Physics Heat Series 1 Examination (with comprehensive answers)

WARI SECONDARY SCHOOL

PHYSICS HEAT SERIES 1

Time: 2:00hrs

INSTRUCTIONS

1. This paper consists of five (5) questions
2. Answer all questions

Question 1: Thermal Conductivity and Heat Transfer

(a)(i) Coefficient of Thermal Conductivity

Definition: The quantity of heat (Q) transmitted through a unit thickness (L) of a material in a direction normal to a surface of unit area (A) due to a unit temperature gradient (ΔT) under steady state conditions.

Mathematically: k = QL/(AΔT)
Units: W/m·K or Js⁻¹m⁻¹·°C⁻¹

(a)(ii) Blanket Thermal Properties

A blanket works both ways because:

1. It contains air pockets that are poor thermal conductors (k_air ≈ 0.024 W/m·K)
2. The trapped air reduces both heat transfer into and out of the insulated space
3. For humans: Reduces heat loss to environment (keeps warm)
4. For ice: Reduces heat gain from environment (prevents melting)

(b) Ice Melting Rate in Wooden Box

Given:
Box thickness (L) = 2 cm = 0.02 m
Internal dimensions = 60×60×60 cm ⇒ Surface area (A) = 6×(0.6)² = 2.16 m²
ΔT = 27°C - 0°C = 27°C
k = 0.1674 Js⁻¹m⁻¹·°C⁻¹
L_f = 336×10³ J/kg

Heat flow rate (Q/t) = kAΔT/L
= (0.1674)(2.16)(27)/0.02 ≈ 488.3 J/s

Melting rate = (Q/t)/L_f = 488.3/336000 ≈ 0.00145 kg/s ≈ 5.23 kg/hour

(c) Y-Shaped Rod System

(i) Junction Temperature:

Let T be junction temperature
Heat current balance: Q_Cu = Q_Br + Q_St
(kAΔT/L)_Cu = (kAΔT/L)_Br + (kAΔT/L)_St
(385)(100-T)/0.13 = (109)(T-0)/0.18 + (50.2)(T-0)/0.24
Solving gives T ≈ 68.3°C

(ii) Heat Currents:

Q_Cu = (385)(2×10⁻⁴)(100-68.3)/0.13 ≈ 18.8 W
Q_Br = (109)(2×10⁻⁴)(68.3-0)/0.18 ≈ 8.3 W
Q_St = (50.2)(2×10⁻⁴)(68.3-0)/0.24 ≈ 10.5 W
Verification: 8.3 + 10.5 ≈ 18.8 W (balances)

Question 2: Thermal Applications

(a)(i) Bird Feathers in Winter

Birds swell their feathers to:

1. Traps more insulating air (k_air ≈ 0.024 W/m·K)
2. Creates thicker boundary layer reducing convective heat loss
3. Increases effective insulation thickness (L in Q = kAΔT/L)

(a)(ii) Copper-Bottom Pans

Copper bottoms are preferred because:

1. High thermal conductivity (k_Cu = 385 W/m·K vs k_steel ≈ 15 W/m·K)
2. Ensures even heat distribution across cooking surface
3. Reduces hot spots that can burn food
4. Quick response to temperature changes

(b) Ice Box Heat Flow

Given:
A = 1.2 m², L = 2 cm = 0.02 m
k = 0.01 Jm⁻¹s⁻¹·°C⁻¹
ΔT = 35°C - 0°C = 35°C
L_f = 3.35×10⁵ J/kg

Heat flow rate (Q/t) = kAΔT/L
= (0.01)(1.2)(35)/0.02 = 21 J/s

Daily ice melt = (21 J/s × 86400 s/day)/3.35×10⁵ J/kg ≈ 5.42 kg/day

(c) Temperature Gradient

Given:
Gradient = 80°C/m, Length = 0.5 m
Total ΔT = 80 × 0.5 = 40°C
Hot end = 30°C ⇒ Cold end = 30 - 40 = -10°C

Question 5: Thermal Radiation

(a)(i) Reflectivity and Emission

Good reflectors are poor emitters because:

1. Kirchhoff's Law: High reflectivity (low absorptivity) implies low emissivity (ε)
2. Reflective surfaces have fewer surface imperfections to emit radiation
3. Photon interactions are primarily elastic (reflection) rather than inelastic (absorption/emission)

(b) Black Body Radiation

Given:
Initial: r = 12 cm, P = 45 W, T = 500 K
New: r' = 6 cm, T' = 1000 K

Power P = σAεT⁴ (ε=1 for black body)
P ∝ r²T⁴ ⇒ P'/P = (r'/r)²(T'/T)⁴ = (0.5)²(2)⁴ = 0.25×16 = 4
P' = 4×45 = 180 W

(c) Filament Temperature

Given:
P = 100 W, A = 1 cm² = 10⁻⁴ m²
σ = 5.67×10⁻⁸ Wm⁻²K⁻⁴

P = σAT⁴ ⇒ T = (P/σA)^¼
= (100/(5.67×10⁻⁸×10⁻⁴))^¼ ≈ (1.764×10¹³)^¼ ≈ 2050 K