JEE Physics Thermal Properties Of Matter Notes Set 04

Heat, Temperature and the Mechanical Equivalent

Heat is energy in transit between bodies (or parts of a body) because of a temperature difference; once absorbed it becomes internal energy, so phrases such as "heat contained in a body" have no meaning. The SI unit is the joule and the common unit is the calorie, with 1 cal = 4.18 J (the heat that warms 1 g of water from 14.5 °C to 15.5 °C).
Work and heat are interchangeable: if work W produces the same temperature rise as heat H, then \( W=JH \), where J is the mechanical equivalent of heat (about 4.18 J/cal). Stirring a liquid or shaking an insulated vessel raises its temperature without any heat flow, which is why a body's specific heat cannot be defined as simply Q/(mΔT) in such cases.
Worked example: a 10 kg mass falls 10 m. Its loss of potential energy is \( mgh=1000\ \text{J}\approx239\ \text{cal} \).

Specific Heat, Heat Capacity and Water Equivalent

The heat needed to change the temperature of a body is \( Q=ms\,\Delta T \), or \( Q=m\int s\,dT \) when s varies with temperature. Specific heat s is measured in J/kg·K or cal/g°C; for water s = 4200 J/kg·K = 1 cal/g°C, and ice and steam each have about half this value.
Heat capacity is \( C=ms \) (J/K or cal/°C). Water equivalent \( W=ms/s_{water} \) is the mass of water that would need the same heat for the same temperature change; in CGS units it equals ms numerically, and it is added to the water mass whenever a calorimeter takes part in the exchange.
Special cases: during melting or boiling \( \Delta T=0 \) so the effective specific heat is infinite; when a liquid is shaken in an insulated flask its temperature rises with Q = 0, so the effective specific heat is zero.

Latent Heat and Heating Curves

A change of state at constant temperature needs \( Q=mL \). For water the latent heat of fusion is 80 cal/g (3.36 × 10⁵ J/kg) and the latent heat of vaporization is 540 cal/g (2.26 × 10⁶ J/kg).
On a temperature-versus-heat graph for a fixed mass, the sloping segments represent single phases and the slope is \( 1/(ms) \): a steeper segment means a smaller specific heat. The flat segments are phase changes and their lengths are proportional to the latent heats.
Worked example: 1 kg of steam at 200 °C cooled to ice at −20 °C releases \( 1\times0.5\times100+540+100+80+0.5\times20=780\ \text{kcal} \).

Calorimetry and the Principle of Mixtures

Calorimetry is the measurement of heat. With no loss to the surroundings, heat lost by the hot bodies equals heat gained by the cold ones: \( m_1s_1(T_1-T_m)=m_2s_2(T_m-T_2) \), where \( T_m \) is the final temperature.
When phase changes are possible, always check by stages instead of assuming a final temperature: compare the heat the hot body can release (while cooling to the transition temperature) with the heat needed for the cold body to reach and complete its transition. If ice survives, the mixture is at 0 °C; if steam survives, it is at 100 °C.
Worked example: 300 g of water at 25 °C and 100 g of ice at 0 °C. The water can give 7500 cal but melting all the ice needs 8000 cal, so the mixture ends at 0 °C with ice left over.
Work-heat problems: a falling body that loses a fraction f of its energy gives \( ms\,\Delta T=f\,mgh \); a fired bullet that stops delivers its kinetic energy \( \tfrac12mv^2 \) as heat (use J = 4.2 J/cal when working in calories).

Thermal Expansion of Solids and Liquids

For a small temperature change the fractional changes are linear \( \Delta\ell/\ell=\alpha\Delta T \), area \( \Delta A/A=2\alpha\Delta T \) and volume \( \Delta V/V=\gamma\Delta T \) with \( \gamma=3\alpha \) for isotropic solids. So the percentage changes of radius, area and volume of a heated ball are in the ratio 1 : 2 : 3.
Every dimension of a body grows by the same factor, including holes and cavities and distances between marks on a plate; a hole expands exactly as if it were filled with the same material, and a hollow vessel's capacity grows by \( 3\alpha \).
Liquids expand with their own γ, and a liquid in a vessel shows an apparent expansion \( \gamma_{app}=\gamma_L-3\alpha_{vessel} \). The density of a substance falls as \( \rho'=\rho/(1+\gamma\Delta T) \). Water is anomalous between 0 °C and 4 °C, where it contracts on heating and is densest at 4 °C.
For an ideal gas at constant pressure \( \gamma=1/T \).

Thermal Stress, Bimetallic Strips and Clocks

A rod clamped so it cannot change length develops stress \( Y\alpha\Delta T \) and force \( YA\alpha\Delta T \). For rods in series between rigid walls, the common force satisfies \( \dfrac{F}{A}\left(\dfrac{L_1}{Y_1}+\dfrac{L_2}{Y_2}\right)=(L_1\alpha_1+L_2\alpha_2)\Delta T \).
Two rods joined end to end behave like one rod with \( \alpha_{eq}=\dfrac{\ell_1\alpha_1+\ell_2\alpha_2}{\ell_1+\ell_2} \).
A bimetallic strip bends towards the metal with the smaller α, with radius of curvature \( R=\dfrac{d}{(\alpha_1-\alpha_2)\Delta T} \), where d is the distance between the central planes of the two strips.
A pendulum clock has period \( T=2\pi\sqrt{L/g} \); heating lengthens L so the period grows by \( \tfrac12\alpha\Delta T \) and the clock loses \( \tfrac12\alpha\Delta T\times86400 \) seconds per day. A metal scale used above its calibration temperature reads low; below it, high.
A moment of inertia changes as \( I\propto L^2 \), so \( \Delta I=2\alpha I\Delta T \).

Temperature Scales and Thermometers

Any linear scale is fixed by two reference points: for a scale X with ice point \( X_0 \) and steam point \( X_{100} \), \( \dfrac{X-X_0}{X_{100}-X_0}=\dfrac{C}{100}=\dfrac{F-32}{180}=\dfrac{K-273}{100} \).
Differences convert by the ratio of degree sizes, so a rise of 25 °C is a rise of 45 °F. The Celsius and Fahrenheit readings coincide at −40.
The constant-volume gas thermometer uses \( T=273.16\,\dfrac{P}{P_{tr}} \) and is the standard reference, because every gas behaves ideally at low pressure. Mercury and alcohol thermometers, resistance thermometers and thermocouples are calibrated against it.

Method Checklist for Problems

Write heat or work in the same unit before equating (J or cal). For mixing problems with possible phase changes, test the extreme outcomes first. For expansion problems, expand all linear dimensions by \( (1+\alpha\Delta T) \) and use the binomial approximation. For floating bodies, set the buoyancy equal to the weight at both temperatures and compare. For anything clamped, replace the prevented expansion by an equal elastic strain and use stress = Y × strain.