Thermal Physics - Power

A 100-watt electric heater element is placed in a pan containing one liter of water. Although the heating element is on for a long time, the water, though close to boiling does not boil. When the heating element is removed, approximately how long will it take the water to cool by 1 degree Celsius? (Assume that the specific heat for water is 4.2 kJ/kg^oC)

A. 20 s
B. 40 s
C. 60 s
D. 130 s
E. 200 s

(GR9677 #13)

Solution:

P = 100 W
V = 1 L = 1 m³ → m = 1 kg (STP)
ΔT = 1^oC
c = 4.2 kJ/kg^oC

Q = mcΔT = Pt
1 × 4200 × 1 = 100t
t = 42 s

Answer: B

Thermal Physics - Specific Heat

Two identical 1 kg blocks of copper metal one initially at a temperature T₁ = 0^oC and the other initially at a temperature T₂ = 100^oC are enclosed in a perfectly insulating container. The two blocks are initially separated. When the blocks are placed in contact, they come to equilibrium at a final temperature T_f. The amount of heat exchanged between the two blocks in this process is equal to which of the following? (The specific heat of copper metal is equal to 0.1 kilocalorie/kg^oK)

A. 50 Kcal
B. 25 Kcal
C. 10 Kcal
D. 5 Kcal
E. 1 Kcal

(GR9677 #14)

Solution:

Q = mcΔT
|Q|_gain = |Q|_lost 
m₁c₁ΔT₁  = m₂c₂ΔT₂

m₁ = m₂ = 1 kg
c₁ = c₂ = c_copper = 0.1 kcal/kg^oK

T₁ = 0^oC = 273^oK

T₂ = 100^oC = 373^oK

ΔT₁ = ΔT₂
T_f   − T₁ = T₂ − T_f 
T_f  = (T₂ + T₁)/2 = (100 + 0)/2 = 50^oC = 323^oK

|Q|_gain = |Q|_lost  = m₁c₁ΔT₁  = 1 × 0.1 × (323 − 273) = 5kcal

Answer: D

Thermal Physics - Isothermal

Suppose one mole of an ideal gas undergoes the reversible cycle ABCA shown in the P-V diagram above, where AB is an isotherm. The molar heat capacities are C_p at constant pressure and C_v at constant volume. The net heat added to the gas during the cycle is equal to

A. RT_h (V₂/V₁)
B. −C_p(T_h − T_c)
C. C_p(T_h − T_c)
D. RT_h ln (V₂/V₁) − C_p(T_h − T_c)
E. RT_h ln (V₂/V₁) − R(T_h − T_c)

(GR9677 #15)

Solution:

AB Isotherm → T Constant = T_h

Ideal Gas: PV = nRT 
P = nRT / V

n = 1 mole,

W_AB = _V1∫^V₂ P dV = RT_h V1∫^V₂ (1/V) dV  = RT_h ln (V₂/V₁)

BC Isobaric → P constant = P₂

W_BC = _V2∫^V₁ P dV = P₂ (V₁− V₂) =  P₂V₁ −  P₂V₂

From the diagram:

P₂V₁ = nRT_c 

P₂V₂ = nRT_h

W_BC = nR(T_c − T_h)  = R(T_c − T_h)

W_CA = 0  since V constant

Total W = W_AB + W_BC = RT_h ln (V₂/V₁) + R(T_c − T_h)

or

W = RT_h ln (V₂/V₁) − R(T_h − T_c)

Answer: E

Thermal Physics - Mean Free Path

The mean free path for the molecules of a gas is approximately given by 1/ησ, where η is the number density and σ is the collision cross section. The mean free path for air molecules at room conditions is approximately

A. 10⁻⁴ m
B. 10⁻⁷ m
C. 10⁻¹⁰ m
D. 10⁻¹³ m
E. 10⁻¹⁶ m

(GR9677 #16)

Solution:

Mean free path = 1/ησ

Number density, η = N/V  
Cross section area, σ = πr² 

For Ideal Gas: PV = NkT

1/ησ = V / Nπr²  = NkT / PNπr² = kT / Pπr²   

k = 1.38 × 10⁻²³ Joule/K
Radius of atom is in order of Angstrom: 10⁻¹⁰ m
P (STP) = 1 atm = 10⁵ Newton/meter²
T (STP) = 0 °C = 32 °F = 273.15 K ≈ 2 × 10² K

kT / Pπr² = (1.38 × 10⁻²³ × 2 ×10²) / (π ×10⁵ × 10⁻²⁰) 
= (1.38 × 2 / π) 10^{−23+2−5+20} ≈ 10⁻⁷

Answer: B

Thermal Physics - Adiabatic

The adiabatic expansion of an ideal gas is described by the equation PV ^γ = C, where γ and C are constants. The work done by the gas is expanding adiabatically from the state (V_i, P_i) to (V_f, P_f) is equal to

A. P_fV_f
B. ½(P_i + P_f)(V_f  − V_i)
C. (P_fV_f   − P_fV_i)/(1 − γ)
D. P_i(V_f ^1+ γ − V_i ^1+ γ )/(1 + γ)
E. P_f (V_f ^1- γ − V_i ^1- γ)/(1 + γ)

(GR9677 #73)

Solution:

PV ^γ = C
P = CV ^−γ

W = ∫ P dV
= C _Vi∫^V_f  V ^−γ dV
= [C/(−γ + 1)] V ^−γ+1 _Vi|^V_f
= [C/(1 − γ)] (V_f ^1−γ− V_i ^1−γ)

= (V_fCV_f ^−γ− V_i CV_i ^1−γ)/(1 − γ)

= (V_fP_f  − V_i P_i )/(1 − γ)

= (P_fV_f  − P_iV_i )/(1 − γ)

Answer: C

Thermal Physics - Entropy

A body of mass m with spesific heat C at temperature 500 K is brought into contact with an identical body at temperature 100 K, and the two are isolated from their surroundings. The change in entropy of the system is equal to

A. ⁴⁄₃ mC
B. mC ln  ⁹⁄₅
C. mC ln 3
D. –mC ln ⁵⁄₃
E. 0

(GR9677 #74)

Solution:

Entropy:




m₁ = m₂ = m
c₁ = c₂ = C
T₁ = 500
T₂ = 100
T_f = (500 + 100)/2 = 300



Answer: B

Thermal Physics - Thermal Conductivities

Window A is a pane of glass 4 millimeters thick, as shown above. Window B is a sandwich consisting of two extremely thin layers of glass separates by an air gap 2 millimeters thick, as shown above. If the thermal conductivities of glass and air are 0.8 watt/meter Window A is a pane of glass 4 millimeters thick, as shown above. Window B is a sandwich consisting of two extremely thin layers of glass separated by an air gap 2 millimeters thick, as shown above. If the thermal conductivities of glass and air are 0.8 watt/meter ^oC and 0.025 watt/meter ^oC, respectively, then the ratio of the heat flow through window A to the heat flow through window B is.

A. 2
B. 4
C. 8
D. 16
E. 32

(GR9677 #75)

Solution:

The thicker the glass, the harder for heat to flow → q ∼ 1/t

Thermal conductivity, κ: property of material that allows the flow of heat through the material. 

Higher κ → higher the flow of heat → q ∼ κ

q ∼ κ/t
q_A / q_B = (κ_A / d_A )( d_B / κ_B) =  (0.8 / 4)(2 /0.025)  =  16

Answer: D

Note:

Fourier's Law of Heat Conduction: q = − κ ∇T  = − κ dT/dx

q = − κ dT/dx (in 1-D),
T = temperature,  
dx  = length or thickness, t

q ∼ κ/t

Thermal Physics - Heat Capacity

For an ideal diatomic gas in thermal equilibrium, the ratio of the molar heat capacity at constant volume at very high temperatures to that at very low temperatures is equal to

A. 1
B. 5/3
C. 2
D. 7/3
E. 3

(GR9677 #79)

Solution:

C_V  for diatomic gas: C_V  = C_Vtrans + C_Vrot  + C_Vvib

For very low T, only translational component contributes
→ C_V(T≪) = ³/₂ Nk

For very high T, all 3 components (translational, vibrational, rotational) contribute
→ C_V(T≫) = (³/₂ + 1 + 1) R = ⁷/₂ Nk

Ratio C_V(T≪) / C_V(T≫)  = 7/3

Answer: D

Thermal Physics - Blackbody Radiation

The total energy of a blackbody radiation source is collected for one minute and used to heat water. The temperature of the water increases from 20.0^oC to 20.5^oC. If the absolute temperature of the blackbody is doubled and the experiment repeated, which of the following statements would be most nearly correct?

A. The temperature of the water would increase from 20^oC to 21^oC 
B. The temperature of the water would increase from 20^oC to 24^oC
C. The temperature of the water would increase from 20^oC to 28^oC
D. The temperature of the water would increase from 20^oC to 36^oC
E. The water would boil within the one-minute time period.

(GR9277 #14)

Solution:

Blackbody Radiation (Stefan-Boltzmann Law): u =  σT⁴  → u  ∝ T⁴
u₁ ∝ T⁴ and u₂ ∝ (2T)⁴

Heat Transfer: Q = mcΔT  → Q ∝ ΔT
Q₁ ∝ ΔT₁ = 20.5 − 20 = 0.5

u₁/u₂ = Q₁/Q₂ 
T⁴/16T⁴  = 0.5/ΔT₂ 
ΔT₂ = 16 × 0.5 = 8

Answer: C

Thermal Physics - Heat Capacity

A Classical model of a diatomic molecule is a springy dumbbell, as shown above, where the dumbbell is free to rotate about axes perpendicular to the spring. In the limit of high temperature, what is the specific heat per mole at constant volume?

A. 3/2 R
B. 5/2 R
C. 7/2 R
D. 9/2 R
E. 11/2 R

(GR9277 #15) 

Solution:

C_V  for diatomic gas: C_V  = C_Vtrans  + C_Vrot  + C_Vvib

At high T, all components contribute and each degree of freedom (DoF) contributes ½R to C_V .

DoF trans = 3 (since it's 3 dimension)
DoF rot = 2 (diatomic atom)
DoF vib = 2 (1 for kinetic energy + 1 potential energy)
→ 3 + 2 + 2 = 7 → 7 DoF contributes ⁷/₂R

Answer: C

Thermal Physics - Carnot Cycle

An engine absorbs heat at a temperature of 727^oC and exhaust heat at a temperature of 527^oC. If the engine operates at maximum possible efficiency, for 2000 joules of heat input the mount of work the engine performs is most nearly 

A. 400 J
B. 1450 J
C. 1600 J
D. 2000 J
E. 2760 J

(GR9277 #16)

Solution:

T_H = 727^oC = 727 + 273 = 1000 K
T_C = 527^oC = 527 + 273 = 800 K

Q_H = 2000 J
W = ?

Efficiency of heat engine cycle: η = W/Q_H

Carnot efficiency: η = (T_H − T_C) /T_H = 1 − T_C/T_H

W/Q_H = 1 − T_C/T_H
W = Q_H (1 − T_C/T_H) =  2000  (1 − 800/1000) = 400 J

Answer: A 

Thermal Physics - Conduction Electron

The Fermi temperature of Cu is about 80,000 K. Which of the following is most nearly equal to the average speed of a conduction electron in Cu? 

A. 2 × 10⁻² m/s
B. 2 m/s
C. 2 × 10² m/s
D. 2 × 10⁴ m/s
E. 2 × 10⁶ m/s

(GR9277 #23)

Solution:

T_F = 8 × 10⁴ K
m_e = 9.11 × 10⁻³¹ kg
k = 1.38 × 10⁻²³ J/K

Fermi energy = Kinetic energy



Answer: E

Thermal Physic - Critical Isotherm

Questions 46-47.

Isotherms and coexistence curves are shown in the PV diagram above for a liquid-gas system. The dashed lines are the boundaries of the labeled regions. Which numbered curve is the critical isotherm? 

A. 1
B. 2
C. 3
D. 4
E. 5

(GR9277 #46)

Solution:

Critical isotherm (critical temperature) → dP/dV = 0

• the derivative of the curve is zero
• the tangent to the curve results in a horizontal line
• the point where the vertical and horizontal dashed lines cross

Answer: B

Thermal Physics - Thermal Equilibrium

See Problem #46

In which region are the liquid and the vapor in equilibrium with each other?

A. A
B. B
C. C
D. D
E. E

(GR9277 #47)

Solution:

Region A: V small  → Liquid
Region E: P large, V small → Liquid
Region D and C: V large →  Gas
Phase equilibrium exists only if the temperature of isotherm lies below critical temperature.

Answer: B

Thermal Physics - Isothermal

A mole of ideal gas initially at temperature T₀ and volume V₀ undergoes a reversible isothermal expansion to volume V₁. If the ratio of specific heats is c_p/c_v = γ and if R is the gas constant, the work done by the gas is

A. Zero
B. RT₀ (V₁/V₀ )^γ
C. RT₀ (V₁/V₀ − 1)
D. c_v T₀ [1 − (V₁/V₀ )^(γ−1)]
E. RT₀ ln (V₁/V₀ )

(GR9277 #62)

Solution:

W = ∫ P dV

Ideal Gas: PV = nRT  → P = nRT / V

W = _V0∫^V₁ nRT (1/V) dV

1 mole → n = 1
Isothermal, constant T = T₀

→ W = RT_0 V0∫^V₁ (1/V) dV = RT₀ ln (V₁/V₀ )

Answer: E  

Thermal Physics - Probability

Which of the following is true if the arrangement of an isolated thermodynamic system is of maximal probability?

A. Spontaneous change to a lower probability occurs
B. The entropy is minimum
C. Botzmann’s constant approaches zero
D. No spontaneous change occurs
E. The entropy is zero

(GR9277 #63)

Solution:

Maximal Probability → highest/maximum entropy → most stable state /equilibrium → No spontaneous change occurs. Spontaneous change occurs when the system is far from equilibrium.

Answer: D

Thermal Physics - Fermi-Dirac Statistics

Questions 71-73

A system in thermal equilibrium at temperature T consists of a large number N₀ of subsystems, each of which can exist only in two states of energy E₁ and E₂, where . In the expressions that follow, k is the Boltzmann constant.

For a system at temperature T, the average number of subsystems in the state of energy E₁ is given by

A. 

B. 

C.

D.

E.

(GR9277 #71)

Solution:

For only 2 states of energy → Fermi-Dirac distribution

Number of states: 

Since , take E₂ = 2ϵ and E₁ = ϵ

For E₁ = ϵ, number of states:

Answer: B

Thermal Physics - Heat Capacity

Questions 71-73

A system in thermal equilibrium at temperature T consists of a large number N₀ of subsystems, each of which can exist only in two states of energy E₁ and E₂, where . In the expressions that follow, k is the Boltzmann constant.

The internal energy of this system at any temperature T is given by . The heat capacity of the system is given by which of the following expressions?

A.

B.

C.

D.

E.

(GR9277 #72)

Solution:

Heat Capacity (at constant  volume), 



Answer: A

Thermal Physics - Entropy

Questions 71-73

A system in thermal equilibrium at temperature T consists of a large number N₀ of subsystems, each of which can exist only in two states of energy E₁ and E₂, where . In the expressions that follow, k is the Boltzmann constant.

Which of the following is true of entropy of the systems?

A. It increases without limit with T from zero at T = 0
B. It decreases with increasing T
C. It increases from zero at T = 0 to N₀k ln 2 at arbitrarily high temperature
D. It is given by N₀k [⁵/₂ ln T − ln p + constant]
E. It cannot be calculated from the information given

(GR9277 #73)

Solution:

At low T, entropy S(T → 0) = 0
At high T, S = kσ

σ = ln (number of states with the same total energy)

For 2  states of energy, σ = ln 2
S = k ln 2

For N₀ subsystems.
S = N₀k ln 2

Answer: C

Thermal Physics - Specific Heat

For an ideal gas, the specific heat at constant pressure C_p is greater than the specific heat at constant volume C_v because the

1. Gas does work on its environment when its pressure remains constant while its temperature is increased.
2. Heat input per degree increase in temperature is the same in processes for which either the pressure or the volume is kept constant.
3. Pressure of the gas remains constant when its temperature remains constant.
4. Increase in the gas’s internal energy is greater when the pressure remains constant than when the volume remains constant
5. Heat needed is greater when the volume remains constant than when the pressure remains constant.

(GR8677 #14)

Solution:

(A) TRUE.
Heat Capacity: C = Q/dT

First law of Thermodynamics: the change in internal energy of a system dU is equal to the heat Q added and the work, W done on or by the system 
→ dU = Q ± W

W done on the system → +W
W done by the system → −W

Gas (the system) does work on its environment
W done by the system
dU = Q − W

At constant V:
Work, W = PdV = 0
Q = dU
C_v = dU/dT

At constant P:
Work, W = PdV ≠ 0
Q = dU + W
C_p = dU/dT + PdV/dT = C_v + PdV/dT
C_p  C_v

(B) FALSE.
This means C_p = C_v, but according to A, C_p  C_v

(C) FALSE.
Ideal gas law: PV = NkT
If T constant, P changes if V changes.

(D) FALSE.
Heat Capacity, C = Q/dT does not depend on the gas’ internal energy, U

(E) FALSE.
See A. At constant V, Q = dU.
At constant P, Q = dU + W.

Answer: A