Thermal Properties of Matter — NCERT Solutions

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 $30^\circ\text{C}$.