Thermal Properties of Matter

Given:
- Triple point of Neon: $T_{Ne} = 24.57\,\text{K}$
- Triple point of Carbon dioxide: $T_{CO_2} = 216.55\,\text{K}$

Formula used:
$$t_C = T - 273.15$$
$$t_F = \frac{9}{5}\,t_C + 32$$

For Neon:

Celsius scale:
$$t_C = 24.57 - 273.15 = -248.58^\circ\text{C}$$

Fahrenheit scale:
$$t_F = \frac{9}{5}\times(-248.58) + 32 = -447.44 + 32 = -415.44^\circ\text{F}$$

For Carbon dioxide:

Celsius scale:
$$t_C = 216.55 - 273.15 = -56.60^\circ\text{C}$$

Fahrenheit scale:
$$t_F = \frac{9}{5}\times(-56.60) + 32 = -101.88 + 32 = -69.88^\circ\text{F}$$

Results:
- Neon: $-248.58^\circ\text{C}$, $-415.44^\circ\text{F}$
- Carbon dioxide: $-56.60^\circ\text{C}$, $-69.88^\circ\text{F}$

Given:
- Triple point of water on scale A = 200 A
- Triple point of water on scale B = 350 B

Concept: The triple point of water is a unique physical state. Both scales measure the same physical temperatures; only the size of their unit intervals differs.

The size of one degree on scale A:
$$\text{1 unit on A} = \frac{273.16\,\text{K}}{200}$$

The size of one degree on scale B:
$$\text{1 unit on B} = \frac{273.16\,\text{K}}{350}$$

For any temperature $T$ (in Kelvin):
$$T_A = \frac{200}{273.16}\times T \quad \text{and} \quad T_B = \frac{350}{273.16}\times T$$

Dividing:
$$\frac{T_A}{T_B} = \frac{200}{350} = \frac{4}{7}$$

$$\boxed{T_A = \frac{4}{7}\,T_B}$$

This is the required relation between the two temperature scales.

Given:
- $R_o = 101.6\,\Omega$ at $T_o = 273.16\,\text{K}$
- $R_1 = 165.5\,\Omega$ at $T_1 = 600.5\,\text{K}$
- Find $T$ when $R = 123.4\,\Omega$

Step 1: Find $\alpha$

Using $R_1 = R_o[1 + \alpha(T_1 - T_o)]$:
$$165.5 = 101.6\,[1 + \alpha(600.5 - 273.16)]$$
$$165.5 = 101.6\,[1 + \alpha \times 327.34]$$
$$\frac{165.5}{101.6} = 1 + 327.34\,\alpha$$
$$1.6289 = 1 + 327.34\,\alpha$$
$$\alpha = \frac{0.6289}{327.34} = 1.921 \times 10^{-3}\,\text{K}^{-1}$$

Step 2: Find $T$ for $R = 123.4\,\Omega$

$$123.4 = 101.6\,[1 + 1.921\times10^{-3}(T - 273.16)]$$
$$\frac{123.4}{101.6} = 1 + 1.921\times10^{-3}(T - 273.16)$$
$$1.2146 = 1 + 1.921\times10^{-3}(T - 273.16)$$
$$0.2146 = 1.921\times10^{-3}(T - 273.16)$$
$$T - 273.16 = \frac{0.2146}{1.921\times10^{-3}} = 111.7\,\text{K}$$
$$T = 273.16 + 111.7 \approx 384.9\,\text{K}$$

The temperature is approximately $384.9\,\text{K}$.

Part (a):

The triple point of water is a unique state where all three phases (solid, liquid, vapour) coexist in equilibrium. It occurs at a unique, reproducible temperature and pressure (273.16 K, 611.2 Pa). It does not depend on external conditions.

The melting point of ice and boiling point of water both depend on atmospheric pressure, which varies from place to place and time to time. Hence they are not reliable fixed points for a universal temperature scale.

Part (b):

The Kelvin scale has only one fixed point — the triple point of water (273.16 K). The other reference is absolute zero (0 K), which is the point of minimum possible molecular activity. Unlike the Celsius scale, the Kelvin scale does not need a second empirically defined fixed point.

Part (c):

The triple point of water is assigned exactly $273.16\,\text{K}$ on the Kelvin scale. However, the melting point of ice (at standard atmospheric pressure) is experimentally found to be $273.15\,\text{K}$ — i.e., $0.01\,\text{K}$ below the triple point. Since the Celsius scale is defined such that $0^\circ\text{C}$ corresponds to the melting point of ice:
$$t_c = T - 273.15$$
Hence we use $273.15$, not $273.16$.

Part (d):

The Fahrenheit scale has 180 divisions between the ice point ($32^\circ\text{F}$) and steam point ($212^\circ\text{F}$), while the Kelvin scale has 100 divisions for the same interval.

So 1 Fahrenheit unit $= \dfrac{100}{180} = \dfrac{5}{9}$ Kelvin unit.

If the new absolute scale has unit size equal to Fahrenheit, then the triple point of water in this scale:
$$T_{new} = 273.16 \times \frac{9}{5} = 491.69 \approx 491.69\,\text{units}$$

The triple point of water on this scale is $491.69$ (in Fahrenheit-sized absolute units).

Given:
- Triple point of water $T_{tr} = 273.16\,\text{K}$

Formula for ideal gas thermometer:
$$T = 273.16 \times \frac{P}{P_{tr}}$$

Part (a):

*Thermometer A (Oxygen):*
$$T_A = 273.16 \times \frac{1.797\times10^5}{1.250\times10^5} = 273.16 \times 1.4376 = 392.69\,\text{K}$$

*Thermometer B (Hydrogen):*
$$T_B = 273.16 \times \frac{0.287\times10^5}{0.200\times10^5} = 273.16 \times 1.435 = 392.07\,\text{K}$$

Thermometer A reads $\approx 392.69\,\text{K}$ and Thermometer B reads $\approx 392.07\,\text{K}$.

Part (b):

The slight difference arises because oxygen and hydrogen are not perfectly ideal gases. Real gases deviate from ideal behaviour, and the deviation is different for different gases. At the pressures used, intermolecular interactions cause small errors.

Further procedure: The experiment should be repeated at lower and lower pressures (so that the gases approach ideal behaviour) and the results extrapolated to zero pressure. At zero pressure, all real gases behave ideally and both thermometers would give the same reading.

Given:
- Calibration temperature: $T_0 = 27.0^\circ\text{C}$
- Measured length of rod: $L_{measured} = 63.0\,\text{cm}$
- Temperature on hot day: $T = 45.0^\circ\text{C}$
- $\alpha_{steel} = 1.20\times10^{-5}\,\text{K}^{-1}$
- $\Delta T = 45.0 - 27.0 = 18.0^\circ\text{C}$

Step 1: Find the actual length of the tape at $45^\circ\text{C}$

The tape itself expands. The true length of 1 m of tape at $45^\circ\text{C}$:
$$L_{tape} = 1\,\text{m}\times[1 + \alpha\,\Delta T] = 1\times[1 + 1.20\times10^{-5}\times18]$$
$$L_{tape} = 1 + 2.16\times10^{-4} = 1.000216\,\text{m}$$

Step 2: Actual length of the rod at $45^\circ\text{C}$

Each centimetre of the tape is actually $1.000216\,\text{cm}$ long.
$$L_{actual} = 63.0 \times 1.000216 = 63.0 + 63.0\times2.16\times10^{-4}$$
$$L_{actual} = 63.0 + 0.0136 = 63.0136\,\text{cm} \approx 63.014\,\text{cm}$$

Step 3: Length of the rod at $27^\circ\text{C}$

The rod itself contracts from $45^\circ\text{C}$ to $27^\circ\text{C}$:
$$L_{27} = L_{actual}\,[1 - \alpha\,\Delta T] = 63.014\times[1 - 1.20\times10^{-5}\times18]$$
$$L_{27} = 63.014\times[1 - 2.16\times10^{-4}]$$
$$L_{27} = 63.014 - 0.01361 \approx 63.0\,\text{cm}$$

Results:
- Actual length of rod at $45^\circ\text{C}$ $\approx 63.014\,\text{cm}$
- Length of rod at $27^\circ\text{C}$ $\approx 63.0\,\text{cm}$

Given:
- Initial temperature: $T_0 = 27^\circ\text{C} = 300\,\text{K}$
- Diameter of shaft at $27^\circ\text{C}$: $d_s = 8.70\,\text{cm}$
- Diameter of hole in wheel: $d_h = 8.69\,\text{cm}$
- $\alpha_{steel} = 1.20\times10^{-5}\,\text{K}^{-1}$

Concept: The shaft must be cooled until its diameter equals the diameter of the hole ($8.69\,\text{cm}$).

Using linear expansion formula:
$$d_s' = d_s\,[1 + \alpha\,\Delta T]$$

For the wheel to slip on:
$$d_s' = d_h$$
$$8.69 = 8.70\,[1 + 1.20\times10^{-5}\times(T - 300)]$$
$$\frac{8.69}{8.70} = 1 + 1.20\times10^{-5}\times(T - 300)$$
$$0.99885 = 1 + 1.20\times10^{-5}\times(T - 300)$$
$$-1.149\times10^{-3} = 1.20\times10^{-5}\times(T - 300)$$
$$T - 300 = \frac{-1.149\times10^{-3}}{1.20\times10^{-5}} = -95.8\,\text{K}$$
$$T = 300 - 95.8 = 204.2\,\text{K}$$
$$t = 204.2 - 273.15 \approx -68.9^\circ\text{C}$$

The shaft must be cooled to approximately $-69^\circ\text{C}$ for the wheel to slip on.

Given:
- Initial diameter: $d_0 = 4.24\,\text{cm}$
- Initial temperature: $T_0 = 27.0^\circ\text{C}$
- Final temperature: $T = 227^\circ\text{C}$
- $\Delta T = 227 - 27 = 200^\circ\text{C}$
- $\alpha_{Cu} = 1.70\times10^{-5}\,\text{K}^{-1}$

Concept: A hole in a material expands just as if it were filled with the same material. The diameter of the hole increases with temperature following the same linear expansion law.

Change in diameter:
$$\Delta d = d_0\,\alpha\,\Delta T$$
$$\Delta d = 4.24 \times 1.70\times10^{-5} \times 200$$
$$\Delta d = 4.24 \times 3.40\times10^{-3}$$
$$\Delta d = 1.4416\times10^{-2}\,\text{cm}$$
$$\boxed{\Delta d \approx 1.44\times10^{-2}\,\text{cm} = 0.0144\,\text{cm}}$$

The diameter of the hole increases by $0.0144\,\text{cm}$.

Given:
- Length: $L = 1.8\,\text{m}$
- Diameter: $d = 2.0\,\text{mm} = 2.0\times10^{-3}\,\text{m}$
- $T_1 = 27^\circ\text{C}$, $T_2 = -39^\circ\text{C}$
- $\Delta T = 27 - (-39) = 66^\circ\text{C}$ (decrease)
- $\alpha = 2.0\times10^{-5}\,\text{K}^{-1}$
- $Y = 0.91\times10^{11}\,\text{Pa}$

Step 1: Cross-sectional area
$$A = \pi\left(\frac{d}{2}\right)^2 = \pi\times(1.0\times10^{-3})^2 = \pi\times10^{-6}\,\text{m}^2$$
$$A = 3.14\times10^{-6}\,\text{m}^2$$

Step 2: Thermal strain

The wire would contract by $\Delta L = L\,\alpha\,\Delta T$ if free, but since it is fixed, this contraction is prevented, creating a tensile strain:
$$\text{Strain} = \alpha\,\Delta T = 2.0\times10^{-5}\times66 = 1.32\times10^{-3}$$

Step 3: Tension (Stress × Area)
$$\text{Stress} = Y\times\text{Strain} = 0.91\times10^{11}\times1.32\times10^{-3}$$
$$\text{Stress} = 1.2012\times10^{8}\,\text{Pa}$$

$$F = \text{Stress}\times A = 1.2012\times10^{8}\times3.14\times10^{-6}$$
$$F = 3.77\times10^{2}\,\text{N} \approx 3.8\times10^{2}\,\text{N}$$

The tension developed in the wire is approximately $377\,\text{N}$.

Given:
- $L_{brass} = L_{steel} = 50\,\text{cm} = 0.50\,\text{m}$
- $\Delta T = 250 - 40 = 210^\circ\text{C}$
- $\alpha_{brass} = 2.0\times10^{-5}\,\text{K}^{-1}$
- $\alpha_{steel} = 1.2\times10^{-5}\,\text{K}^{-1}$

Change in length of brass rod:
$$\Delta L_{brass} = L_{brass}\,\alpha_{brass}\,\Delta T = 0.50\times2.0\times10^{-5}\times210$$
$$\Delta L_{brass} = 0.50\times4.2\times10^{-3} = 2.1\times10^{-3}\,\text{m} = 0.21\,\text{cm}$$

Change in length of steel rod:
$$\Delta L_{steel} = L_{steel}\,\alpha_{steel}\,\Delta T = 0.50\times1.2\times10^{-5}\times210$$
$$\Delta L_{steel} = 0.50\times2.52\times10^{-3} = 1.26\times10^{-3}\,\text{m} = 0.126\,\text{cm}$$

Total change in length of combined rod:
$$\Delta L_{total} = \Delta L_{brass} + \Delta L_{steel} = 0.21 + 0.126 = 0.336\,\text{cm} \approx 0.34\,\text{cm}$$

Thermal stress at the junction:

Since the ends of the rod are free to expand, there is no constraint on the total expansion. Each rod expands freely. However, the two rods have different coefficients of expansion, so they expand by different amounts. Since they are joined at the junction, the junction experiences a differential expansion, but because the ends are free, no compressive or tensile thermal stress is developed in the bulk of the rods.

The total change in length is $0.34\,\text{cm}$. No thermal stress is developed at the junction since the ends are free to expand.

Given:
- $\alpha_V = 49\times10^{-5}\,\text{K}^{-1}$
- $\Delta T = 30^\circ\text{C}$

Concept:

Density $\rho = \dfrac{m}{V}$. Since mass is constant:
$$\rho' = \frac{m}{V'} = \frac{m}{V(1 + \alpha_V\,\Delta T)} = \frac{\rho}{1 + \alpha_V\,\Delta T}$$

Fractional change in density:
$$\frac{\Delta\rho}{\rho} = \frac{\rho - \rho'}{\rho} = 1 - \frac{1}{1 + \alpha_V\,\Delta T}$$

For small $\alpha_V\,\Delta T$:
$$\frac{\Delta\rho}{\rho} \approx -\alpha_V\,\Delta T$$

(Negative sign indicates decrease in density)

$$\left|\frac{\Delta\rho}{\rho}\right| = 49\times10^{-5}\times30 = 1470\times10^{-5} = 1.47\times10^{-2}$$

The fractional change in density is $1.47\times10^{-2}$ (decrease), i.e., density decreases by about $1.47\%$.

Given:
- Power of machine: $P = 10\,\text{kW} = 10^4\,\text{W}$
- Mass of aluminium block: $m = 8.0\,\text{kg} = 8000\,\text{g}$
- Time: $t = 2.5\,\text{min} = 150\,\text{s}$
- Useful power (50% of total): $P_{useful} = 0.50\times10^4 = 5\times10^3\,\text{W}$
- Specific heat: $s = 0.91\,\text{J g}^{-1}\text{K}^{-1} = 910\,\text{J kg}^{-1}\text{K}^{-1}$

Step 1: Heat supplied to the block
$$Q = P_{useful}\times t = 5\times10^3\times150 = 7.5\times10^5\,\text{J}$$

Step 2: Rise in temperature
$$Q = m\,s\,\Delta T$$
$$\Delta T = \frac{Q}{m\,s} = \frac{7.5\times10^5}{8.0\times910}$$
$$\Delta T = \frac{7.5\times10^5}{7280} \approx 103\,\text{K}$$

The rise in temperature of the aluminium block is approximately $103^\circ\text{C}$.

Given:
- Mass of copper: $m_{Cu} = 2.5\,\text{kg} = 2500\,\text{g}$
- Initial temperature of copper: $T_i = 500^\circ\text{C}$
- Final temperature (ice at $0^\circ\text{C}$): $T_f = 0^\circ\text{C}$
- $s_{Cu} = 0.39\,\text{J g}^{-1}\text{K}^{-1}$
- $L_f = 335\,\text{J g}^{-1}$

Step 1: Heat released by copper block
$$Q = m_{Cu}\times s_{Cu}\times(T_i - T_f)$$
$$Q = 2500\times0.39\times(500 - 0)$$
$$Q = 2500\times0.39\times500 = 4.875\times10^5\,\text{J}$$

Step 2: Mass of ice melted
$$Q = m_{ice}\times L_f$$
$$m_{ice} = \frac{Q}{L_f} = \frac{4.875\times10^5}{335}$$
$$m_{ice} = 1455\,\text{g} \approx 1.455\,\text{kg}$$

The maximum amount of ice that can melt is approximately $1.45\,\text{kg}$.

Given:
- Mass of metal: $m = 0.20\,\text{kg} = 200\,\text{g}$
- Initial temperature of metal: $T_m = 150^\circ\text{C}$
- Water equivalent of calorimeter: $W = 0.025\,\text{kg} = 25\,\text{g}$
- Volume of water: $150\,\text{cm}^3 \Rightarrow$ mass of water $= 150\,\text{g}$
- Initial temperature of water + calorimeter: $T_w = 27^\circ\text{C}$
- Final temperature: $T_f = 40^\circ\text{C}$
- $s_{water} = 1\,\text{cal g}^{-1}\text{K}^{-1} = 4.18\,\text{J g}^{-1}\text{K}^{-1}$

Applying principle of calorimetry (heat lost = heat gained):

$$m\,s_{metal}\,(T_m - T_f) = (m_{water} + W)\,s_{water}\,(T_f - T_w)$$

$$200\times s_{metal}\times(150 - 40) = (150 + 25)\times4.18\times(40 - 27)$$

$$200\times s_{metal}\times110 = 175\times4.18\times13$$

$$22000\,s_{metal} = 9509.5$$

$$s_{metal} = \frac{9509.5}{22000} = 0.4322\,\text{J g}^{-1}\text{K}^{-1}$$

$$\boxed{s_{metal} \approx 0.43\,\text{J g}^{-1}\text{K}^{-1}}$$

Effect of heat losses:

If heat is lost to the surroundings, the water and calorimeter absorb less heat than what the metal actually released. This means the calculated value of $s_{metal}$ (based on heat gained by water) will be smaller than the actual value. So our answer is smaller than the actual specific heat.

Explanation of the difference:

A monatomic gas (like Ar, He) has only 3 translational degrees of freedom, so its molar specific heat at constant volume:
$$C_V = \frac{3}{2}R \approx 2.98\,\text{cal mol}^{-1}\text{K}^{-1} \approx 2.92\,\text{cal mol}^{-1}\text{K}^{-1}$$

The gases listed (H₂, N₂, O₂, NO, CO) are diatomic molecules. They have:
- 3 translational degrees of freedom
- 2 rotational degrees of freedom

Total degrees of freedom = 5, so:
$$C_V = \frac{5}{2}R \approx \frac{5}{2}\times1.987 \approx 4.97\,\text{cal mol}^{-1}\text{K}^{-1}$$

This matches the observed values (~4.87 to 5.02 cal mol⁻¹ K⁻¹) very well. The extra degrees of freedom (rotational) allow these molecules to absorb more energy per mole per degree rise in temperature compared to monatomic gases.

Inference from Chlorine's larger value ($6.17\,\text{cal mol}^{-1}\text{K}^{-1}$):

Chlorine (Cl₂) is also a diatomic molecule, but its value is significantly higher than the expected $\approx 5\,\text{cal mol}^{-1}\text{K}^{-1}$. This suggests that at room temperature, vibrational degrees of freedom are also partially active in Cl₂. A diatomic molecule has 1 vibrational mode contributing $R$ to $C_V$ when fully active, giving $C_V = \frac{7}{2}R \approx 6.95\,\text{cal mol}^{-1}\text{K}^{-1}$. The value 6.17 is between 5 and 7, indicating partial excitation of vibrational modes in Cl₂ at room temperature. This is because Cl₂ has a lower vibrational frequency (heavier atoms, weaker bond relatively), making vibrational modes accessible at room temperature.

Given:
- Initial temperature: $T_1 = 101^\circ\text{F}$
- Final temperature: $T_2 = 98^\circ\text{F}$
- Fall in temperature: $\Delta t_F = 3^\circ\text{F}$
- Time: $t = 20\,\text{min}$
- Mass of child: $m = 30\,\text{kg} = 30000\,\text{g}$
- $s_{body} = s_{water} = 1\,\text{cal g}^{-1}\text{K}^{-1}$
- $L = 580\,\text{cal g}^{-1}$

Step 1: Convert temperature fall to Celsius
$$\Delta t_C = \frac{5}{9}\times\Delta t_F = \frac{5}{9}\times3 = \frac{5}{3}^\circ\text{C}$$

Step 2: Heat lost by the body
$$Q = m\times s\times\Delta t_C = 30000\times1\times\frac{5}{3}$$
$$Q = 30000\times\frac{5}{3} = 50000\,\text{cal}$$

Step 3: Mass of sweat evaporated
$$Q = m_{sweat}\times L$$
$$m_{sweat} = \frac{Q}{L} = \frac{50000}{580} = 86.2\,\text{g}$$

Step 4: Rate of evaporation
$$\text{Rate} = \frac{m_{sweat}}{t} = \frac{86.2\,\text{g}}{20\,\text{min}} = 4.31\,\text{g/min}$$

The average rate of extra evaporation is approximately $4.3\,\text{g min}^{-1}$.

Given:
- Side of cubical box: $a = 30\,\text{cm} = 0.30\,\text{m}$
- Thickness: $L = 5.0\,\text{cm} = 0.05\,\text{m}$
- Initial mass of ice: $m_0 = 4.0\,\text{kg}$
- Time: $t = 6\,\text{h} = 6\times3600 = 21600\,\text{s}$
- Outside temperature: $T_2 = 45^\circ\text{C}$
- Temperature of ice: $T_1 = 0^\circ\text{C}$
- $\Delta T = 45^\circ\text{C}$
- $K = 0.01\,\text{J s}^{-1}\text{m}^{-1}\text{K}^{-1}$
- $L_f = 335\times10^3\,\text{J kg}^{-1}$

Step 1: Total surface area of the box
$$A = 6\times a^2 = 6\times(0.30)^2 = 6\times0.09 = 0.54\,\text{m}^2$$

Step 2: Rate of heat flow into the box
$$H = \frac{KA\,\Delta T}{L} = \frac{0.01\times0.54\times45}{0.05}$$
$$H = \frac{0.243}{0.05} = 4.86\,\text{J/s}$$

Step 3: Total heat flowing in 6 hours
$$Q = H\times t = 4.86\times21600 = 1.049\times10^5\,\text{J}$$

Step 4: Mass of ice melted
$$m_{melted} = \frac{Q}{L_f} = \frac{1.049\times10^5}{335\times10^3} = 0.313\,\text{kg}$$

Step 5: Ice remaining
$$m_{remaining} = 4.0 - 0.313 = 3.687\,\text{kg} \approx 3.687\,\text{kg}$$

The amount of ice remaining after 6 hours is approximately $3.687\,\text{kg}$.

Given:
- Base area: $A = 0.15\,\text{m}^2$
- Thickness: $L = 1.0\,\text{cm} = 0.01\,\text{m}$
- Rate of boiling: $\dot{m} = 6.0\,\text{kg/min} = \dfrac{6.0}{60} = 0.1\,\text{kg/s}$
- $K_{brass} = 109\,\text{J s}^{-1}\text{m}^{-1}\text{K}^{-1}$
- $L_v = 2256\times10^3\,\text{J kg}^{-1}$
- Boiling point of water: $T_2 = 100^\circ\text{C}$

Step 1: Rate of heat required to boil water
$$H = \dot{m}\times L_v = 0.1\times2256\times10^3 = 2.256\times10^5\,\text{J/s}$$

Step 2: Using Fourier's law of heat conduction
$$H = \frac{KA(T_1 - T_2)}{L}$$
$$T_1 - T_2 = \frac{H\times L}{K\times A}$$
$$T_1 - T_2 = \frac{2.256\times10^5\times0.01}{109\times0.15}$$
$$T_1 - T_2 = \frac{2256}{16.35} = 137.98^\circ\text{C} \approx 138^\circ\text{C}$$

$$T_1 = 100 + 138 = 238^\circ\text{C}$$

The temperature of the part of the flame in contact with the boiler is approximately $238^\circ\text{C}$.

Part (a): Body with large reflectivity is a poor emitter

By Kirchhoff's law of radiation, at thermal equilibrium, a good absorber is also a good emitter. A body with large reflectivity absorbs very little radiation (since absorptivity + reflectivity = 1 for an opaque body, so high reflectivity means low absorptivity). Therefore, by Kirchhoff's law, it is also a poor emitter.

Part (b): Brass tumbler feels colder than wooden tray

Brass is a good conductor of heat, while wood is a poor conductor (insulator). On a chilly day, both are at the same temperature (below body temperature). When we touch brass, heat flows rapidly from our hand to the brass (due to high thermal conductivity), making it feel very cold. Wood conducts heat away from our hand much more slowly, so it does not feel as cold. The sensation of coldness depends on the rate of heat transfer, not just the temperature.

Part (c): Optical pyrometer gives too low a temperature in open, correct in furnace

An optical pyrometer is calibrated assuming the object is a perfect black body (emissivity $e = 1$). A red hot iron piece in the open is not a black body (e < 1); it emits less radiation than a black body at the same temperature. The pyrometer interprets this lower emission as a lower temperature, giving a reading that is too low.

When the same piece is inside a furnace, the furnace cavity acts approximately as a black body enclosure. The radiation coming out of the furnace hole is black body radiation regardless of the emissivity of the iron piece. Hence the pyrometer gives the correct temperature.

Part (d): Earth without atmosphere would be inhospitably cold

The atmosphere acts like a blanket around the Earth. It absorbs the infrared radiation emitted by the Earth's surface and re-radiates part of it back to the surface (greenhouse effect). Without the atmosphere, all the heat radiated by the Earth would escape into space, and the average surface temperature would drop to about $-18^\circ\text{C}$ (instead of the current $+15^\circ\text{C}$), making it inhospitably cold.

Part (e): Steam-based heating is more efficient than hot water

When steam at $100^\circ\text{C}$ condenses to water at $100^\circ\text{C}$, it releases the latent heat of vaporisation ($\approx 2256\,\text{kJ/kg}$), which is very large. Hot water, on the other hand, only releases sensible heat as it cools down. For the same mass of fluid circulated, steam delivers far more heat to the building. Hence steam-based systems are more efficient in warming a building.

Given:
- Case 1: Body cools from $80^\circ\text{C}$ to $50^\circ\text{C}$ in $t_1 = 5\,\text{min}$
- Case 2: Body cools from $60^\circ\text{C}$ to $30^\circ\text{C}$ in time $t_2 = ?$
- Surrounding temperature: $T_0 = 20^\circ\text{C}$

Using Newton's Law of Cooling:
$$\frac{\Delta T}{t} = K\left(\bar{T} - T_0\right)$$
where $\bar{T}$ is the average temperature of the body.

Case 1:
- Average temperature: $\bar{T}_1 = \dfrac{80 + 50}{2} = 65^\circ\text{C}$
- Excess temperature: $\bar{T}_1 - T_0 = 65 - 20 = 45^\circ\text{C}$
- Temperature drop: $\Delta T_1 = 80 - 50 = 30^\circ\text{C}$

$$\frac{30}{5} = K\times45 \quad \Rightarrow \quad K = \frac{6}{45} = \frac{2}{15}\,\text{min}^{-1}$$

Case 2:
- Average temperature: $\bar{T}_2 = \dfrac{60 + 30}{2} = 45^\circ\text{C}$
- Excess temperature: $\bar{T}_2 - T_0 = 45 - 20 = 25^\circ\text{C}$
- Temperature drop: $\Delta T_2 = 60 - 30 = 30^\circ\text{C}$

$$\frac{30}{t_2} = K\times25 = \frac{2}{15}\times25 = \frac{50}{15} = \frac{10}{3}$$

$$t_2 = \frac{30\times3}{10} = 9\,\text{min}$$

The body takes $9\,\text{minutes}$ to cool from $60^\circ\text{C}$ to $30^\circ\text{C}$.