Thermal Properties Of MatterClass 11 Physics Notes

Temperature and Heat

Everything around us is made of matter, and one of the key properties of matter is its temperature.

Temperature is a measure of the 'hotness' or 'coldness' of an object. It's a relative term; a cup of hot tea has a higher temperature than a glass of ice water. Our sense of touch gives us a basic idea of temperature, but it's not reliable enough for science.

Heat is the form of energy that is transferred from one object to another because of a difference in their temperatures. Heat always flows from a hotter object (higher temperature) to a colder object (lower temperature). This transfer continues until both objects reach the same temperature, a state we call thermal equilibrium.

• The SI unit of heat is the joule (J).
• The SI unit of temperature is the kelvin (K).
• A common unit for temperature is degree Celsius (°C).

When an object is heated, it can undergo several changes: its temperature might rise, it might expand, or it might even change its state (like ice melting into water).

Measurement of Temperature

To measure temperature accurately, we use a device called a thermometer. Thermometers work based on a simple principle: some physical properties of materials change consistently with temperature. The most common property used is the expansion and contraction of a liquid.

• Liquid-in-glass thermometers (like those using mercury or alcohol) rely on the fact that the liquid's volume changes linearly with temperature. As the thermometer gets hotter, the liquid expands and rises up the narrow tube.

Temperature Scales

To give a numerical value to temperature, we need a scale. A scale is defined by two fixed reference points. The most common reference points are:

1. Ice Point: The temperature at which pure water freezes at standard pressure ($0^\circ\text{C}$).
2. Steam Point: The temperature at which pure water boils at standard pressure ($100^\circ\text{C}$).

The two most familiar scales are Celsius and Fahrenheit.

• Celsius Scale: Ice point is $0^\circ\text{C}$ and steam point is $100^\circ\text{C}$. There are 100 intervals between these points.
• Fahrenheit Scale: Ice point is $32^\circ\text{F}$ and steam point is $212^\circ\text{F}$. There are 180 intervals between these points.

The relationship between Fahrenheit ($t_F$) and Celsius ($t_C$) is linear and can be expressed with the following formula: $\frac{t_F - 32}{180} = \frac{t_C}{100}$

Ideal-Gas Equation and Absolute Temperature

While liquid-in-glass thermometers are common, they can give slightly different readings because different liquids expand differently. A gas thermometer, however, is more consistent, especially with low-density gases.

The behavior of a gas is described by its pressure ($P$), volume ($V$), and temperature ($T$). For low-density gases, these are related by simple laws:

• Boyle's Law: At a constant temperature, $PV = \text{constant}$.
• Charles' Law: At a constant pressure, $V/T = \text{constant}$.

Combining these laws gives us the ideal gas law: $PV/T = \text{constant}$. This can be written in a more general form known as the ideal-gas equation: $PV = \mu R T$ Where:

• $\mu$ is the number of moles of the gas.
• $R$ is the universal gas constant, which has a value of $8.31 \text{ J mol}^{-1} \text{K}^{-1}$.
• $T$ is the temperature in kelvin.

Absolute Zero and the Kelvin Scale

If we plot the pressure of a gas versus its temperature (at constant volume), we get a straight line. If we extend this line backwards, it seems that the pressure would become zero at a temperature of $-273.15^\circ\text{C}$. This temperature is called absolute zero. It's the theoretical point where a gas would have zero pressure and is the lowest possible temperature.

This concept is the foundation of the Kelvin temperature scale, named after Lord Kelvin.

• Absolute zero is defined as $0 \text{ K}$.
• The size of one kelvin is the same as the size of one degree Celsius.
• The relationship between Kelvin ($T$) and Celsius ($t_C$) is: $T = t_C + 273.15$

Thermal Expansion

Most substances expand when heated and contract when cooled. This change in the dimensions of a body due to a change in its temperature is called thermal expansion.

• Linear Expansion: Expansion in length (e.g., a metal rod getting longer).
• Area Expansion: Expansion in area (e.g., a metal sheet getting larger).
• Volume Expansion: Expansion in volume (e.g., the mercury in a thermometer expanding).

Coefficient of Linear Expansion

For a small change in temperature, $\Delta T$, the fractional change in length ($\Delta l / l$) of a rod is directly proportional to $\Delta T$. $\frac{\Delta l}{l} = \alpha_l \Delta T$ Here, $\alpha_l$ is the coefficient of linear expansion. It is a property of the material and tells us how much it expands for each degree of temperature change. Metals generally have higher values of $\alpha_l$ than materials like glass.

Coefficient of Volume Expansion

Similarly, the fractional change in volume ($\Delta V / V$) is proportional to the temperature change $\Delta T$. $\alpha_V = \left(\frac{\Delta V}{V}\right) \frac{1}{\Delta T}$ Here, $\alpha_V$ is the coefficient of volume expansion. For an isotropic solid (one that expands equally in all directions), the relationship between the volume and linear expansion coefficients is simple: $\alpha_V = 3\alpha_l$

Anomalous Behavior of Water

Water behaves unusually. Instead of expanding when heated from $0^\circ\text{C}$, it actually contracts between $0^\circ\text{C}$ and $4^\circ\text{C}$. This means water has its maximum density at $4^\circ\text{C}$.

This property is crucial for aquatic life. In winter, as a lake cools, the surface water gets denser and sinks. This process stops once the entire lake reaches $4^\circ\text{C}$. As the surface water cools further (below $4^\circ\text{C}$), it becomes less dense and stays on top, where it eventually freezes. Because the ice forms on the surface, it insulates the water below, allowing fish and other organisms to survive in the liquid water at the bottom.

Thermal Stress

What if you try to prevent an object from expanding? If you heat a metal rod but fix its ends so it can't get longer, a large compressive force will build up inside it. This is called thermal stress.

Given

• A rectangular sheet with initial length $a$ and breadth $b$.
• Initial area $A = ab$.
• Coefficient of linear expansion is $\alpha_l$.
• Temperature increases by $\Delta T$.

To Find

The relationship between the coefficient of area expansion and $\alpha_l$.

Formula

• Change in length: $\Delta a = \alpha_l a \Delta T$
• Change in breadth: $\Delta b = \alpha_l b \Delta T$
• New area $A' = (a + \Delta a)(b + \Delta b)$
• Change in area $\Delta A = A' - A$

Solution

The new area is $A' = (a + \Delta a)(b + \Delta b) = ab + a\Delta b + b\Delta a + (\Delta a)(\Delta b)$. The increase in area is $\Delta A = A' - A = a\Delta b + b\Delta a + (\Delta a)(\Delta b)$.

Substitute the expressions for $\Delta a$ and $\Delta b$: $\Delta A = a(\alpha_l b \Delta T) + b(\alpha_l a \Delta T) + (\alpha_l a \Delta T)(\alpha_l b \Delta T)$ $\Delta A = 2 \alpha_l ab \Delta T + (\alpha_l)^2 ab (\Delta T)^2$ Since $A=ab$, $\Delta A = 2 \alpha_l A \Delta T + (\alpha_l)^2 A (\Delta T)^2$ The value of $\alpha_l$ is very small (around $10^{-5} \text{ K}^{-1}$), so the term $(\alpha_l)^2$ is extremely small and can be neglected. $\Delta A \approx 2 \alpha_l A \Delta T$ Rearranging the formula to find the coefficient of area expansion, which we can call $\alpha_A$: $\alpha_A = \frac{\Delta A}{A \Delta T} \approx 2\alpha_l$

Final Answer The coefficient of area expansion is approximately twice the coefficient of linear expansion.

Given

• Initial temperature, $T_1 = 27^\circ\text{C}$
• Initial diameter of the iron ring, $L_{T1} = 5.231 \text{ m}$
• Required final diameter of the iron ring, $L_{T2} = 5.243 \text{ m}$ (to match the wheel rim)
• Coefficient of linear expansion for iron, $\alpha_l = 1.20 \times 10^{-5} \text{ K}^{-1}$

To Find

The final temperature, $T_2$, to which the ring must be heated.

Formula

$L_{T2} = L_{T1}[1 + \alpha_l(T_2 - T_1)]$

Solution

Substitute the given values into the formula: $5.243 \text{ m} = 5.231 \text{ m} [1 + 1.20 \times 10^{-5} \text{ K}^{-1} (T_2 - 27^\circ\text{C})]$ First, divide both sides by $5.231$: $\frac{5.243}{5.231} = 1 + 1.20 \times 10^{-5} (T_2 - 27)$ $1.002294 = 1 + 1.20 \times 10^{-5} (T_2 - 27)$ Subtract 1 from both sides: $0.002294 = 1.20 \times 10^{-5} (T_2 - 27)$ Solve for $(T_2 - 27)$: $T_2 - 27 = \frac{0.002294}{1.20 \times 10^{-5}} \approx 191.17$ Solve for $T_2$: $T_2 = 191.17 + 27 = 218.17^\circ\text{C}$

Final Answer The ring should be heated to approximately $218^\circ\text{C}$.

Specific Heat Capacity

The amount of heat required to raise the temperature of a substance depends on three factors:

1. The mass of the substance ($m$).
2. The desired change in temperature ($\Delta T$).
3. The nature of the substance itself.

Heat Capacity (S) is the amount of heat ($\Delta Q$) needed to raise the temperature of a substance by one unit ($\Delta T$). $S = \frac{\Delta Q}{\Delta T}$

Different materials require different amounts of heat to change their temperature. This unique property is called specific heat capacity.

Specific Heat Capacity (s) is the amount of heat required to raise the temperature of a unit mass of a substance by one degree. $s = \frac{1}{m} \frac{\Delta Q}{\Delta T}$ The SI unit for specific heat capacity is J kg⁻¹ K⁻¹.

Water has a very high specific heat capacity ($4186 \text{ J kg}^{-1} \text{K}^{-1}$). This means it takes a lot of energy to heat water up, and it releases a lot of energy as it cools down.

• Coolant: Water is used in car radiators because it can absorb a large amount of heat from the engine without its temperature rising too quickly.
• Climate: Coastal areas have milder climates because the ocean absorbs huge amounts of heat in the summer and releases it slowly in the winter, moderating the temperature. Deserts, with sandy soil (low specific heat), heat up very quickly during the day and cool down very quickly at night.

Molar Specific Heat Capacity (C) is used when the amount of substance is measured in moles ($\mu$) instead of mass. $C = \frac{1}{\mu} \frac{\Delta Q}{\Delta T}$ The SI unit is J mol⁻¹ K⁻¹. For gases, we specify whether the heat is added at constant pressure ($C_p$) or constant volume ($C_v$).

Calorimetry

Calorimetry is the science of measuring heat. It works on a fundamental principle: in an isolated system, heat lost by a hot object is equal to the heat gained by a cold object.

A calorimeter is a device used to measure heat transfer. It typically consists of an insulated container (often made of copper) with a stirrer and a thermometer. The insulation minimizes heat exchange with the surroundings, so we can assume that all the heat transfer happens between the objects inside.

Given

• Mass of aluminium sphere, $m_{Al} = 0.047 \text{ kg}$
• Initial temperature of sphere, $T_{Al, initial} = 100^\circ\text{C}$
• Mass of water, $m_w = 0.25 \text{ kg}$
• Mass of copper calorimeter, $m_{cu} = 0.14 \text{ kg}$
• Initial temperature of water and calorimeter, $T_{initial} = 20^\circ\text{C}$
• Final temperature of the mixture, $T_{final} = 23^\circ\text{C}$
• Specific heat of water, $s_w = 4.18 \times 10^3 \text{ J kg}^{-1} \text{K}^{-1}$
• Specific heat of copper, $s_{cu} = 0.386 \times 10^3 \text{ J kg}^{-1} \text{K}^{-1}$

To Find

Specific heat capacity of aluminium, $s_{Al}$.

Formula

• Heat lost or gained: $Q = ms\Delta T$
• Principle of calorimetry: Heat Lost = Heat Gained

Solution

1. Calculate the heat lost by the aluminium sphere: The temperature of the sphere drops from $100^\circ\text{C}$ to $23^\circ\text{C}$. $\Delta T_{Al} = 100^\circ\text{C} - 23^\circ\text{C} = 77^\circ\text{C}$ $\text{Heat Lost} = m_{Al} s_{Al} \Delta T_{Al} = (0.047 \text{ kg}) \times s_{Al} \times (77^\circ\text{C})$

2. Calculate the heat gained by the water and calorimeter: The temperature of the water and calorimeter rises from $20^\circ\text{C}$ to $23^\circ\text{C}$. $\Delta T_{gain} = 23^\circ\text{C} - 20^\circ\text{C} = 3^\circ\text{C}$ Heat gained by water: $Q_w = m_w s_w \Delta T_{gain} = (0.25 \text{ kg})(4.18 \times 10^3 \text{ J kg}^{-1} \text{K}^{-1})(3^\circ\text{C})$ Heat gained by calorimeter: $Q_{cu} = m_{cu} s_{cu} \Delta T_{gain} = (0.14 \text{ kg})(0.386 \times 10^3 \text{ J kg}^{-1} \text{K}^{-1})(3^\circ\text{C})$ $\text{Total Heat Gained} = (m_w s_w + m_{cu} s_{cu}) \Delta T_{gain}$ $\text{Total Heat Gained} = [(0.25 \times 4.18 \times 10^3) + (0.14 \times 0.386 \times 10^3)] \times 3$ $\text{Total Heat Gained} = (1045 + 54.04) \times 3 = 1099.04 \times 3 = 3297.12 \text{ J}$

3. Equate Heat Lost and Heat Gained: $(0.047) \times s_{Al} \times (77) = 3297.12$ $3.619 \times s_{Al} = 3297.12$ $s_{Al} = \frac{3297.12}{3.619} \approx 911.05 \text{ J kg}^{-1} \text{K}^{-1}$

Final Answer The specific heat capacity of aluminium is approximately $0.911 \text{ kJ kg}^{-1} \text{K}^{-1}$.

Change of State

Matter typically exists in three states: solid, liquid, and gas. A transition from one state to another is called a change of state or a phase transition. These changes happen when a substance absorbs or releases heat.

• Melting (or Fusion): Solid to liquid.
• Freezing: Liquid to solid.
• Vaporization: Liquid to gas (vapor).
• Condensation: Gas (vapor) to liquid.
• Sublimation: Solid directly to gas (e.g., dry ice).

A crucial point about phase changes is that the temperature of the substance remains constant while the change is happening, even though heat is being added or removed. The energy is used to change the state, not to raise the temperature.

Melting Point and Boiling Point

• Melting Point: The constant temperature at which a solid and its liquid form can coexist in thermal equilibrium.
• Boiling Point: The constant temperature at which a liquid and its vapor form can coexist in thermal equilibrium.

These points are affected by pressure:

• Effect of Pressure on Boiling: Increasing the pressure on a liquid raises its boiling point. This is the principle behind a pressure cooker, which cooks food faster because the water boils at a temperature higher than $100^\circ\text{C}$. At high altitudes, the atmospheric pressure is lower, so water boils at a lower temperature, making cooking more difficult.
• Effect of Pressure on Melting: For most substances, increased pressure raises the melting point. For water, however, increased pressure lowers the melting point. This leads to a phenomenon called regelation: if you press a wire down on a block of ice, the ice under the wire melts (due to pressure), allowing the wire to pass through. The water then refreezes above the wire.

Triple Point

The Triple Point of a substance is the specific combination of temperature and pressure at which all three phases (solid, liquid, and gas) can coexist in thermal equilibrium. For water, the triple point occurs at $273.16 \text{ K}$ ($0.01^\circ\text{C}$) and a very low pressure of $6.11 \times 10^{-3} \text{ Pa}$.

Latent Heat

The heat energy required to change the state of a substance is called latent heat. The word "latent" means hidden, because this energy does not cause a temperature change.

The amount of heat ($Q$) required for a mass ($m$) to change state is given by: $Q = m L$ where L is the latent heat of the substance for that process, measured in J kg⁻¹.

• Latent Heat of Fusion ($L_f$): The heat per unit mass required to change a substance from solid to liquid at its melting point. For water, $L_f = 3.33 \times 10^5 \text{ J kg}^{-1}$.
• Latent Heat of Vaporization ($L_v$): The heat per unit mass required to change a substance from liquid to vapor at its boiling point. For water, $L_v = 22.6 \times 10^5 \text{ J kg}^{-1}$.

Given

• Mass of ice, $m_{ice} = 0.15 \text{ kg}$
• Mass of water, $m_{water} = 0.30 \text{ kg}$
• Initial temperature of ice, $T_{ice, initial} = 0^\circ\text{C}$
• Initial temperature of water, $T_{water, initial} = 50^\circ\text{C}$
• Final temperature of mixture, $T_{final} = 6.7^\circ\text{C}$
• Specific heat of water, $s_w = 4186 \text{ J kg}^{-1} \text{K}^{-1}$

To Find

The latent heat of fusion of ice, $L_f$.

Formula

• Heat lost by water = $m_w s_w (T_{water, initial} - T_{final})$
• Heat gained by ice = (Heat to melt ice) + (Heat to warm the melted water)
• Heat gained = $m_{ice} L_f + m_{ice} s_w (T_{final} - T_{ice, initial})$
• Heat Lost = Heat Gained

Solution

1. Calculate the heat lost by the hot water: $\text{Heat Lost} = (0.30 \text{ kg})(4186 \text{ J kg}^{-1} \text{K}^{-1})(50.0^\circ\text{C} - 6.7^\circ\text{C})$ $\text{Heat Lost} = (0.30)(4186)(43.3) = 54376.14 \text{ J}$

2. Calculate the heat gained by the ice: The ice first melts at $0^\circ\text{C}$ and then the resulting water warms up to $6.7^\circ\text{C}$.

• Heat to melt ice: $Q_{melt} = m_{ice} L_f = (0.15 \text{ kg}) L_f$
• Heat to warm melted water: $Q_{warm} = m_{ice} s_w (6.7^\circ\text{C} - 0^\circ\text{C})$ $Q_{warm} = (0.15 \text{ kg})(4186 \text{ J kg}^{-1} \text{K}^{-1})(6.7^\circ\text{C}) = 4206.93 \text{ J}$ Total heat gained = $Q_{melt} + Q_{warm} = (0.15)L_f + 4206.93 \text{ J}$

3. Equate Heat Lost and Heat Gained: $54376.14 \text{ J} = (0.15)L_f + 4206.93 \text{ J}$ $(0.15)L_f = 54376.14 - 4206.93 = 50169.21$ $L_f = \frac{50169.21}{0.15} = 334461.4 \text{ J kg}^{-1}$

Final Answer The heat of fusion of ice is approximately $3.34 \times 10^5 \text{ J kg}^{-1}$.

Given

• Mass, $m = 3 \text{ kg}$
• $s_{ice} = 2100 \text{ J kg}^{-1} \text{K}^{-1}$
• $s_{water} = 4186 \text{ J kg}^{-1} \text{K}^{-1}$
• $L_{fice} = 3.35 \times 10^5 \text{ J kg}^{-1}$
• $L_{steam} = 2.256 \times 10^6 \text{ J kg}^{-1}$

To Find

Total heat ($Q$) required for the entire process.

Formula

• Heat to change temperature: $Q = ms\Delta T$
• Heat to change state: $Q = mL$

Solution

This process happens in four steps. We calculate the heat for each step and add them together.

1. Heat to raise ice from $-12^\circ\text{C}$ to $0^\circ\text{C}$ ($Q_1$): $\Delta T_1 = 0^\circ\text{C} - (-12^\circ\text{C}) = 12^\circ\text{C}$ $Q_1 = m s_{ice} \Delta T_1 = (3 \text{ kg})(2100 \text{ J kg}^{-1} \text{K}^{-1})(12^\circ\text{C}) = 75,600 \text{ J}$

2. Heat to melt ice at $0^\circ\text{C}$ to water at $0^\circ\text{C}$ ($Q_2$): $Q_2 = m L_{fice} = (3 \text{ kg})(3.35 \times 10^5 \text{ J kg}^{-1}) = 1,005,000 \text{ J}$

3. Heat to raise water from $0^\circ\text{C}$ to $100^\circ\text{C}$ ($Q_3$): $\Delta T_2 = 100^\circ\text{C} - 0^\circ\text{C} = 100^\circ\text{C}$ $Q_3 = m s_{water} \Delta T_2 = (3 \text{ kg})(4186 \text{ J kg}^{-1} \text{K}^{-1})(100^\circ\text{C}) = 1,255,800 \text{ J}$

4. Heat to vaporize water at $100^\circ\text{C}$ to steam at $100^\circ\text{C}$ ($Q_4$): $Q_4 = m L_{steam} = (3 \text{ kg})(2.256 \times 10^6 \text{ J kg}^{-1}) = 6,768,000 \text{ J}$

Total Heat Required ($Q$): $Q = Q_1 + Q_2 + Q_3 + Q_4$ $Q = 75,600 + 1,005,000 + 1,255,800 + 6,768,000 = 9,104,400 \text{ J}$

Final Answer The total heat required is approximately $9.1 \times 10^6 \text{ J}$.

Heat Transfer

Heat energy moves from a hotter place to a colder place through three distinct modes: conduction, convection, and radiation.

Conduction

Conduction is the transfer of heat through a material without any actual movement of the material itself. It happens through molecular collisions.

Metals are good thermal conductors, while materials like wood, plastic, and air are poor conductors (insulators).

The rate of heat flow ($H$) through a uniform bar is given by: $H = K A \frac{T_C - T_D}{L}$ Where:

• H is the heat current (rate of heat flow, in J/s or Watts).
• K is the thermal conductivity of the material (a measure of how well it conducts heat).
• A is the cross-sectional area of the bar.
• L is the length of the bar.
• ($T_C - T_D$) is the temperature difference between the hot end ($T_C$) and the cold end ($T_D$).

Given

• Steel Rod (1): $L_1 = 15.0 \text{ cm} = 0.15 \text{ m}$, $K_1 = 50.2 \text{ J s}^{-1} \text{m}^{-1} \text{K}^{-1}$, $T_1 = 300^\circ\text{C}$
• Copper Rod (2): $L_2 = 10.0 \text{ cm} = 0.10 \text{ m}$, $K_2 = 385 \text{ J s}^{-1} \text{m}^{-1} \text{K}^{-1}$, $T_2 = 0^\circ\text{C}$
• Area relationship: $A_1 = 2A_2$

To Find

The temperature of the junction, $T$.

Formula

In steady state, the rate of heat flow through the steel rod ($H_1$) is equal to the rate of heat flow through the copper rod ($H_2$). $H_1 = \frac{K_1 A_1 (T_1 - T)}{L_1}$ $H_2 = \frac{K_2 A_2 (T - T_2)}{L_2}$ $H_1 = H_2$

Solution

Set the two expressions for heat flow equal to each other: $\frac{K_1 A_1 (300 - T)}{L_1} = \frac{K_2 A_2 (T - 0)}{L_2}$ Substitute the known values and the area relationship ($A_1 = 2A_2$): $\frac{50.2 \times (2A_2) \times (300 - T)}{0.15} = \frac{385 \times A_2 \times T}{0.10}$ The area $A_2$ cancels out from both sides: $\frac{50.2 \times 2 \times (300 - T)}{0.15} = \frac{385 \times T}{0.10}$ $669.33 \times (300 - T) = 3850 \times T$ $200800 - 669.33 T = 3850 T$ $200800 = 3850 T + 669.33 T$ $200800 = 4519.33 T$ $T = \frac{200800}{4519.33} \approx 44.43^\circ\text{C}$

Final Answer The temperature of the steel-copper junction is $44.4^\circ\text{C}$.

Given

• Iron bar (1): $L_1 = 0.1 \text{ m}$, $A_1 = 0.02 \text{ m}^2$, $K_1 = 79 \text{ W m}^{-1} \text{K}^{-1}$, $T_1 = 373 \text{ K}$
• Brass bar (2): $L_2 = 0.1 \text{ m}$, $A_2 = 0.02 \text{ m}^2$, $K_2 = 109 \text{ W m}^{-1} \text{K}^{-1}$, $T_2 = 273 \text{ K}$
• Since $L_1 = L_2 = L = 0.1 \text{ m}$ and $A_1 = A_2 = A = 0.02 \text{ m}^2$.

To Find

(i) Junction temperature, $T_0$ (ii) Equivalent thermal conductivity, $K'$ (iii) Heat current, $H$

Formula

• Heat current: $H = \frac{KA(T_{hot} - T_{cold})}{L}$
• In steady state: $H_{iron} = H_{brass}$
• For a compound bar in series: $H = \frac{K'A(T_1 - T_2)}{L_1 + L_2}$
• Equivalent conductivity for two bars of equal length and area in series: $K' = \frac{2K_1 K_2}{K_1 + K_2}$

Solution

(i) Temperature of the junction ($T_0$)

In steady state, the heat current through both bars is the same. $H = \frac{K_1 A (T_1 - T_0)}{L} = \frac{K_2 A (T_0 - T_2)}{L}$ The terms $A$ and $L$ cancel out. $K_1 (T_1 - T_0) = K_2 (T_0 - T_2)$ $K_1 T_1 - K_1 T_0 = K_2 T_0 - K_2 T_2$ $K_1 T_1 + K_2 T_2 = T_0 (K_1 + K_2)$ $T_0 = \frac{K_1 T_1 + K_2 T_2}{K_1 + K_2}$ Substitute the values: $T_0 = \frac{(79)(373) + (109)(273)}{79 + 109} = \frac{29467 + 29757}{188} = \frac{59224}{188} = 315 \text{ K}$ Answer for part (i) = $315 \text{ K}$

(ii) Equivalent thermal conductivity ($K'$)

For two bars of equal length and area connected end-to-end (in series), the equivalent thermal conductivity is: $K' = \frac{2 K_1 K_2}{K_1 + K_2}$ $K' = \frac{2 \times 79 \times 109}{79 + 109} = \frac{17222}{188} \approx 91.6 \text{ W m}^{-1} \text{K}^{-1}$ Answer for part (ii) = $91.6 \text{ W m}^{-1} \text{K}^{-1}$

(iii) Heat current through the compound bar ($H$)

We can calculate the heat current using the properties of the compound bar. The total length is $2L = 0.2 \text{ m}$. $H = \frac{K' A (T_1 - T_2)}{2L}$ $H = \frac{(91.6 \text{ W m}^{-1} \text{K}^{-1}) (0.02 \text{ m}^2) (373 \text{ K} - 273 \text{ K})}{2 \times 0.1 \text{ m}}$ $H = \frac{91.6 \times 0.02 \times 100}{0.2} = \frac{183.2}{0.2} = 916 \text{ W}$ Answer for part (iii) = $916.1 \text{ W}$ (using the more precise value for K')

Convection

Convection is the transfer of heat by the actual motion of matter. It only occurs in fluids (liquids and gases).

• Natural Convection: Happens due to differences in density. When you heat a pot of water on a stove, the water at the bottom gets hot, expands, and becomes less dense. This hot water rises, and the cooler, denser water from the top sinks to take its place. This creates a circulating flow called a convection current.
• Forced Convection: Occurs when a fluid is forced to move by a pump or fan. Examples include a forced-air heating system in a house or the cooling system in a car engine.

Radiation

Radiation is the transfer of heat in the form of electromagnetic waves. Unlike conduction and convection, it does not require a medium and can travel through a vacuum.

This is how the Earth receives heat from the Sun across the vast emptiness of space. When you stand near a fire, you feel its warmth primarily through radiation.

The amount of heat a body can absorb or emit by radiation depends on its color and texture.

• Black, dull surfaces are good absorbers and good emitters of radiation.
• White, shiny surfaces are poor absorbers and poor emitters (they are good reflectors).

• We wear light-colored clothes in summer to reflect the sun's heat and dark-colored clothes in winter to absorb it.
• The bottom of a cooking pot is often blackened to absorb maximum heat from the stove.
• A thermos (Dewar flask) has silvered inner walls to reflect heat radiation, keeping hot things hot and cold things cold.

Blackbody Radiation

A blackbody is an idealized object that absorbs all radiation that falls on it. It is also a perfect emitter of radiation. The radiation it emits is called blackbody radiation.

Two important laws describe blackbody radiation:

1. Wien's Displacement Law: This law relates the peak wavelength ($\lambda_m$) of the emitted radiation to the absolute temperature ($T$) of the blackbody. $\lambda_m T = \text{constant}$ The constant is about $2.9 \times 10^{-3} \text{ m K}$. This law explains why a heated piece of iron first glows dull red, then brighter yellow, and finally white-hot as its temperature increases (the peak wavelength shifts to shorter, more energetic wavelengths).

2. Stefan-Boltzmann Law: This law states that the total energy radiated per unit time ($H$) from a blackbody is proportional to the fourth power of its absolute temperature ($T$). $H = A \sigma T^4$

   • $A$ is the surface area.
   • $\sigma$ is the Stefan-Boltzmann constant ($5.67 \times 10^{-8} \text{ W m}^{-2} \text{K}^{-4}$).

For real objects, which are not perfect radiators, we introduce a factor called emissivity (e), which is a number between 0 and 1. $H = A e \sigma T^4$ For a perfect blackbody, $e=1$. A shiny object might have an emissivity close to 0.

If a body at temperature $T$ is in surroundings at temperature $T_s$, it both radiates and absorbs heat. The net rate of heat loss is: $H_{net} = e \sigma A (T^4 - T_s^4)$

Newton's Law of Cooling

For small temperature differences between an object and its surroundings, the rate at which the object loses heat is directly proportional to the temperature difference. This is Newton's Law of Cooling.

The rate of loss of heat ($-\frac{dQ}{dt}$) is proportional to the temperature difference between the body ($T_2$) and its surroundings ($T_1$). $-\frac{dQ}{dt} = k(T_2 - T_1)$ where $k$ is a positive constant that depends on the area and nature of the object's surface.

Since the heat lost is also related to the change in temperature of the object itself ($dQ = ms dT_2$), the law can be written in terms of the rate of temperature change: $-ms \frac{dT_2}{dt} = k(T_2 - T_1)$ This shows that an object cools faster when the temperature difference is large and the rate of cooling slows down as the object's temperature approaches the surrounding temperature.

Given

• Room temperature, $T_1 = 20^\circ\text{C}$
• Case 1: Cools from $94^\circ\text{C}$ to $86^\circ\text{C}$ in $t_1 = 2 \text{ min}$.
• Case 2: Cools from $71^\circ\text{C}$ to $69^\circ\text{C}$.

To Find

The time it takes for Case 2, $t_2$.

Formula

Newton's Law of Cooling (approximate form): $\frac{\text{Change in temperature}}{\text{Time}} = K \times (\text{Average Temperature} - \text{Surrounding Temperature})$

Solution

Case 1:

• Change in temperature = $94 - 86 = 8^\circ\text{C}$.
• Average temperature = $\frac{94 + 86}{2} = 90^\circ\text{C}$.
• Temperature difference with surroundings = $90 - 20 = 70^\circ\text{C}$.

Using the formula: $\frac{8^\circ\text{C}}{2 \text{ min}} = K (70^\circ\text{C}) \quad \implies \quad 4 = 70K \quad \text{(Equation 1)}$

Case 2:

• Change in temperature = $71 - 69 = 2^\circ\text{C}$.
• Average temperature = $\frac{71 + 69}{2} = 70^\circ\text{C}$.
• Temperature difference with surroundings = $70 - 20 = 50^\circ\text{C}$.

Let the time taken be $t_2$. Using the formula: $\frac{2^\circ\text{C}}{t_2} = K (50^\circ\text{C}) \quad \text{(Equation 2)}$

Solve for $t_2$: Divide Equation 1 by Equation 2: $\frac{4}{2/t_2} = \frac{70K}{50K}$ $\frac{4t_2}{2} = \frac{70}{50} = \frac{7}{5}$ $2t_2 = \frac{7}{5}$ $t_2 = \frac{7}{10} = 0.7 \text{ min}$ To convert this to seconds: $0.7 \text{ min} \times 60 \text{ s/min} = 42 \text{ s}$.

Final Answer It will take $0.7$ minutes, or $42$ seconds, to cool from $71^{\circ}\text{C}$ to $69^{\circ}\text{C}$.