cb
Tb
mw
cw
Tw
= 0.489 kg ,
= 376 J/kg ·◦ C ,
= 114◦ C ,
= 0.265 kg ,
= 4180 J/kg ·◦ C ,
= 32◦ C .
1. The pressure decreases three times.
2. The pressure decreases to two thirds of its
original value.
3. The pressure increases three times. correct
Explanation:
Let :
Concept 14 51
006 10.0 points
What change in pressure occurs in a party
balloon that is squeezed to one third of its
original volume with no change in temperature?
4. The pressure increases one and a half
times.
5. There is no change in pressure.
and
Explanation:
According to Boyle’s law, P V is a constant,
so the pressure should increase three times.
Version 001 – Thermo Quest – tubman – (IBII201516)
3
gas.
Ideal Gas Path
007 10.0 points
Identify the parameter paths for an ideal gas
that are isovolumetric / isobaric / isothermal.
P
4. The collisions between molecules are inelastic. correct
5. The only appreciable forces on the
molecules are those that occur during collisions.
T1 > T2 > T3
C
A
3. The molecules are in random motion.
B
T1
T2
T3
V
1. C / A / B
2. A / C / B correct
3. B / C / A
4. B / A / C
5. A / B / C
6. C / B / A
Explanation:
A is isovolumetric since volume is constant.
B is isothermal since temperature is constant.
C is isobaric since pressure is constant.
Explanation:
The classical model of ideal gas makes the
following assumptions:
1. The molecules are in random motion.
2. The molecules obey Newton’s laws of
motion.
3. The volume of the molecules is negligible
compared with the volume of occupied by the
gas.
4. The only appreciable forces on the
molecules are those that occur during collisions.
5. The collisions between molecules are
elastic.
Thus, the statement that “The collisions
between molecules are inelastic” is NOT correct.
Air Leaving a Room
009 10.0 points
A room of volume 90 m3 contains air having
an average molar mass of 15.5 g/mol.
If the temperature of the room is raised
from 17.5 ◦ C to 33 ◦ C, what mass of air will
leave the room? Assume that the air pressure
in the room is maintained at 146.3 kPa.
Correct answer: 4.28002 kg.
AP B 1998 MC 61
008 10.0 points
Which of the following statements is NOT a
correct assumption of the classical model of
an ideal gas?
1. The molecules obey Newton’s laws of motion.
2. The volume of the molecules is negligible
compared with the volume of occupied by the
Explanation:
Let :
V = 90 m3 ,
M = 15.5 g/mol ,
R = 8.31451 J/K mol ,
T1 = 17.5◦ C = 290.5 K ,
T2 = 33◦ C = 306 K , and
P = 146.3 kPa = 1.463 × 105 Pa .
Version 001 – Thermo Quest – tubman – (IBII201516)
Pf Vf
Pi Vi
=
Ti
Tf
Tf Pi
(435 K)(101 kPa)
Pf =
=
Ti
293 K
P V = nRT
PV
= n , so
RT
PV
∆n = n2 − n1 =
R
1
1
−
T2 T1
= 149.949 kPa .
and the change of mass is
1
1
PV
∆m = ∆n M =
−
M
R
T2 T1
(1.463 × 105 Pa) 90 m3
=
8.31451 J/K mol
1
1
×
−
306 K 290.5 K
1 kg
× (15.5 g/mol)
1000 g
= −4.28002 kg ,
Double the RMS Speed
011 10.0 points
A gas is at 20 ◦ C.
To what temperature must it be heated to
double the rms speed of its molecule?
Correct answer: 899 ◦ C.
Explanation:
Let : Ti = 20 ◦ C = 293 K .
vf,rms = 2 vi,rms
At a given temperature, the kinetic energy is
a mass loss of 4.28002 kg .
Heating a Sealed Bottle
010 10.0 points
A sealed glass bottle containing air at atmospheric pressure (101 kPa) and having a
volume of 50 cm3 is at 20◦ C. It is then tossed
into an open fire.
When the temperature of the air in the bottle reaches 162◦ C, what is the pressure inside
the bottle? Assume any volume changes of
the bottle are negligible.
Correct answer: 149.949 kPa.
Explanation:
Let :
Pi
Vi
Ti
Tf
4
= 101 kPa ,
= 50 cm3 ,
= 20◦ C = 293 K , and
= 162◦ C = 435 K .
If no gas escapes from the tank, the number
of moles remains constant; so from the ideal
gas law P V = n R T ,
K=
1 2
3
vrms = kB T
2
2
Thus
Tf
=
Ti
1 2
2 vf,rms
1 2
2 vi,rms
=
(2 vi,rms )2
= 4.
vi,rms
Tf = 4 Ti
= 4 (293 K)
= 1172 K = 899 ◦ C