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

Nuclear & Particle Physics - Uranium Decay

A uranium nucleus decays at rest into a thorium nucleus and a helium nucleus, as shown above. Which of the following is true?

A. Each decay product has the same kinetic energy.
B. Each decay product has the same speed.
C. The decay products tend to go in the same direction.
D. The thorium nucleus has more momentum than the helium nucleus.
E. The helium nucleus has more kinetic energy than the thorium nucleus.

(GR9277 #75)

Solution:

Conservation of momentum:

Uranium nucleus decays at rest:

C. FALSE


Thus, Helium has more momentum and kinetic energy than Thorium.
A, B, D are FALSE

Answer: E

Classical Mechanics - Kepler's Law

The period of a hypothetical Earth satellite orbiting at sea level would be 80 minutes. In terms of the Earth’s radius R_e, the radius of a synchronous satellite orbit (period 24 hours) is most nearly

A. 3 R_e
B. 7 R_e
C. 18 R_e
D. 320 R_e
E. 5800 R_e

(GR8677 #75)

Solution:

Kepler’s 3rd Law:

R³ / T²  = constant    

Given:
T_earth = 80 minutes
T_satellite = 24 × 60 minutes

R_e³ / T_e²  = R_s³ / T_s²  

R_s³ = ( T_s² / T_e² ) R_e³ 

R_s³ = (24 × 60 / 80)² R_e³ = (18)² R_e³ 

R_s = (18)^2/3 R_e = (324)^1/3 R_e ≈ 7 R_e

Answer: B

Classical Mechanics - Rotational Motion

The matrix shown above transforms the components of a vector in one coordinate frame S to the components of the same vector in a second coordinates frame S'. This matrix represents a rotation of the reference frame S by

A. 30^o clockwise about the x-axis
B. 30^o counterclockwise about the z-axis
C. 45^o clockwise about the z-axis
D. 60^o clockwise about the y-axis
E. 60^o counterclockwise about the z-axis

(GR0177 #75)

Solution:

Rotation about z-axis has the form:



Since    and    E is the right answer.

Answer: E