Examples #1: Introduction to thermodynamics
1. A physicist and an engineer nd themselves in a mountain lodge where the only heat is provided by
a large wood stove. The physicist argues that they cannot increase the total energy of the molecules
in the cabin, and therefore it makes no sense to continue putting logs into the stove. The engineer
strongly disagrees, referring to the laws of thermodynamics and common sense. Who is right? Why
do we heat the room?
• The physicist is right that one cannot change the total energy of all molecules in the cabin: burning
logs will only convert the chemical energy of log molecules (which are disassembled in the process)
into heat - the kinetic energy that gets distributed to many more molecules of the stove and air.
However, this physicist obviously spends too much time reading books, so he lost his touch with
reality. The engineer understands that the heat obtained from burning logs (guarantied by the
conservation of energy, 1st law of thermodynamics) will increase the temperature in the cabin,
which is what they actually care about.
• Ocial: The cabin is not hermetically closed, so air pressure in the cabin is stays the same and
equal to the outside pressure after burning logs. The volume of the cabin obviously does not
change, so pV = N kB T implies that N T is kept xed when logs are burned. Consequently, the
internal energy of the air in the cabin E ∝ N kB T stays the same (as stated by the physicist).
However, temperature rises at expense of reducing the number of molecules N inside the cabin.
2. An immersion heater of power J = 500 W is used to heat water in a bowl. After 2 minutes, the
temperature increases from T1 = 85o C to T2 = 90o C. The heater is then switched o for an additional
minute, and the temperature drops by ∆T = 1o C. Estimate the mass m of the water in the bowl.
The thermal capacity of water is c = 4.2 × 103 J/kg K.
• In the rst t1 = 2 minutes, the heater dumps heat Q1 = Jt1 to the water. Water is incompressible,
so its volume will not change. The heat is partially converted to the water's internal energy, and
partially lost to the environment: Q1 = CV (T2 − T1 ) + ∆Q1 . After the heater is turned o, water
loses energy by transferring heat ∆Q2 = CV (T3 − T2 ) within t2 = 1 minute to the environment.
We may assume that the rate of heat loss J 0 to the environment (bowl and air) is constant, so
that:
∆Q1 = J 0 t1
We must eliminate J 0 :
,
∆Q2 = J 0 t2
∆Q2
∆Q1
=
t1
t2
Jt1 − CV (T2 − T1 )
CV (T3 − T2 )
=
t1
t2
−1
T2 − T1
T3 − T2
CV = J
+
t1
t2
The relationship between heat capacity Cv and thermal capacity c can be guessed from the units
of c:
CV = mc
where m is the total mass of the water. Hence,
m=
Cv
J
=
c
c
T2 − T1
T3 − T2
+
t1
t2
1
−1
≈ 2 kg
3. A long cylindrical tank is placed on a carriage that can slide without friction on horizontal rails. The
mass of the empty tanker is M = 180 kg. Initially, the tank is lled with an ideal gas of total mass
m = 120 kg at a pressure p0 = 150 atm at an ambient temperature T0 = 300 K. Then, one end of the
tank is heated to T0 + ∆T = 335 K, while the other end is kept at T0 . Find the pressure in the tank
and the new position of the center of mass of the tanker when the system reaches equilibrium.
• From the equation of state p0 = n̄kB T0 and the initial condition, we nd the mean concentration
of the gas molecules:
n̄ =
p0
N
=
V
kB T0
After one end of the tank is heated and equilibrium is established, temperature in the tank varies
linearly with the distance x from its cooler end:
T (x) = T0 +
x
∆T
L
where L is the tank length. The equilibrium pressure must, however, be constant in the tank or
else gas would be moving around. The equation of state, then, predicts that the gas concentration
must change with x:
⇒
p = n(x)kB T (x)
n(x) =
−1
p
p x
=
T0 + ∆T
kB T (x)
kB
L
The total number of particles of the gas and the gas volume did not change in the heating process
(let A be the cross-section area of the tank):
ˆL
N = ALn̄ = A
dx n(x)
=
0
Ap
kB
ˆL
dx
1
T0 +
0
=
Ap
1
×
kB
T0
ˆL
dx
0
=
=
x
L ∆T
1
1+
x ∆T
L T0
∆T
ˆ /T0
Ap
1
1
LT0
dξ
×
×
kB
T0
∆T
1+ξ
0
∆T
Ap L
× log 1 +
kB ∆T
T0
After canceling the tank volume AL on both sides, we nd:
∆T
k ∆T
n̄ =
B
T0
p=
∆T
log 1 + T0
log 1 +
∆T T0
∆T
T0
p0 −−−−−→
∆T
1+
2T0
p0 = 159 atm
The gas' center-of-mass was initially at the center of the tank, but eventually shifts to the position:
1
x0 =
Ln̄
ˆL
dx x n(x) = · · ·
0
4. Consider a cylinder L = 1 m long with a thin massless piston clamped in such a way that it divides
the cylinder into two equal parts. The cylinder is in a large heat bath at T = 300 K. The left side
of the cylinder contains nl = 1 mole of helium gas at pl = 4 atm. The right contains helium gas at
2
pressure pr = 1 atm. Let the piston be released.
(a) What is the nal equilibrium position?
(b) How much heat will be transmitted to the bath in the process of equilibration? (note R =
8.3 J/mole K).
• (a) In equilibrium, the pressure p and temperature T (xed by the bath) will be the same on both
sides of the piston:
p=
nr RT
nl RT
=
ALl
ALr
where nl , nr are molar amounts of the gas and Ll , Lr are the lengths of the cylinder sections to
the left and right of the piston respectively (A is the cylinder cross-section, and Ll + Lr = L).
We are given nl and we don't know nr . To determine it, consider the initial state:
pl =
pr =
nl RT
AL/2
nr RT
AL/2
⇒
⇒
A=
nr =
2nl RT
pl L
pr AL
pr
= nl
2RT
pl
Therefore, we nd the new position of the piston:
nr
nl
=
Ll
Lr
,
Ll + Lr = L
Ll
Lr
L − Ll
=
=
nl
nr
nr
nr
pl
4
pr
1
nl
,
Lr =
L=
L= L
L=
L= L
Ll =
nl + nr
pl + pr
5
nl + nr
pl + pr
5
• (b) The temperature of the gas does not change (it's xed by the bath), but some work is obviously
done by pushing the piston. This energy is ultimately exchanged with the bath in the form of
heat, in order to maintain the same temperature. So, the amount of heat Q transfered to the
bath equals the work done to move the piston (to the right, the gas on the left is initially at a
higher pressure):
ˆ
Q=W
=
=
=
=
=
ˆ nl RT
nr RT
(pl dVl + pr dVr ) =
dVl +
dVr
Vl
Vr
ˆ nl
nr
RT
dLl +
dLr
Ll
Lr
Ll
Lr
RT nl log
+ nr log
L/2
L/2
2pl
pr
2pr
nl RT log
+
log
pl + pr
pl
pl + pr
600 J
3