THERMAL PROPERTIES OF MATTER-Notes

Heat is the form of energy transferred between two (or more) systems or a system and its surroundings by virtue of temperature difference.

The SI unit of heat energy transferred is expressed in joule (J) while SI unit of temperature is Kelvin (K), and degree Celsius \(\mathrm{(^\circ C)}\) is a commonly used unit of temperature.

Measurement of temperature is the process of assigning a numerical value to the degree of hotness or coldness of a body by bringing it into thermal equilibrium with a thermometer and correlating the temperature with a physical property that varies uniformly with temperature, in accordance with the zeroth law of thermodynamics.

The two familiar temperature scales are the Fahrenheit temperature scale and the Celsius temperature scale. The ice and steam point have values \(\mathrm{32^\circ\ F}\) and \(\mathrm{212^\circ\ F}\), respectively, on the Fahrenheit scale and \(\mathrm{0^\circ\ C}\) and \(\mathrm{100^\circ\ C}\) on the Celsius scale. On the Fahrenheit scale, there are 180 equal intervals between two reference points, and on the Celsius scale, there are 100.

Relationship between Fahrenheit temperature \(\mathrm{(t_f)}\) versus celsius temperature \(\mathrm{(t_c)}\)

Important Aspects

• Temperature is not measured directly; it is inferred from a thermometric property (such as volume, pressure, resistance, or emf).
• Measurement is possible only after thermal equilibrium is established between the body and the thermometer.
• The numerical value depends on the temperature scale used, which is defined through fixed reference points and calibration.
• The gas thermometer provides the basis of the absolute (Kelvin) temperature scale, which is independent of the nature of the substance.

Boyle’s law states that for a fixed mass of a given gas kept at constant temperature, the pressure of the gas is inversely proportional to its volume.

Mathematically,

• Boyle’s law is a law of isothermal transformation.
• It is valid for gases at low pressure and high temperature, where real gases behave nearly ideally.
• The law implies that when volume decreases, molecular collisions with container walls increase, resulting in higher pressure.
• A plot of \(P \text{ versus }\mathbf{1}/V\)is a straight line, while a plot of \(P \text{ versus } V\)is a rectangular hyperbola.

Charles’ law states that for a fixed mass of a given gas kept at constant pressure, the volume of the gas is directly proportional to its absolute temperature.

Mathematically,

• Charles’ law represents an isobaric transformation.
• Temperature must always be measured on the absolute (Kelvin) scale for the law to hold.
• The law explains that increase in temperature increases the average kinetic energy of gas molecules, causing them to occupy a larger volume at constant pressure.
• A graph of \(V\) versus \(T\) (in kelvin) is a straight line passing through the origin.
• Extrapolation of the \(V-T\) graph to zero volume leads to the concept of absolute zero.

Charles’ law is a fundamental gas law that, together with Boyle’s law, forms the basis of the ideal-gas equation.

Ideal-gas equation is the mathematical relation that connects the macroscopic variables of a gas—pressure, volume, and absolute temperature—and is expressed as \[\boxed{\;\mathrm{PV=\mu RT}\;}\] where P is the pressure, V is the volume, \(\mu\) is the number of moles, R is the universal gas constant, and T is the absolute temperature of the gas.

Absolute temperature is the temperature measured on the Kelvin scale, which is independent of the nature of the gas and is directly proportional to the average kinetic energy of the gas molecules. It is defined such that \[T(\mathrm{K})=t(^\circ\mathrm{C})+273.15\] and \(T=0\,\mathrm{K}\) corresponds to absolute zero, the theoretical state where molecular thermal motion is minimum.

• The ideal-gas equation is valid when a real gas behaves ideally, i.e., at low pressure and high temperature.
• In the relation \(PV\propto T\), temperature must always be expressed in the absolute (Kelvin) scale.
• Absolute temperature provides a natural thermodynamic scale, derived from gas behaviour and consistent with kinetic theory.
• The ideal-gas equation establishes the physical significance of absolute temperature by linking it to molecular motion.

Absolute zero is the lowest possible temperature on the thermodynamic (Kelvin) scale, at which an ideal gas would have zero volume on extrapolation and the random thermal motion of its molecules is minimum.

Numerically,

• Absolute zero is the zero point of the Kelvin scale.
• It is obtained by extrapolating the volume–temperature graph (Charles’ law) or pressure–temperature graph (gas thermometer) of an ideal gas.
• At absolute zero, molecular motion does not completely cease, but thermal kinetic energy becomes minimum, consistent with physical laws.
• Absolute zero is a theoretical limit and cannot be attained in practice.
• The concept gives physical meaning to absolute temperature, linking it to molecular motion.

Thermal expansion is the phenomenon by which the dimensions of a substance increase on heating due to an increase in the average separation between its constituent particles, caused by enhanced molecular vibration.

• Thermal expansion occurs because interatomic distances increase with temperature, not because particles themselves expand.
• It is observed in solids, liquids, and gases, with magnitude generally: \[\text{Gases}\gt\text{Liquids}\gt\text{Solids}\]
• In solids, thermal expansion is classified as:
  
  
  • Linear expansion (change in length) \[\frac{\Delta l}{l}=\alpha_l\Delta T\]
  • Areal expansion (change in area) \[\frac{\Delta A}{A}=2\alpha_l\Delta T\]
  • Volumetric expansion (change in volume) \[\frac{\Delta V}{V}=3\alpha_l\Delta T\]
• Thermal expansion has important practical applications such as expansion joints in bridges, fitting of metal rims, and bimetallic strips.

The coefficient of linear expansion of a solid is defined as the fractional increase in length per unit original length per unit rise in temperature, provided the expansion is small and the material remains within its elastic limit.

Mathematically,

The coefficient of volume expansion, also called volume expansivity, of a substance is defined as the fractional increase in volume per unit original volume per unit rise in temperature, when the substance is heated uniformly and the temperature change is small.

Mathematically,

Imagine a cube of length, \(l\), that expands equally in all directions, when its temperature increases by \(\Delta T\). We have

\(\Delta l\lt l\text{ and }{\Delta l}^2 \text{ and } {\Delta l}^3\lt\lt l\) therfore, neglecting higher order of \(\Delta l\)

Thermal stress is the stress developed in a body when its natural thermal expansion or contraction is prevented due to external constraints, while its temperature is changed.

When a solid is heated or cooled:

• it tends to expand or contract freely,
• but if its ends are rigidly fixed or motion is restricted, strain cannot occur,
• as a result, internal restoring forces develop, producing thermal stress.

Mathematical expression (for a rod with ends rigidly fixed):

where
\(Y\) = Young’s modulus of the material,
\(\alpha\) = coefficient of linear expansion,
\(\Delta T\) = change in temperature.

Important Points

• Thermal stress arises only when expansion or contraction is constrained.
• If a body is allowed to expand freely, no thermal stress is produced.
• Thermal stress is directly proportional to:
  • Young’s modulus,
  • coefficient of linear expansion,
  • temperature change
• It explains practical phenomena such as:
  • cracking of rails and bridges,
  • breaking of glass on sudden heating or cooling.

Specific heat capacity of a substance is defined as the amount of heat required to raise the temperature of unit mass of the substance by one unit, without any change in its physical state.

Mathematically,

or equivalently,

where
\(Q\) = heat supplied to the substance,
\(m\) = mass of the substance,
\(\Delta T\) = rise in temperature.

If the amount of substance is specified in terms of moles \(\mu\), instead of mass \(m\) in kg, we can define heat capacity per mole of the substance by \[\Delta Q=\mu C \Delta T\]

where,
\(C\) is known as molar specific heat capacity

Important Point

• Specific heat capacity is a characteristic property of the material.
• Its SI unit is \(J\ kg^{-1}\ K^{-1}\).
• A substance with high specific heat capacity requires more heat to produce the same temperature rise.
• Water has a high specific heat capacity, which plays an important role in climate moderation and biological processes.
• The formula \(Q=mC\Delta T\) is valid only when no phase change occurs.

The molar specific heat capacity at constant pressure, denoted by \(C_p\), is defined as the amount of heat required to raise the temperature of one mole of a substance by one unit, when the pressure of the system is kept constant.

Mathematically,

or, for a finite temperature change,

where,
\(Q\) = heat supplied,
\(n\) = number of moles,
\(\Delta T\) = rise in temperature.

The molar specific heat capacity at constant volume, denoted by \(C_v\), is defined as the amount of heat required to raise the temperature of one mole of a substance by one unit, when the volume of the system is kept constant.

Mathematically,

or, for a finite temperature change,

where,
\(Q\) = heat supplied,
\(n\) = number of moles,
\(\Delta T\) = rise in temperature.

For an ideal gas

where \(R\) is the universal gas constant.

Calorimetry is the branch of thermal physics concerned with the measurement of heat exchanged by a body or a system during a physical or chemical process, based on the principle of conservation of energy.

Principle of Calorimetry

In an isolated system, when bodies at different temperatures are brought into thermal contact, the total heat lost by the hotter body is equal to the total heat gained by the colder body, until thermal equilibrium is established.

Mathematically,

A calorimeter is a well-insulated device used to measure the amount of heat exchanged between bodies or systems during a thermal process, such as heating, cooling, or change of state.

Important Points

• A calorimeter is designed to minimise heat exchange with the surroundings, so that all heat transfer occurs only between the bodies placed inside it.
• It is usually made of a thin metallic container (often copper or aluminium) with a known heat capacity.
• The calorimeter may contain:
  • an inner vessel,
  • a stirrer for uniform temperature,
  • a thermometer for accurate measurement.
• In calorimetric calculations, the heat absorbed by the calorimeter itself is taken into account using its water equivalent.

Physical significance:

The calorimeter provides an experimental means to apply the principle of calorimetry, enabling accurate determination of specific heat, latent heat, and thermal properties of substances.

The melting point of a solid is the definite temperature at which the solid changes into the liquid state at a given pressure, with solid and liquid phases coexisting in thermal equilibrium.

• At the melting point, the temperature of the substance remains constant even though heat is continuously supplied.
• The heat supplied at this temperature is used as latent heat of fusion to overcome intermolecular forces, not to raise temperature.
• The melting point of a substance is a characteristic physical property and depends on pressure.
• For most solids, the melting point increases with increase in pressure, whereas for ice it decreases with pressure.
• At the melting point, solid and liquid have the same temperature but different internal energies.

Regelation is the phenomenon in which ice melts under the application of pressure and refreezes when the pressure is removed, provided the temperature is at or below the melting point of ice.

• Regelation occurs because the melting point of ice decreases with increase in pressure.
• When pressure is applied on ice, it melts locally; when the pressure is removed, the water formed freezes again, as the melting point rises back.
• This phenomenon is explained using the pressure dependence of melting point and is a special property of substances like ice whose solid phase is less dense than the liquid phase.
• Regelation explains practical observations such as:
  • a wire passing through a block of ice without cutting it into two parts,
  • motion of glaciers over rocks.

Vaporisation is the process by which a substance changes from the liquid state to the gaseous state at a given temperature and pressure, by absorbing heat energy.

• Vaporisation can occur in two ways:
  • Evaporation:
    occurs at all temperatures from the free surface of a liquid.
  • Boiling:
    occurs at a fixed temperature (boiling point) throughout the liquid.
• During boiling, the temperature of the liquid remains constant even though heat is continuously supplied.
• The heat absorbed during vaporisation is used as latent heat of vaporisation to overcome intermolecular attractions.
• Vaporisation is accompanied by a large increase in volume.
• At the boiling point, the vapour pressure of the liquid equals the atmospheric pressure.

The boiling point of a liquid is the definite temperature at which the vapour pressure of the liquid becomes equal to the external (atmospheric) pressure, causing the liquid to change into vapour throughout its entire mass.

• At the boiling point, the temperature of the liquid remains constant even though heat is continuously supplied.
• Boiling occurs throughout the liquid, not just at the surface.
• The heat supplied at the boiling point is used as latent heat of vaporisation to overcome intermolecular forces.
• The boiling point of a liquid depends on external pressure:
  • it increases with increase in pressure,
  • it decreases with decrease in pressure.
• Boiling differs from evaporation, which can occur at all temperatures and only from the surface.

Sublimation is the process in which a substance changes directly from the solid state to the gaseous state without passing through the liquid state, at a given temperature and pressure.

• During sublimation, the solid absorbs heat in the form of latent heat of sublimation.
• The temperature of the substance remains constant during the phase change.
• Sublimation occurs in substances for which the vapour pressure of the solid becomes equal to atmospheric pressure at temperatures below their melting point.
• Common examples include camphor, naphthalene, ammonium chloride, and dry ice.
• Sublimation involves a large increase in volume and requires energy to overcome intermolecular forces

Latent heat is the amount of heat absorbed or released by a substance during a change of state at constant temperature, without any change in its temperature.

• Latent heat is associated with phase change, not with temperature change.
• The heat supplied or released is used in changing the intermolecular separation and potential energy, not in increasing kinetic energy.
• Latent heat is of two main types:
  • Latent heat of fusion:
    heat required to change unit mass of a solid into liquid at its melting point.
  • Latent heat of vaporisation:
    heat required to change unit mass of a liquid into vapour at its boiling point.
• The relation used in numerical problems is: \[Q=mL\] where \(L\) is the specific latent heat.

Conduction is the mode of heat transfer in which heat flows from the hotter region to the colder region of a body or between bodies in contact, without any macroscopic movement of matter.

The rate of flow of heat (or heat current) H is proportional to the temperature difference \((T_C – T_D)\) and the area of cross-section A and is inversely proportional to the length L :

Mathematically,

The constant of proportionality\(K\) is called the thermal conductivity of the material.

The greater the value of K for a material, the more rapidly will it conduct heat.

The SI unit of \(K\) is \(\mathrm{J\ s^{–1}\ m^{–1}\ K^{–1}}\) or \(\mathrm{W\ m^{–1}\ K^{–1}}\).

• Conduction occurs due to energy transfer between neighbouring atoms or molecules through collisions and, in metals, also due to free electrons.
• It is the dominant mode of heat transfer in solids.
• Heat always flows along the temperature gradient, from higher to lower temperature.
• The rate of heat conduction depends on:
  • nature of the material,
  • temperature difference,
  • cross-sectional area,
  • length of the conductor.
• Good conductors (e.g., metals) allow heat to flow easily, while insulators (e.g., wood, glass) oppose heat flow.

Convection is the mode of heat transfer in which heat is carried from one place to another by the actual bulk motion of a fluid (liquid or gas), from a region of higher temperature to a region of lower temperature.

• Convection occurs only in fluids (liquids and gases), not in solids.
• It is caused by density differences produced due to temperature variation in the fluid.
• Heated fluid becomes less dense and rises, while cooler, denser fluid sinks, setting up convection currents.
• Convection plays an important role in:
  • heating of liquids and gases,
  • land and sea breezes,
  • atmospheric circulation.
• There are two types:
  • Natural convection:
    due to buoyancy forces,
  
  
  • Forced convection:
    due to external agents like fans or pumps.

Radiation is the mode of heat transfer in which thermal energy is transmitted in the form of electromagnetic waves, without requiring any material medium.

• Radiation can occur through vacuum as well as through transparent media.
• All bodies at a temperature above absolute zero emit thermal radiation.
• The amount of heat radiated depends on:
  • temperature of the body,
  • nature of its surface (black or polished),
  • surface area.
• Radiation does not involve motion of matter, unlike conduction and convection.
• It is the only mode of heat transfer possible in outer space (e.g., heat received from the Sun).

Blackbody radiation refers to the thermal radiation emitted by an ideal blackbody, which is a body that absorbs all incident radiation of all wavelengths and re-emits energy solely as a function of its temperature, independent of the nature or shape of the body.

Wien’s Displacement Law

Wien’s displacement law states that for a blackbody, the wavelength corresponding to the maximum intensity of emitted radiation is inversely proportional to its absolute temperature.

Mathematically,

where
\(\lambda_{\mathrm{max}}\) = wavelength of maximum emission,
\(T\) = absolute temperature of the blackbody,
The value of the constant (Wien’s constant) is \(\mathrm{2.9 × 10^{–3}\ m\ K}\).

• A blackbody is an idealized concept; it is a perfect absorber and perfect emitter of radiation.
• The radiation emitted by a blackbody depends only on its absolute temperature, not on the material.
• As the temperature of a blackbody increases:
  • the total energy radiated increases,
  • the wavelength corresponding to maximum intensity shifts towards shorter wavelengths.
• Blackbody radiation provides the theoretical basis for understanding thermal radiation and is foundational to later developments in modern physics.
• Practical approximations of a blackbody include a small aperture in a hollow cavity.

Stefan–Boltzmann law states that the total radiant energy emitted per unit area per unit time by a blackbody is directly proportional to the fourth power of its absolute temperature.

Mathematically

where
\(E\) = radiant energy emitted per unit area per unit time,
\(T\) = absolute temperature of the blackbody,
\(\sigma\) = Stefan–Boltzmann constant. Its value in SI units is \(\mathrm{5.67 × 10^{–8}\ W\ m^{–2}\ K^{–4}}\).
\(A\) = Area

A body at temperature \(T\), with surroundings at temperatures \(T_s\), emits, as well as, receives energy. For a perfect radiator, the net rate of loss of radiant energy is \[H=A\sigma (T^4-T_s^4)\]

A blacksmith fixes iron ring on the rim of the wooden wheel of a horse cart. The diameter of the rim and the iron ring are 5.243 m and 5.231 m, respectively at 27 °C. To what temperature should the ring be heated so as to fit the rim of the wheel?

Solution

Given: Initial temperature of the iron ring and wooden rim is \(T_{1} = 27^{\circ}\text{C}\).
Diameter of wooden rim \(D_{\text{rim}} = 5.243 \text{ m}\).
Diameter of iron ring \(D_{\text{ring}} = 5.231 \text{ m}\).
Coefficient of linear expansion of iron is taken as \(\alpha_{l} = 1.2 \times 10^{-5} \text{ }^{\circ}\text{C}^{-1}\).

To fit the ring on the rim, the diameter of the iron ring must increase from \(5.231 \text{ m}\) to \(5.243 \text{ m}\).
So, the increase in diameter is
\[ \Delta l = D_{\text{rim}} - D_{\text{ring}} = 5.243 - 5.231 = 0.012 \text{ m} \] The fractional change in length is given by the linear expansion formula
\[ \frac{\Delta l}{l} = \alpha_{l}\left(T_{2} - T_{1}\right) \]

Rearranging for the rise in temperature,
\[ T_{2} - T_{1} = \frac{\Delta l}{l \, \alpha_{l}} \] Substituting the values,
\[ T_{2} - T_{1} = \frac{5.243 - 5.231}{5.231 \times 1.2 \times 10^{-5}} \] \[ T_{2} - T_{1} = \frac{0.012}{5.231 \times 1.2 \times 10^{-5}} \approx 191^{\circ}\text{C} \]

Therefore,
\[ T_{2} = T_{1} + 191^{\circ}\text{C} = 27^{\circ}\text{C} + 191^{\circ}\text{C} = 218^{\circ}\text{C} \] Hence, the iron ring should be heated to about \(218^{\circ}\text{C}\) so that it just fits on the wooden rim.

A sphere of 0.047 kg aluminium is placed for sufficient time in a vessel containing boiling water, so that the sphere is at 100 °C. It is then immediately transfered to 0.14 kg copper calorimeter containing 0.25 kg water at 20 °C. The temperature of water rises and attains a steady state at 23 °C. Calculate the specific heat capacity of aluminium.

Solution

Given: Mass of aluminium sphere \(m_{\text{Al}} = 0.047 \, \text{kg}\),
initial temperature \(T_{\text{Al}} = 100^\circ \text{C}\).
Mass of copper calorimeter \(m_{\text{Cu}} = 0.14 \, \text{kg}\),
mass of water \(m_{\text{w}} = 0.25 \, \text{kg}\),
initial temperature \(T_{\text{w}} = T_{\text{Cu}} = 20^\circ \text{C}\).
Final equilibrium temperature \(T_{\text{f}} = 23^\circ \text{C}\).
Specific heat capacity of water \(s_{\text{w}} = 4.18 \times 10^3 \, \text{J kg}^{-1} \text{K}^{-1}\),
specific heat capacity of copper \(s_{\text{Cu}} = 0.386 \times 10^3 \, \text{J kg}^{-1} \text{K}^{-1}\).

Heat lost by aluminium sphere equals heat gained by water and copper calorimeter.
Heat lost by aluminium sphere,
\[ Q_{\text{Al}} = m_{\text{Al}} \, s_{\text{Al}} \, (T_{\text{Al}} - T_{\text{f}}) = 0.047 \times s_{\text{Al}} \times (100 - 23) = 0.047 \times s_{\text{Al}} \times 77 \] Heat gained by water and calorimeter,
\[ Q_{\text{w} + \text{Cu}} = m_{\text{w}} \, s_{\text{w}} \, (T_{\text{f}} - T_{\text{w}}) + m_{\text{Cu}} \, s_{\text{Cu}} \, (T_{\text{f}} - T_{\text{Cu}}) \] \[ = (23 - 20) \left[ 0.25 \times 4.18 \times 10^3 + 0.14 \times 0.386 \times 10^3 \right] \]

Calculate the heat gained,
\[ \begin{aligned} 0.25 \times 4.18 \times 10^3 &= 1045 \\ 0.14 \times 0.386 \times 10^3 &= 54.04 \\ \text{Total } &= 1045 + 54.04 = 1099.04 \, \text{J kg}^{-1} \text{K}^{-1} \\ Q_{\text{w} + \text{Cu}} &= 3 \times 1099.04 = 3297.12 \, \text{J} \end{aligned} \] By principle of calorimetry, \(Q_{\text{Al}} = Q_{\text{w} + \text{Cu}}\).

Thus,
\[ 0.047 \times s_{\text{Al}} \times 77 = 3297.12 \] \[ \begin{aligned} s_{\text{Al}} &= \frac{3297.12}{0.047 \times 77} \\ &= \frac{3297.12}{3.619} \\ &\approx 911 \, \text{J kg}^{-1} \text{K}^{-1} \end{aligned} \] The specific heat capacity of aluminium is approximately \(911 \, \text{J kg}^{-1} \text{K}^{-1}\).

When 0.15 kg of ice at 0 °C is mixed with 0.30 kg of water at 50 °C in a container, the resulting temperature is 6.7 °C. Calculate the heat of fusion of ice. (swater = \(\mathrm{4186\ J\ kg^{–1}\ K^{–1}}\))

Solution

Given: Mass of ice \(m_{\text{ice}} = 0.15 \, \text{kg}\) at \(T_{\text{ice}} = 0^\circ \text{C}\),
mass of water \(m_{\text{w}} = 0.30 \, \text{kg}\) at \(T_{\text{w}} = 50^\circ \text{C}\).
Final equilibrium temperature \(T_{\text{f}} = 6.7^\circ \text{C}\).
Specific heat capacity of water \(s_{\text{w}} = 4186 \, \text{J kg}^{-1} \text{K}^{-1}\).

Heat lost by hot water equals heat gained by ice to melt and then raise temperature of resulting water to final temperature.
Heat lost by hot water,
\[ H_{\text{lost}} = m_{\text{w}} \, s_{\text{w}} \, (T_{\text{w}} - T_{\text{f}}) = 0.30 \times 4186 \times (50 - 6.7) \] \[ = 0.30 \times 4186 \times 43.3 = 54376.14 \, \text{J} \] Heat gained by ice: heat to melt + heat to raise temperature of melt water to \(T_{\text{f}}\).

Latent heat of fusion required to melt ice,
\[ Q_{\text{melt}} = m_{\text{ice}} \, L_{\text{f}} = 0.15 \, L_{\text{f}} \] Heat to raise temperature of \(0.15 \, \text{kg}\) water from \(0^\circ \text{C}\) to \(6.7^\circ \text{C}\),
\[ Q_{\text{raise}} = m_{\text{ice}} \, s_{\text{w}} \, (T_{\text{f}} - 0) = 0.15 \times 4186 \times 6.7 = 4206.93 \, \text{J} \] Total heat gained by ice system,
\[ Q_{\text{gain}} = 0.15 \, L_{\text{f}} + 4206.93 \]

By principle of calorimetry, heat lost = heat gained,
\[ \begin{aligned} 54376.14 &= 0.15 \, L_{\text{f}} + 4206.93 \\ 0.15 \, L_{\text{f}} &= 54376.14 - 4206.93 \\ &= 50169.21 \\ L_{\text{f}} &= \frac{50169.21}{0.15} \\ &= 334461.4 \, \text{J kg}^{-1} \\ &\approx 3.34 \times 10^5 \, \text{J kg}^{-1} \end{aligned} \] The heat of fusion of ice is \(3.34 \times 10^5 \, \text{J kg}^{-1}\).

Calculate the heat required to convert 3 kg of ice at –12 °C kept in a calorimeter to steam at 100 °C at atmospheric pressure. Given specific heat capacity of ice = \(\mathrm{2100\ J\ kg^{–1}\ K^{–1}}\), specific heat capacity of water = \(\mathrm{4186\ J\ kg^{– 1}\ K^{–1}}\), latent heat of fusion of ice = \(\mathrm{3.35 × 10^5\ J\ kg^{–1}}\) and latent heat of steam = \(\mathrm{2.256 ×10^6\ J\ kg^{–1}}\).

Solution

Given: Mass of ice \(m = 3 \, \text{kg}\),
initial temperature \(T_{\text{i}} = -12^\circ \text{C}\),
final state steam at \(100^\circ \text{C}\).
Specific heat capacity of ice \(s_{\text{ice}} = 2100 \, \text{J kg}^{-1} \text{K}^{-1}\),
specific heat capacity of water \(s_{\text{w}} = 4186 \, \text{J kg}^{-1} \text{K}^{-1}\).
Latent heat of fusion of ice \(L_{\text{f}} = 3.35 \times 10^5 \, \text{J kg}^{-1}\),
latent heat of vaporization \(L_{\text{v}} = 2.256 \times 10^6 \, \text{J kg}^{-1}\).

Total heat required \(Q\) to convert ice at \(-12^\circ \text{C}\) to steam at \(100^\circ \text{C}\) is sum of heat for four processes.
\(Q_1\): Raise temperature of ice from \(-12^\circ \text{C}\) to \(0^\circ \text{C}\),
\[ \begin{aligned} Q_1 &= m \, s_{\text{ice}} \, \Delta T \\ &= 3 \times 2100 \times 12 \\ &= 75600 \, \text{J} \end{aligned} \] \(Q_2\): Latent heat to melt ice at \(0^\circ \text{C}\) to water at \(0^\circ \text{C}\),
\[ \begin{aligned} Q_2 &= m \, L_{\text{f}} \\ &= 3 \times 3.35 \times 10^5 \\ &= 1.005 \times 10^6 \, \text{J} \end{aligned} \]

\(Q_3\): Raise temperature of water from \(0^\circ \text{C}\) to \(100^\circ \text{C}\),
\[ \begin{aligned} Q_3 &= m \, s_{\text{w}} \, \Delta T \\ &= 3 \times 4186 \times 100 \\ &= 1.2558 \times 10^6 \, \text{J} \end{aligned} \] \(Q_4\): Latent heat to convert water at \(100^\circ \text{C}\) to steam at \(100^\circ \text{C}\),
\[ \begin{aligned} Q_4 &= m \, L_{\text{v}} \\ &= 3 \times 2.256 \times 10^6 \\ &= 6.768 \times 10^6 \, \text{J} \end{aligned} \]

Total heat required,
\[ \begin{aligned} Q &= Q_1 + Q_2 + Q_3 + Q_4 \\ &= 75600 + 1005000 + 1255800 + 6768000 \\ &= 75600 + 1005000 = 1080600 \\ &\quad + 1255800 = 2336400 \\ &\quad + 6768000 = 9104400 \, \text{J} \\ &= 9.10 \times 10^6 \, \text{J} \end{aligned} \] The total heat required is \(9.10 \times 10^6 \, \text{J}\).

What is the temperature of the steel-copper junction in the steady state of the system shown in Fig. 10.15. Length of the steel rod = 15.0 cm, length of the copper rod = 10.0 cm, temperature of the furnace = 300 °C, temperature of the other end = 0 °C. The area of cross section of the steel rod is twice that of the copper rod. (Thermal conductivity of steel = \(\mathrm{50.2\ J\ s^{–1}\ m^{–1}\ K^{–1}}\); and of copper = \(\mathrm{385\ J\ s^{–1}\ m^{–1}\ K^{–1}}\)).

Solution

Given: Length of steel rod \(L_{\text{s}} = 15.0 \, \text{cm} = 0.15 \, \text{m}\),
length of copper rod \(L_{\text{c}} = 10.0 \, \text{cm} = 0.10 \, \text{m}\).
Furnace temperature \(T_{\text{H}} = 300^\circ \text{C}\),
cold end temperature \(T_{\text{C}} = 0^\circ \text{C}\).
Cross-sectional area of steel rod \(A_{\text{s}} = 2A\),
copper rod \(A_{\text{c}} = A\).
Thermal conductivity of steel \(K_{\text{s}} = 50.2 \, \text{J s}^{-1} \text{m}^{-1} \text{K}^{-1}\),
Thermal conductivity of copper \(K_{\text{c}} = 385 \, \text{J s}^{-1} \text{m}^{-1} \text{K}^{-1}\).

In steady state, rate of heat flow through steel rod equals rate of heat flow through copper rod.
Let junction temperature be \(T^\circ \text{C}\).
Heat current through steel,
\[ H_{\text{s}} = \frac{K_{\text{s}} A_{\text{s}} (300 - T)}{L_{\text{s}}} \] Heat current through copper,
\[ H_{\text{c}} = \frac{K_{\text{c}} A_{\text{c}} (T - 0)}{L_{\text{c}}} \] At steady state, \(H_{\text{s}} = H_{\text{c}}\).

Thus,
\[ \begin{aligned} \frac{K_{\text{s}} A_{\text{s}} (300 - T)}{L_{\text{s}}} &= \frac{K_{\text{c}} A_{\text{c}} T}{L_{\text{c}}} \\ \frac{50.2 \times 2A \times (300 - T)}{0.15} &= \frac{385 \times A \times T}{0.10} \\ \frac{50.2 \times 2 (300 - T)}{0.15} &= \frac{385 T}{0.10} \\ \frac{100.4 (300 - T)}{0.15} &= 3850 T \\ 100.4 (300 - T) &= 3850 T \times 0.15 \\ 100.4 (300 - T) &= 577.5 T \end{aligned} \]

Solving for \(T\),
\[ \begin{aligned} 100.4 \times 300 - 100.4 T &= 577.5 T \\ 30120 &= 577.5 T + 100.4 T \\ 30120 &= 677.9 T \\ T &= \frac{30120}{677.9} \\ &\approx 44.4^\circ \text{C} \end{aligned} \] The temperature of the steel-copper junction is \(44.4^\circ \text{C}\).

An iron bar \(\mathrm{L1 = 0.1\ m,\ A1 =0.02\ m^2,\ K1 = 79\ W\ m^{–1}\ K^{–1}}\) and a brass bar \(\mathrm{(L2 = 0.1\ m,\ A2 = 0.02\ m^2,\ K_2 = 109\ W\ m^{–1}\ K^{–1})}\) are soldered end to end as shown in Fig. 10.16. The free ends of the iron bar and brass bar are maintained at 373 K and 273 K respectively. Obtain expressions for and hence compute (i) the temperature of the junction of the two bars, (ii) the equivalent thermal conductivity of the compound bar, and (iii) the current through the compound bar.

Solution

Given for iron bar: length \(L_1 = 0.1 \, \text{m}\),
area \(A_1 = 0.02 \, \text{m}^2\),
thermal conductivity \(K_1 = 79 \, \text{W m}^{-1} \text{K}^{-1}\).
For brass bar: length \(L_2 = 0.1 \, \text{m}\),
area \(A_2 = 0.02 \, \text{m}^2\),
thermal conductivity \(K_2 = 109 \, \text{W m}^{-1} \text{K}^{-1}\).
Free end of iron bar at \(T_1 = 373 \, \text{K}\),
free end of brass bar at \(T_2 = 273 \, \text{K}\).

(i) Temperature of the junction \(T_0\): In steady state, heat current through iron equals heat current through brass.
\[ \frac{K_1 A_1 (373 - T_0)}{L_1} = \frac{K_2 A_2 (T_0 - 273)}{L_2} \] Since \(L_1 = L_2\) and \(A_1 = A_2\), this simplifies to,
\[ K_1 (373 - T_0) = K_2 (T_0 - 273) \] \[ 79 (373 - T_0) = 109 (T_0 - 273) \]

Solving for \(T_0\),
\[ \begin{aligned} 79 \times 373 - 79 T_0 &= 109 T_0 - 109 \times 273 \\ 79 \times 373 + 109 \times 273 &= 109 T_0 + 79 T_0 \\ 29467 + 29757 &= 188 T_0 \\ 59224 &= 188 T_0 \\ T_0 &= \frac{59224}{188} \\ &\approx 315 \, \text{K} \end{aligned} \]

(ii) Equivalent thermal conductivity \(K_{\text{eq}}\) of the compound bar (total length \(2L = 0.2 \, \text{m}\), total area \(A = 0.02 \, \text{m}^2\)):
For two bars in series with equal length and area,
\[ K_{\text{eq}} = \frac{2 K_1 K_2}{K_1 + K_2} \] \[ \begin{aligned} K_{\text{eq}} &= \frac{2 \times 79 \times 109}{79 + 109} \\ &= \frac{17222}{188} \\ &\approx 91.6 \, \text{W m}^{-1} \text{K}^{-1} \end{aligned} \]

(iii) Heat current \(H\) through the compound bar,
\[ H = \frac{K_{\text{eq}} A (T_1 - T_2)}{2L} \] \[ \begin{aligned} H &= \frac{91.6 \times 0.02 \times (373 - 273)}{0.2} \\ &= \frac{91.6 \times 0.02 \times 100}{0.2} \\ &= \frac{183.2}{0.2} \\ &= 916 \, \text{W} \end{aligned} \]

A pan filled with hot food cools from 94 °C to 86 °C in 2 minutes when the room temperature is at 20 °C. How long will it take to cool from 71 °C to 69 °C?

Solution

Given: A pan cools from \(94^\circ \text{C}\) to \(86^\circ \text{C}\) in \(\Delta t_1 = 2 \, \text{min} = 120 \, \text{s}\),
room temperature \(T_0 = 20^\circ \text{C}\).
Find time to cool from \(71^\circ \text{C}\) to \(69^\circ \text{C}\).

Newton's law of cooling states that the rate of cooling is proportional to temperature difference with surroundings: \(\frac{dT}{dt} = -k (T - T_0)\).
For small temperature changes, approximate rate of heat loss \(\frac{dQ}{dt} \propto \Delta T\), where \(\Delta T\) is average temperature excess.
First interval: temperature drop \(\Delta \theta_1 = 94 - 86 = 8^\circ \text{C}\),
average temperature \((94 + 86)/2 = 90^\circ \text{C}\),
average excess \(\Delta T_1 = 90 - 20 = 70^\circ \text{C}\).

Cooling rate for first interval,
\[ \frac{\Delta \theta_1}{\Delta t_1} = k' \, \Delta T_1 \] \[ \frac{8}{120} = k' \times 70 \] \[ k' = \frac{8}{120 \times 70} \] Second interval: temperature drop \(\Delta \theta_2 = 71 - 69 = 2^\circ \text{C}\), average temperature \((71 + 69)/2 = 70^\circ \text{C}\), average excess \(\Delta T_2 = 70 - 20 = 50^\circ \text{C}\).

Cooling rate for second interval,
\[ \frac{\Delta \theta_2}{t} = k' \, \Delta T_2 \] \[ \frac{2}{t} = \frac{8}{120 \times 70} \times 50 \] \[ \frac{2}{t} = \frac{8 \times 50}{120 \times 70} \] \[ \begin{aligned} t &= \frac{2 \times 120 \times 70}{8 \times 50} \\ &= \frac{120 \times 70 \times 2}{8 \times 50} \\ &= \frac{16800}{400} \\ &= 42 \, \text{s} \end{aligned} \] The pan will take \(42 \, \text{s}\) to cool from \(71^\circ \text{C}\) to \(69^\circ \text{C}\).