Thermal Properties of Matter

Formulas:
Celsius: $T_C = T_K – 273.15$
Fahrenheit: $T_F = \frac{9}{5}T_C + 32$

(a) Neon ($T_K = 24.57 \text{ K}$):
$T_C = 24.57 – 273.15 = \mathbf{-248.58^\circ \text{C}}$.
$T_F = \frac{9}{5}(-248.58) + 32 = -447.44 + 32 = \mathbf{-415.44^\circ \text{F}}$.

(b) Carbon Dioxide ($T_K = 216.55 \text{ K}$):
$T_C = 216.55 – 273.15 = \mathbf{-56.60^\circ \text{C}}$.
$T_F = \frac{9}{5}(-56.60) + 32 = -101.88 + 32 = \mathbf{-69.88^\circ \text{F}}$.

The triple point of water is a fixed physical state ($T_K = 273.16 \text{ K}$).
Given:
Scale A: $200 \text{ A}$ corresponds to $273.16 \text{ K}$.
Scale B: $350 \text{ B}$ corresponds to $273.16 \text{ K}$.

Since both are absolute scales, $T \propto T_K$.
Value of 1 K on Scale A = $200 / 273.16$.
Value of 1 K on Scale B = $350 / 273.16$.
For any temperature $T$:
$\frac{T_A}{200} = \frac{T_B}{350} \Rightarrow \frac{T_A}{4} = \frac{T_B}{7}$.
Relation: $T_A = \frac{4}{7} T_B$.

Formula: $R = R_0 [1 + \alpha(T – T_0)]$.
$T_0 = 273.16 \text{ K}$ (Triple point), $R_0 = 101.6 \Omega$.
$T_1 = 600.5 \text{ K}$ (Lead melt), $R_1 = 165.5 \Omega$.

Step 1: Find $\alpha$:
$165.5 = 101.6 [1 + \alpha(600.5 – 273.16)]$
$\frac{165.5}{101.6} – 1 = \alpha(327.34)$
$0.6289 = 327.34 \alpha \Rightarrow \alpha \approx 1.921 \times 10^{-3} \text{ K}^{-1}$.

Step 2: Find T for $R = 123.4 \Omega$:
$123.4 = 101.6 [1 + \alpha(T – 273.16)]$
$\frac{123.4 – 101.6}{101.6} = \alpha(T – 273.16)$
$0.2145 = (1.921 \times 10^{-3})(T – 273.16)$
$T – 273.16 = \frac{0.2145}{0.001921} \approx 111.66 \text{ K}$.
$T = 273.16 + 111.66 \approx \mathbf{384.8 \text{ K}}$.

(a) Why Triple Point?
The triple point of water is a unique point ($P, T$) where ice, water, and vapor coexist. Melting and boiling points depend on pressure, which varies. The triple point is pressure-independent (defined at unique P).

(b) Other Fixed Point on Kelvin Scale:
Absolute Zero ($0 \text{ K}$).

(c) Why 273.15 and not 273.16?
$0^\circ \text{C}$ is the ice point at 1 atm. The triple point ($0.01^\circ \text{C}$) is slightly higher. Since $T_{ice} \approx 273.15 \text{ K}$ and $t_c$ starts from there, the relation is $t_c = T – 273.15$.

(d) Triple point on “Fahrenheit-sized” absolute scale:
Interval size of F = $180$ divisions (32 to 212). C/K has 100 divisions.
Ratio of unit size = $180/100 = 1.8$.
Value of triple point = $273.16 \times 1.8 \approx \mathbf{491.69}$.

For constant volume gas thermometer, $T \propto P \Rightarrow T = T_{tr} \times \frac{P}{P_{tr}}$.

(a) Absolute Temp of Sulphur Point:
Thermometer A: $T_A = 273.16 \times \frac{1.797 \times 10^5}{1.250 \times 10^5} \approx \mathbf{392.69 \text{ K}}$.
Thermometer B: $T_B = 273.16 \times \frac{0.287 \times 10^5}{0.200 \times 10^5} \approx \mathbf{391.98 \text{ K}}$.

(b) Reason for Difference:
Real gases (Oxygen, Hydrogen) are not perfectly ideal. They deviate from Boyle’s law at finite pressures. To reduce discrepancy, measurements should be taken at lower pressures and extrapolated to zero pressure ($P \to 0$) where all gases behave ideally.

Problem: The tape is calibrated at $27^\circ \text{C}$. On a hot day ($45^\circ \text{C}$), the tape expands. The “1 cm” mark on the tape is physically further than 1 cm away from zero. If it reads 63.0 cm, the actual length is longer.

Calculation:
$\Delta T = 45 – 27 = 18^\circ \text{C}$.
Reading $l_{measured} = 63.0 \text{ cm}$.
Actual Length = $l_{measured} \times (1 + \alpha_{steel} \Delta T)$.
$L = 63.0 [1 + (1.20 \times 10^{-5})(18)]$
$L = 63.0 [1 + 0.000216] = 63.0136 \text{ cm}$.

Length of steel rod on $27^\circ \text{C}$ day:
Since the rod is also made of steel, it contracts by the same factor as the tape when cooled back to 27. So, the reading would be the same? No. The tape is correct at 27. So if we measure it at 27, we get the true length at 27.
Length at 45 = 63.0136 cm. Length at 27 = $63.0136 / (1 + \alpha \Delta T) = \mathbf{63.0 \text{ cm}}$.

Shaft diameter at $27^\circ \text{C} (T_1) = 8.70 \text{ cm}$.
Hole diameter at $27^\circ \text{C} = 8.69 \text{ cm}$.
The shaft must be cooled so its diameter shrinks to $8.69 \text{ cm}$.

$L_2 = L_1 (1 + \alpha \Delta T)$
$8.69 = 8.70 (1 + 1.20 \times 10^{-5} \Delta T)$
$\frac{8.69}{8.70} – 1 = 1.2 \times 10^{-5} \Delta T$
$-0.001149 = 1.2 \times 10^{-5} \Delta T$
$\Delta T = \frac{-0.001149}{1.2 \times 10^{-5}} \approx -95.8^\circ \text{C}$.
$T_{final} = 27 – 95.8 = \mathbf{-68.8^\circ \text{C}}$.

Concept: A hole in a sheet expands exactly as if it were filled with the same material (photographic expansion).

$D_1 = 4.24 \text{ cm}$. $T_1 = 27^\circ \text{C}$. $T_2 = 227^\circ \text{C}$. $\Delta T = 200 \text{ K}$.
$\alpha_{cu} = 1.70 \times 10^{-5} \text{ K}^{-1}$.

$\Delta D = 4.24 \times 1.70 \times 10^{-5} \times 200$
$\Delta D = 4.24 \times 3.4 \times 10^{-3} \approx \mathbf{0.0144 \text{ cm}}$.
New Diameter = $4.2544 \text{ cm}$.

Wire held taut at $27^\circ \text{C}$. Cooled to $-39^\circ \text{C}$. $\Delta T = -66 \text{ K}$.
Since ends are fixed, the wire cannot contract. This creates tensile stress.

Thermal Strain $\varepsilon = \frac{\Delta L}{L} = \alpha \Delta T$.
Young’s Modulus $Y = \frac{\text{Stress}}{\text{Strain}} \Rightarrow \text{Stress} = Y \alpha \Delta T$.
Tension $F = \text{Stress} \times A = Y \alpha \Delta T (\pi r^2)$.

Values: $Y = 0.91 \times 10^{11}$, $\alpha = 2.0 \times 10^{-5}$, $\Delta T = 66$, $r = 1 \text{ mm} = 10^{-3} \text{ m}$.
$F = (0.91 \times 10^{11}) (2.0 \times 10^{-5}) (66) (3.14 \times 10^{-6})$
$F = 0.91 \times 2 \times 66 \times 3.14 \times 10^0 \approx \mathbf{377.2 \text{ N}}$.

Brass: $L_1 = 50 \text{ cm}$. Steel: $L_2 = 50 \text{ cm}$.
$\Delta T = 250 – 40 = 210^\circ \text{C}$.
$\alpha_B = 2.0 \times 10^{-5}$, $\alpha_S = 1.2 \times 10^{-5}$.

Total Change in Length:
$\Delta L = \Delta L_{brass} + \Delta L_{steel}$
$\Delta L = L_1 \alpha_B \Delta T + L_2 \alpha_S \Delta T = L \Delta T (\alpha_B + \alpha_S)$ (Since $L_1=L_2=L=50$)
$\Delta L = 50 \times 210 \times (2.0 + 1.2) \times 10^{-5}$
$\Delta L = 10500 \times 3.2 \times 10^{-5} = \mathbf{0.336 \text{ cm}}$.

Thermal Stress?
Since the ends are free to expand, no internal stress is developed at the junction.

$\gamma = 49 \times 10^{-5} \text{ K}^{-1}$. $\Delta T = 30^\circ \text{C}$.
Volume expansion: $V’ = V(1 + \gamma \Delta T)$.
Density $\rho = M/V$. $\rho’ = M/V’ = \frac{M}{V(1 + \gamma \Delta T)} = \rho (1 + \gamma \Delta T)^{-1}$.
Using Binomial approx: $\rho’ \approx \rho (1 – \gamma \Delta T)$.
Fractional change $\frac{\Delta \rho}{\rho} = \frac{\rho’ – \rho}{\rho} = -\gamma \Delta T$.

Magnitude = $49 \times 10^{-5} \times 30 = \mathbf{1.47 \times 10^{-2}}$.

Power $P = 10 \text{ kW} = 10^4 \text{ J/s}$.
Time $t = 2.5 \text{ min} = 150 \text{ s}$.
Total Energy $E = P \times t = 1.5 \times 10^6 \text{ J}$.
Energy heating the block ($50\%$) $Q = 0.5 E = 7.5 \times 10^5 \text{ J}$.
Mass $m = 8.0 \text{ kg}$. Sp. heat $c = 0.91 \text{ J/g K} = 910 \text{ J/kg K}$.

$\Delta T = \frac{7.5 \times 10^5}{8.0 \times 910} = \frac{750000}{7280} \approx \mathbf{103^\circ \text{C}}$.

Copper: $m_{cu} = 2.5 \text{ kg}$, $T = 500^\circ \text{C}$. $c_{cu} = 390 \text{ J/kg K}$.
Final temp will be $0^\circ \text{C}$ (ice melts).
Heat released by copper: $Q = m c \Delta T = 2.5 \times 390 \times (500 – 0) = 487,500 \text{ J}$.

Mass of ice melted ($m_{ice}$): $Q = m_{ice} L_f$.
$L_f = 335 \text{ J/g} = 335 \times 10^3 \text{ J/kg}$.
$m_{ice} = \frac{487500}{335000} \approx \mathbf{1.45 \text{ kg}}$.

Metal: $m_1 = 0.20 \text{ kg}$, $T_1 = 150^\circ \text{C}$. $c_1 = ?$.
Calorimeter + Water: Equiv Mass $m_2 = 0.025 + 0.150 = 0.175 \text{ kg}$ (taking density of water 1g/cc).
$T_2 = 27^\circ \text{C}$. Final $T_f = 40^\circ \text{C}$. $c_w = 4186 \text{ J/kg K}$.

Heat Lost by Metal = Heat Gained by Water+Calorimeter.
$m_1 c_1 (150 – 40) = m_2 c_w (40 – 27)$
$0.20 \times c_1 \times 110 = 0.175 \times 4186 \times 13$
$22 c_1 = 9523.15 \Rightarrow c_1 \approx \mathbf{433 \text{ J/kg K}}$.

If heat losses occurred, the actual heat gained by water would be less than what metal lost? No, heat lost to surroundings means the water didn’t get all the energy. So we underestimated the denominator or overestimated the temperature change? Actually, if heat is lost, $T_f$ is lower than ideal. The temperature drop of metal $(150-T_f)$ is larger, and rise of water is smaller. Calculating $c$ without accounting for loss usually gives a smaller value than actual.

Explanation:
Monatomic gases have only translational degrees of freedom ($f=3$), giving $C_v = \frac{3}{2}R \approx 2.98 \text{ cal/mol K}$.
Diatomic gases (like $H_2, N_2, O_2$) have rotational degrees of freedom too ($f=5$), giving $C_v = \frac{5}{2}R \approx 4.96 \text{ cal/mol K}$. This matches the data for most gases listed.

Chlorine Anomaly (6.17):
Chlorine has a value significantly higher than 5. This implies that at room temperature, vibrational modes are also active in Chlorine, adding more degrees of freedom, increasing the heat capacity.

Mass $m = 30 \text{ kg}$. $c_{body} \approx c_{water} = 4186 \text{ J/kg K}$.
Temp drop: $101^\circ \text{F} \to 98^\circ \text{F}$. $\Delta T_F = 3^\circ \text{F}$.
$\Delta T_C = 3 \times \frac{5}{9} = \frac{5}{3}^\circ \text{C}$.

Heat to be lost: $Q = mc\Delta T = 30 \times 4186 \times \frac{5}{3} = 209,300 \text{ J}$.
Time $t = 20 \text{ min}$.

Evaporation Latent Heat $L_v = 580 \text{ cal/g} = 580 \times 4.186 \times 10^3 \text{ J/kg} = 2.428 \times 10^6 \text{ J/kg}$.
Mass of sweat $m_{evap} = \frac{Q}{L_v} = \frac{209300}{2.428 \times 10^6} \approx 0.0862 \text{ kg} = 86.2 \text{ g}$.

Rate = $86.2 \text{ g} / 20 \text{ min} \approx \mathbf{4.3 \text{ g/min}}$.

Area $A = 6 \times (0.3)^2 = 0.54 \text{ m}^2$. Thickness $d = 0.05 \text{ m}$.
$\Delta T = 45 – 0 = 45^\circ \text{C}$. $t = 6 \text{ h} = 21600 \text{ s}$.
$K = 0.01 \text{ W/m K}$.

Heat Flow $Q = \frac{KA \Delta T t}{d} = \frac{0.01 \times 0.54 \times 45 \times 21600}{0.05} = 104,976 \text{ J}$.

Mass melted $m = Q/L = 104976 / (335 \times 10^3) \approx 0.313 \text{ kg}$.
Remaining Ice = $4.0 – 0.313 = \mathbf{3.69 \text{ kg}}$.

Rate of boiling = $6.0 \text{ kg/min} = 0.1 \text{ kg/s}$.
Power required $P = \text{Rate} \times L_v = 0.1 \times 2256 \times 10^3 = 2.256 \times 10^5 \text{ W}$.

Conduction Formula: $P = \frac{KA \Delta T}{d}$.
$2.256 \times 10^5 = \frac{109 \times 0.15 \times (T – 100)}{0.01}$
$2256 = 16.35 (T – 100)$
$T – 100 = 137.9 \Rightarrow T \approx \mathbf{238^\circ \text{C}}$.

(a) Reflectivity vs Emissivity: Kirchhoff’s law states good absorbers are good emitters. A body with large reflectivity is a poor absorber, hence a poor emitter.

(b) Brass vs Wood: Brass is a good conductor; it conducts heat away from your hand rapidly, making it feel cold. Wood is an insulator.

(c) Optical Pyrometer Reading: In the open, the red hot iron reflects some sky light and emits less radiation than a black body (emissivity < 1). The pyrometer assumes black body ($e=1$) so it reads low. In a furnace, the iron and walls reach thermal equilibrium, approximating a black body cavity, giving a correct reading.

(d) Atmosphere: The atmosphere acts like a blanket (Greenhouse effect), reflecting infrared radiation back to Earth, keeping it warm.

(e) Steam vs Hot Water: Steam at $100^\circ \text{C}$ releases huge latent heat of vaporization ($2256 \text{ kJ/kg}$) upon condensing, whereas water only releases specific heat.

Formula: $\frac{T_1 – T_2}{t} = K \left[ \frac{T_1 + T_2}{2} – T_s \right]$. $T_s = 20$.

Step 1 (80 to 50):
$\frac{80-50}{5} = K [65 – 20] \Rightarrow 6 = 45K \Rightarrow K = \frac{6}{45} = \frac{2}{15}$.

Step 2 (60 to 30):
$\frac{60-30}{t} = \frac{2}{15} [45 – 20]$
$\frac{30}{t} = \frac{2}{15} (25) = \frac{10}{3}$
$t = \frac{30 \times 3}{10} = \mathbf{9 \text{ min}}$.