Thermal Properties of Matter — JEE Mains Physics Formulas and Traps

Temperature Scales, Expansion and Thermal Stress

Converting between linear temperature scales

Two linear scales

\frac{X - X_{ice}}{X_{steam} - X_{ice}} = \frac{C}{100} = \frac{F - 32}{180} = \frac{K - 273}{100}

Linear, area and volume expansion of solids and gases

Thermal expansion

\Delta L = L\alpha\Delta T \qquad \Delta A = A(2\alpha)\Delta T \qquad \Delta V = V(3\alpha)\Delta T \qquad \gamma_{gas} = \frac{1}{V}\frac{dV}{dT}

Thermal stress in a rod that cannot expand

Thermal stress

\sigma = Y\alpha\Delta T \qquad F = YA\alpha\Delta T \qquad u = \tfrac{1}{2}Y(\alpha\Delta T)^{2}

Common traps

Adding 32 to a temperature change

The 32 shifts a reading, not a difference. A rise of 25 Celsius degrees is a rise of 45 Fahrenheit degrees, not 77.

Using 0 and 100 for a faulty thermometer

A faulty thermometer has its own ice and steam readings. Measure the fraction between those readings, then put it on the true scale.

Using α for an area or a volume

An area grows by 2αΔT and a volume by 3αΔT. Using α alone gives an answer two or three times too small, and that value is usually an option.

Thinking a hole shrinks when the plate is heated

A hole grows exactly as a disc of the same metal would. The metal around it expands outward, carrying the edge of the hole with it.

Reporting the rise instead of the final temperature

ΔL = LαΔT gives the rise ΔT. If the question asks for the temperature to heat to, add the starting temperature. Options often include both.

Measuring the temperature change from the wrong state

Thermal tension is proportional to the change from the stress-free temperature, not from the last temperature mentioned. To raise the tension by 40%, the total cooling from the stress-free state must rise by 40%.

Putting the length into the force

Force = YAαΔT. The length appears in both the forced extension and the strain, and cancels. A rod twice as long pushes on its clamps with the same force.

Calorimetry and Latent Heat

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Heat from a heater, a fuel or lost kinetic energy

Energy into heat

Q = ms\Delta T \qquad \eta Pt = ms\Delta T \qquad f\cdot\tfrac{1}{2}mv^{2} = ms\Delta T + mL

Latent heat and the heating curve

Latent heat and heating-curve slope

Q = mL \qquad \frac{dT}{dQ} = \frac{1}{ms}

Mixing ice and water: heat lost equals heat gained

Heat balance

m_ws_w(T_w - T) = m_is_i(0 - T_i) + m_iL + m_is_w(T - 0)

Common traps

Grams and kilograms in one equation

Specific heats are given per gram or per kilogram, and heats of combustion per gram. Put every quantity in one system before dividing, or the answer is off by a thousand.

Stopping at the melting point

A body that heats up AND melts needs msΔT to reach its melting point and then mL more. Leaving out either term gives a mass or speed that is wrong by a large factor.

Which g the options use

Answers built on g = 9.8 and g = 10 differ by 2%, and both may appear as options. Use the value the question gives. If it gives none, work with 9.8 and check which option the result matches.

Temperature does not rise during melting

Heat supplied while ice melts goes into the change of phase. A graph that keeps rising through 0°C, or an answer that adds a temperature rise there, is wrong.

Forgetting to warm the ice to 0°C first

Ice below 0°C must reach 0°C before it can melt. That term, ms_ice times the degrees below zero, is small but changes the answer between close options.

Latent heat in kJ

Latent heats are often given in kJ/kg or cal/g while specific heats are in J kg⁻¹ K⁻¹. Convert to one unit before adding the stages.

Assuming all the ice melts

Solving the balance without the test can give a final temperature below 0°C, which is impossible with water present. Compare the two heats first; if the water's is smaller, the answer is 0°C.

Forgetting that melted ice also warms

Once the ice has melted it is water at 0°C, and it must be warmed to the final temperature too. Leave out m_i s_w T and the final temperature comes out too high.

Heat Conduction

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Slabs and rods in series: thermal resistance and junction temperature

Conduction in series

H = \frac{KA\Delta T}{L} \qquad R = \frac{L}{KA} \qquad \theta = \frac{\theta_1R_2 + \theta_2R_1}{R_1 + R_2} \qquad K_{eq} = \frac{L_1 + L_2}{L_1/K_1 + L_2/K_2}

Parallel paths, junctions, box walls and spherical shells

Parallel paths, junctions and shells

\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} \qquad \sum \frac{\theta_i - \theta}{R_i} = 0 \qquad \frac{dm}{dt} = \frac{H}{L_f} \qquad R_{shell} = \frac{r_2 - r_1}{4\pi Kr_1r_2}

Common traps

Adding conductivities in series

In series it is the RESISTANCES L/(KA) that add. Averaging or adding the conductivities gives an equivalent conductivity that is too large.

Forgetting that the area goes as r²

For rods, R = L/(Kπr²). Doubling the radius cuts the resistance to a quarter, not a half. Using a diameter for the radius makes the same slip.

Using one face of a box

Heat enters through every wall, so the area is the total of all six faces, 2(lb + bh + hl). Using one face, or forgetting the factor 2, gives a melting rate several times too small.

Treating a spherical shell as a flat slab

The area of a shell grows as r², so L/(KA) with one area is wrong. The resistance is (r₂ − r₁)/(4πK r₁r₂).

Losing the sign at a junction

Write every rod's current as flowing INTO the junction, (θᵢ − θ)/Rᵢ, and set the sum to zero. Mixing in- and out-directions doubles one term or cancels another.

Radiation and Newton's Law of Cooling

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Stefan's law, Wien's law and Newton's law of cooling

Radiation and cooling

P = e\sigma AT^{4} \qquad \lambda_mT = b \qquad \frac{T_1 - T_2}{t} = k\left(\frac{T_1 + T_2}{2} - T_s\right) \qquad T - T_s = (T_0 - T_s)e^{-kt}

Common traps

Celsius in Stefan's law

P = eσAT⁴ needs the kelvin temperature. A body at 127°C against one at 27°C radiates (400/300)⁴ ≈ 3.2 times as much, not (127/27)⁴ ≈ 490 times.

Averaging the wrong quantity

In the average form, average the two TEMPERATURES and then subtract the surroundings. The rate on the left is the drop divided by the time, not the excess divided by the time.

Minutes and seconds

k comes out per minute if the times are in minutes. A time of 10/3 minutes is 200 s; check the unit the options use before choosing.

Doubling the excess does not quadruple the rate

Newton's law is linear in the excess temperature: double the excess, double the rate. The fourth power belongs to Stefan's law and the absolute temperature.

Thermal Properties of Matter formulas

Temperature Scales, Expansion and Thermal Stress

Converting between linear temperature scales

Linear, area and volume expansion of solids and gases

Thermal stress in a rod that cannot expand

Common traps

Calorimetry and Latent Heat

Heat from a heater, a fuel or lost kinetic energy

Latent heat and the heating curve

Mixing ice and water: heat lost equals heat gained

Common traps

Heat Conduction

Slabs and rods in series: thermal resistance and junction temperature

Parallel paths, junctions, box walls and spherical shells

Common traps

Radiation and Newton's Law of Cooling

Stefan's law, Wien's law and Newton's law of cooling

Common traps

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