JEE Physics Thermal Properties Of Matter Notes Set 03

Temperature Scales and Thermal Expansion

1. Celsius, Fahrenheit and kelvin readings are linked by \( \dfrac{C}{5}=\dfrac{F-32}{9}=\dfrac{K-273}{5} \). A change of 1 °C equals a change of 1 K.
2. On heating, a solid grows by \( \Delta l=l\alpha\Delta\theta \), \( \Delta A=A\beta\Delta\theta \), \( \Delta V=V\gamma\Delta\theta \), and \( \alpha:\beta:\gamma=1:2:3 \).
3. A pendulum clock depends on \( T=2\pi\sqrt{l/g} \). If the rod lengthens by heating, T rises and the clock loses time: \( \dfrac{\Delta T}{T}=\dfrac{1}{2}\alpha\Delta\theta \).
4. A bimetallic strip bends toward the metal that expands less, because the metal that expands more ends up on the outer, longer side of the curve.

Heat, Specific Heat and Calorimetry

1. Heat needed to change the temperature: \( Q=ms\Delta\theta \). Thermal (heat) capacity \( =ms \) and water equivalent \( =ms/s_{water} \).
2. Heat needed for a change of state at constant temperature: \( Q=mL \). For ice \( L_f=80 \) cal/g and for steam \( L_v=540 \) cal/g.
3. Principle of calorimetry: heat lost by hot bodies equals heat gained by cold bodies if no heat escapes.
4. Before assuming a final temperature of a mixture with ice, check whether the available heat can melt all the ice; if not, the final temperature is 0 °C.

Conduction

In a rod with its sides lagged, the heat flow per second is \( H=\dfrac{KA(\theta_1-\theta_2)}{l} \). In the steady state the same H passes every section, whatever the area. Thermal resistance is \( R=\dfrac{l}{KA} \); resistances add in series (same H) and conductances add in parallel (same temperature difference). For two rods of equal size end to end, the junction temperature is \( \theta=\dfrac{K_1\theta_1+K_2\theta_2}{K_1+K_2} \).

Newton's Law of Cooling

For a small excess temperature, the rate of fall is proportional to the excess over the surroundings: \( -\dfrac{d\theta}{dt}=k(\theta-\theta_0) \). For short intervals use the average temperature of the interval, \( \dfrac{\theta_1-\theta_2}{t}=k\left(\dfrac{\theta_1+\theta_2}{2}-\theta_0\right) \). The rate of cooling also depends on surface area, mass and specific heat of the body.

Radiation

Solar constant: the energy received per unit area per unit time at the top of the atmosphere, about \( 1.4\ \text{kW m}^{-2} \).

Worked Examples

1. A black body at 400 K emits \( 10^{6} \) J s\(^{-1}\) m\(^{-2}\). It emits 16 times as much at temperature T. Then \( (T/400)^{4}=16 \), so \( T=800 \) K.
2. A body cools from 40 °C to 36 °C in 2 minutes in surroundings at 20 °C. Average excess in the first interval is 18 °C, giving \( k=\dfrac{2}{18}=\dfrac{1}{9}\ \text{min}^{-1} \). For 36 °C to 32 °C the average excess is 14 °C, so the rate is \( \dfrac{14}{9} \) °C/min and the time is \( \dfrac{4}{14/9}\approx2.57 \) min.
3. Blocks with \( K_1:K_2=1:3 \) join end to end between 100 °C and 0 °C. \( \theta=\dfrac{1\times100+3\times0}{4}=25 \) °C.
4. The Wien peak of a body is at 14.46 micron. \( T=\dfrac{2892\times10^{-6}}{14.46\times10^{-6}}=200 \) K.