Physics Heat Series 2 Examination (Step-by-step working for each problem)

WARI SECONDARY SCHOOL

PHYSICS HEAT SERIES 2

Question 1: Thermometry

(a)(i) Thermometric Property

Definition: A physical property that changes measurably and reproducibly with temperature and can be used to measure temperature.

Examples: Electrical resistance, volume of liquid, gas pressure, thermoelectric emf.

(a)(ii) Platinum Resistance Thermometer

Platinum is used because:

1. High melting point (1768°C) allows wide temperature range
2. Resistance varies linearly with temperature over wide range
3. Chemically stable and doesn't oxidize easily
4. Reproducible resistance-temperature relationship

(b)(i) Platinum Thermometer Characteristics

1°C Change Meaning: Corresponds to a specific change in resistance (ΔR = R₀α where α is temperature coefficient)

Advantages:

1. High accuracy (±0.001°C possible)
2. Wide range (-200°C to +850°C)

Limitation for Varying Temperatures: Thermal inertia of platinum wire causes slow response to rapid temperature changes

(b)(ii) Temperature Scale Conversion

Given: R_θ = R₀(1 + 800αθ - αθ²)
At θ = 400°C:
R₄₀₀ = R₀(1 + 800α×400 - α×400²)
= R₀(1 + 320000α - 160000α) = R₀(1 + 160000α)

For platinum scale (linear): R_Pt = R₀(1 + α_Ptθ_Pt)
Equating at 400°C: 1 + 160000α = 1 + α_Ptθ_Pt
Assuming α values are consistent: θ_Pt ≈ 400°C (requires more data for exact calculation)

(c)(i) Scale Differences

The scales differ because:

1. Platinum resistance varies quadratically with temperature while gas thermometers are more linear
2. Different thermometric properties have different temperature dependencies
3. Only at fixed points (ice/steam points) are the scales defined to agree

(c)(ii) Platinum Temperature Calculation

Given: R_θ = R₀(1 + 3.8×10⁻³θ - 5.6×10⁻⁷θ²)
At θ = 200°C:
R₂₀₀/R₀ = 1 + 3.8×10⁻³×200 - 5.6×10⁻⁷×40000
= 1 + 0.76 - 0.0224 = 1.7376

Platinum scale is linear: R_Pt = R₀(1 + αθ_Pt)
Assuming α = 3.8×10⁻³/°C:
1.7376 = 1 + 3.8×10⁻³θ_Pt ⇒ θ_Pt ≈ 194°C

Question 2: Cooling and Calorimetry

(a) Newton's Law of Cooling

Statement: The rate of heat loss of a body is directly proportional to the difference in temperatures between the body and its surroundings, provided the difference is small.

dQ/dt = -k(T - T_s)
Where:
• k = cooling constant
• T = body temperature
• T_s = surrounding temperature

(b) Cooling Problem

(i) Surrounding Temperature:

From 90°C→80°C in 10 min: (90-80)/10 = k[(90+80)/2 - T_s]
From 80°C→70°C in 12 min: (80-70)/12 = k[(80+70)/2 - T_s]
Solving simultaneous equations gives T_s ≈ 25°C

(ii) Future Temperature:

Using cooling rate equation with T_s = 25°C:
From 70°C in 5 min: (70-T)/5 = k[(70+T)/2 - 25]
Solving gives T ≈ 65.5°C

(c)(i) Convective Heat Loss

Rate of convective heat loss: Q/t = hA(T_surface - T_fluid)
Where:
• h = convective heat transfer coefficient
• A = surface area
• T = temperatures

(c)(ii) Specific Heat Capacity

For glycerin:
Mass = 100cm³ × 1.2g/cm³ = 120g = 0.12kg
Heat lost = (30 + 0.12c_g)×10 = 300 + 1.2c_g J in 210s

For water:
Heat lost = (30 + 0.1×4200)×10 = 4230 J in 990s

Assuming same cooling conditions:
(300 + 1.2c_g)/210 = 4230/990 ⇒ c_g ≈ 2400 J/kg·K

Question 4: Thermometric Properties

(a) Thermometer Qualities

1. Linearity: Property should vary linearly with temperature
2. Sensitivity: Significant change in property over temperature range
3. Reproducibility: Consistent measurements under same conditions

(b)(i) Unknown Temperature Calculation

Resistance Thermometer:

θ = (R_θ - R₀)/(R₁₀₀ - R₀) × 100
= (64.992 - 63)/(75 - 63) × 100 ≈ 16.6°C

Gas Thermometer:

θ = (P_θ - P₀)/(P₁₀₀ - P₀) × 100
= (8.51 - 8.00)/(11.0 - 8.00) × 100 ≈ 17°C

(b)(ii) Scale Differences

The slight difference (16.6°C vs 17°C) occurs because different thermometric properties have slightly different temperature dependencies between fixed points.

(c) Thermocouple Calculations

Given E = 1 + aθ - bθ²
At 100°C: 4.5 = 1 + 100a - 10000b
At 200°C: 10.5 = 1 + 200a - 40000b
Solving gives a ≈ 0.06, b ≈ 0.0002

Neutral temp (dE/dθ=0): θ_n = a/2b ≈ 150°C
Inversion temp (E=0): θ_i ≈ 300°C