JEE Physics Thermal Properties Of Matter Notes Set 01

Temperature Scales

Temperature tells us how hot a body is. Two fixed points define a scale: the ice point (melting ice) and the steam point (boiling water at normal pressure). The gap between them is the fundamental interval, which is divided into 100 parts on the Celsius scale, 180 parts on the Fahrenheit scale and 100 parts on the Kelvin scale.
Any linear thermometer obeys \( \dfrac{C}{100} = \dfrac{F-32}{180} = \dfrac{K-273}{100} = \dfrac{R - R_{ice}}{R_{steam}-R_{ice}} \). In words: the fractional position of the reading between the two fixed points is the same on every scale.
A temperature difference of 1 Celsius degree equals 1 kelvin and \( \dfrac{9}{5} \) Fahrenheit degrees. The two scales read the same at \( -40 \). The lowest possible temperature is absolute zero, 0 K or \( -273.15 \)°C. A pyrometer is used for very high temperatures such as that of a furnace or the sun.

Thermal Expansion

Every linear dimension of an isotropic solid grows by the same fraction, so a hole in a plate becomes bigger on heating, and a bimetallic strip bends towards the metal with the smaller coefficient of expansion. For a liquid the apparent expansion in a container is \( \gamma_{app} = \gamma_{liquid} - \gamma_{vessel} \).
If a rod is clamped so that it cannot expand, a thermal stress appears: stress \( = Y\alpha\,\Delta T \) and force \( F = YA\alpha\,\Delta T \). A rod free to expand has no stress. Density changes with temperature as \( \rho' = \dfrac{\rho}{1+\gamma\,\Delta T} \). A scale that is calibrated at a higher temperature and used at a lower one reads more than the true length, because the scale itself has contracted.

Heat, Specific Heat and Calorimetry

Heat is energy that flows because of a temperature difference. The heat needed to change the temperature of a mass m by \( \Delta T \) is \( Q = ms\,\Delta T \), where s is the specific heat. The product ms is the thermal (heat) capacity of the body. During a change of state the temperature stays constant and \( Q = mL \), where L is the latent heat.
Useful values: 1 cal = 4.18 J (often 4.2 J), s of water = 1 cal/g°C = 4200 J/kg K, s of ice = 0.5 cal/g°C, latent heat of fusion of ice = 80 cal/g (3.36 × 10⁵ J/kg), latent heat of vaporisation of water = 540 cal/g (about 2.26 × 10⁶ J/kg).
Principle of calorimetry: in an isolated system, heat lost by the hot bodies equals heat gained by the cold bodies. When ice is mixed with water, first check whether all the ice can melt. If the heat available from the water cooling to 0°C is less than that needed to melt the ice, the final temperature is 0°C and some ice remains.

Heating curve of ice: the temperature rises along a slope while ice warms (slope inversely proportional to ms), stays flat during melting, rises again while the water warms, stays flat at the boiling point, and then rises for steam. The flat parts are the latent heats and the lengths of the flat parts are in the ratio of the latent heats.

Conduction

In conduction, heat passes from particle to particle without bulk movement of matter. In steady state the heat current through a uniform rod is \[ \dfrac{Q}{t} = \dfrac{KA(T_1 - T_2)}{L} = \dfrac{T_1 - T_2}{R}, \qquad R = \dfrac{L}{KA}. \]
Thermal resistances combine like electrical resistances. In series \( R = R_1 + R_2 \) and the same heat current flows; for two slabs of equal area, \( K_{eff} = \dfrac{L_1+L_2}{L_1/K_1 + L_2/K_2} \). In parallel the conductances add, \( \dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2} \), and the temperature difference is the same. The interface temperature of two slabs in series follows from equal heat current: \( \dfrac{K_1(T_1 - T)}{d_1} = \dfrac{K_2(T - T_2)}{d_2} \).
In steady state the temperature along a uniform rod falls linearly and does not change with time, although it differs from point to point. In the unsteady state some heat is stored in the rod and its temperature still changes. Metals are good conductors and gases, wood and air-filled materials such as wool and mud are poor ones, which explains why a metal feels colder than wood at the same temperature.

Convection and Radiation

Convection is heat transfer by the actual movement of a fluid, driven by density differences under gravity. Sea breeze, land breeze, boiling water and wind systems are examples. Natural convection cannot occur in a place without gravity.
Radiation is carried by electromagnetic waves and needs no medium. A good absorber is also a good emitter, and a perfect black body absorbs all radiation falling on it.

Because \( \lambda_m \propto 1/T \), a hotter star looks bluer and a cooler one looks redder. In cooling problems the rate of fall of temperature is \( \dfrac{\text{rate of heat loss}}{ms} \), so of two bodies losing heat at the same rate, the one with the smaller heat capacity cools faster. Temperatures in the Stefan and Wien laws must always be in kelvin.

Worked Examples

Example 1. A steel rod of length 2 m at 20°C is heated to 120°C. Find the increase in length if \( \alpha = 1.2\times10^{-5} \) per °C.
Answer:
\( \Delta\ell = \ell_0\alpha\,\Delta T = 2\times1.2\times10^{-5}\times100 = 2.4\times10^{-3} \) m.
The rod becomes 2.4 mm longer.

Example 2. How much heat is needed to change 50 g of ice at −10°C into water at 20°C? (s of ice = 0.5 cal/g°C, L = 80 cal/g)
Answer:
Warm the ice: \( 50\times0.5\times10 = 250 \) cal.
Melt it: \( 50\times80 = 4000 \) cal.
Warm the water: \( 50\times1\times20 = 1000 \) cal.
Total \( = \) 5250 cal.

Example 3. Two slabs of equal area, thickness 4 cm and 6 cm, with conductivities 2 and 3 units respectively, are in contact. The outer faces are at 100°C and 0°C. Find the interface temperature.
Answer:
Series: equal heat currents, \( \dfrac{2(100 - T)}{4} = \dfrac{3(T - 0)}{6} \).
\( 0.5(100 - T) = 0.5T \), so \( T = \) 50°C.

Example 4. A black body at 727°C radiates power P. At what temperature will it radiate 16P?
Answer:
\( P \propto T^4 \), so \( T' = T\times16^{1/4} = 2T \).
\( T = 1000 \) K, so \( T' = 2000 \) K, that is 1727°C.

Example 5. A liquid cools from 80°C to 70°C in 5 minutes in a room at 30°C. How long will it take to cool from 70°C to 60°C?
Answer:
First interval: mean 75°C, excess 45°C: \( \dfrac{10}{5} = k\times45 \), so \( k = \dfrac{2}{45} \) per min.
Second interval: mean 65°C, excess 35°C: \( \dfrac{10}{t} = \dfrac{2}{45}\times35 \).
\( t = \dfrac{450}{70} = \) 6.4 minutes.