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 - 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 - 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

Nuclear & Particle Physics - Gamma Rays Detector

An 8-cm diameter by 8-cm long NAI(TI) detector detects gamma rays of a specific energy from a point source of radioactivity. When the source is placed just next to the detector at the center of the circular face, 50 percents of all emitted gamma rays at that energy are detected. If the detector is moved to 1 meter away, the fraction of detected gamma rays drops to

A. 10⁻⁴
B. 2 ×10⁻⁴
C. 4 ×10⁻⁴
D. 8 ×10⁻⁴
E. 16 ×10⁻⁴

(GR0177 #14)

Solution:

The net power radiated by the source: P = I A
I = intensity
A = area of the surface

No power loss mentioned: P₁ = P₂
I₁ A₁ = I₂ A₂

A₁ = area of circle with diameter 8 cm
r₁  = 4 × 10⁻² m

A₂ = area of sphere with radius 1 m

The problem asks for the fraction, not how much the intensity is detected.
Thus, the fraction is just the ratio of the areas:



Answer: C