Class 11 Physics Heat Numericals: Thermal Expansion, Quantity of Heat and Rate of Heat Flow With Formulas and Solved Examples

The heat section of NEB Class 11 Physics is four chapters long: Heat and Temperature, Thermal Expansion, Quantity of Heat and Rate of Heat Flow. The first is conceptual. The other three are where the numericals come from, and a board paper almost always carries at least one numerical from this group, plus a derivation such as the relation between the expansion coefficients or the specific heat by the method of mixtures.

This guide takes the three numerical chapters in order, states the formulas, shows the derivation the examiner expects, and works one example per formula with the arithmetic written out. The full chapter notes, with more worked examples and the solved board questions, are on the Thermal Expansion, Quantity of Heat and Rate of Heat Flow chapter pages, with the conceptual groundwork in Heat and Temperature.

Thermal expansion

When a solid is heated its molecules vibrate with larger amplitude, their mean separation grows, and the solid expands in length, area and volume.

Each coefficient is the fractional change per unit rise in temperature, with unit $\text{K}^{-1}$.

The derivation to remember: $\beta = 2\alpha$ and $\gamma = 3\alpha$

Take a cube of side $l_1$. After heating, each side is $l_1(1 + \alpha,\Delta\theta)$, so the volume is

$$V_2 = l_1^3(1 + \alpha,\Delta\theta)^3 = V_1\left(1 + 3\alpha,\Delta\theta + 3\alpha^2\Delta\theta^2 + \alpha^3\Delta\theta^3\right)$$

Since $\alpha$ is of the order of $10^{-5}$, the squared and cubed terms are negligible, leaving $V_2 = V_1(1 + 3\alpha,\Delta\theta)$. Comparing with $V_2 = V_1(1 + \gamma,\Delta\theta)$ gives $\gamma = 3\alpha$. The same argument on a square plate gives $\beta = 2\alpha$, so

$$\alpha : \beta : \gamma = 1 : 2 : 3$$

Worked example 1: expansion of a rail

A steel rail is 20 m long at 10 °C. Find its increase in length at 50 °C if $\alpha$ for steel is $1.2 \times 10^{-5}\ \text{K}^{-1}$.

$$\Delta L = \alpha L_1 \Delta\theta = 1.2 \times 10^{-5} \times 20 \times 40 = 9.6 \times 10^{-3}\ \text{m}$$

The rail grows by 9.6 mm, which is why expansion gaps are left between rail sections.

Worked example 2: volume expansion from the linear coefficient

A brass vessel has a volume of 500 cm³ at 0 °C. Find its volume at 100 °C if $\alpha_{\text{brass}} = 1.9 \times 10^{-5}\ \text{K}^{-1}$.

$$\gamma = 3\alpha = 5.7 \times 10^{-5}\ \text{K}^{-1}$$

$$\Delta V = \gamma V_1 \Delta\theta = 5.7 \times 10^{-5} \times 500 \times 100 = 2.85\ \text{cm}^3$$

So the volume at 100 °C is 502.85 cm³. Questions on liquids add one more step: the apparent expansion of a liquid in a container is the real expansion minus the container's expansion, $\gamma_{\text{real}} = \gamma_{\text{apparent}} + \gamma_{\text{container}}$.

The conceptual question from this chapter is the anomalous expansion of water: between 0 °C and 4 °C water contracts on heating, so its density is greatest at 4 °C, which is why ponds freeze from the top and fish survive below.

Quantity of heat

The heat needed to change the temperature of a body is proportional to its mass and to the temperature change:

$$Q = m s,\Delta T$$

where $s$ is the specific heat capacity (J kg⁻¹ K⁻¹; 4200 for water). Heat capacity is $ms$, the heat needed to raise the whole body by one kelvin. When a substance changes state at constant temperature, the heat involved is the latent heat:

$$Q = mL$$

The principle of calorimetry

When a hot body and a cold body exchange heat with no loss to the surroundings, heat lost by the hot body equals heat gained by the cold body. This single line solves every method-of-mixtures problem.

Worked example 3: final temperature of a mixture

A 0.2 kg copper block at 100 °C is dropped into 0.5 kg of water at 20 °C. Find the final temperature, taking $s_{\text{Cu}} = 400$ and $s_w = 4200\ \text{J kg}^{-1}\text{K}^{-1}$ and neglecting the container.

Let the final temperature be $\theta$.

$$0.2 \times 400 \times (100 - \theta) = 0.5 \times 4200 \times (\theta - 20)$$

$$8000 - 80\theta = 2100\theta - 42000$$

$$50000 = 2180\theta \quad\Rightarrow\quad \theta = 22.9\ ^{\circ}\text{C}$$

The small rise is the point: water's specific heat is more than ten times copper's.

Worked example 4: melting ice in warm water

0.1 kg of ice at 0 °C is added to 0.5 kg of water at 40 °C. Find the final temperature, with $L_f = 3.36 \times 10^5\ \text{J kg}^{-1}$.

Heat to melt the ice: $0.1 \times 3.36 \times 10^5 = 33600\ \text{J}$. The melted ice, now water at 0 °C, then warms to $\theta$.

$$0.5 \times 4200 \times (40 - \theta) = 33600 + 0.1 \times 4200 \times \theta$$

$$84000 - 2100\theta = 33600 + 420\theta$$

$$50400 = 2520\theta \quad\Rightarrow\quad \theta = 20\ ^{\circ}\text{C}$$

Always check first whether there is enough heat to melt all the ice; if there is not, the final temperature is 0 °C and the question becomes "how much ice melts".

Newton's law of cooling

For a small temperature difference, the rate of loss of heat is proportional to the excess of the body's temperature over its surroundings: $\dfrac{dQ}{dt} = -k(T - T_s)$. The derivation gives $\log_e(T - T_s) = -\dfrac{k}{ms}t + C$, so a graph of $\log_e(T - T_s)$ against time is a straight line, which is the graph question.

Worked example 5: cooling time

A body cools from 80 °C to 70 °C in 5 minutes when the surroundings are at 30 °C. How long will it take to cool from 70 °C to 60 °C?

Using the average temperature in each interval:

$$\frac{80 - 70}{5} = k(75 - 30) \quad\Rightarrow\quad k = \frac{2}{45}$$

$$\frac{70 - 60}{t} = k(65 - 30) = \frac{2 \times 35}{45} \quad\Rightarrow\quad t = \frac{10 \times 45}{70} = 6.4\ \text{min}$$

The second interval takes longer because the body is closer to room temperature.

Rate of heat flow

Heat conducted through a slab of area $A$ and thickness $x$ whose faces are held at $T_1$ and $T_2$ is

$$\frac{Q}{t} = \frac{k A (T_1 - T_2)}{x}$$

where $k$ is the thermal conductivity (W m⁻¹ K⁻¹) and $(T_1 - T_2)/x$ is the temperature gradient. Define $k$ as the rate of heat flow through unit area of a slab of unit thickness per unit temperature difference.

Worked example 6: heat through a window

A glass window of area 2 m² and thickness 4 mm has its inner face at 20 °C and outer face at 5 °C. Find the rate of heat loss, with $k_{\text{glass}} = 0.8\ \text{W m}^{-1}\text{K}^{-1}$.

$$\frac{Q}{t} = \frac{0.8 \times 2 \times 15}{0.004} = 6000\ \text{W}$$

A thin pane conducts a surprising amount of heat; this is the numerical behind double glazing.

Radiation and Stefan's law

A perfectly black body absorbs all radiation falling on it and is the best emitter at any temperature. Its emissive power is proportional to the fourth power of its absolute temperature:

$$E = \sigma T^4, \qquad \sigma = 5.67 \times 10^{-8}\ \text{W m}^{-2}\text{K}^{-4}$$

For a real body of emissivity $e$ (between 0 and 1) in surroundings at $T_2$, the net rate of emission per unit area is $e\sigma(T_1^4 - T_2^4)$.

Worked example 7: power radiated by a black body

Find the power radiated by a black body of surface area 0.01 m² at 500 K.

$$P = \sigma A T^4 = 5.67 \times 10^{-8} \times 0.01 \times (500)^4 = 5.67 \times 10^{-8} \times 0.01 \times 6.25 \times 10^{10} = 35.4\ \text{W}$$

Remember to convert to kelvin before raising to the fourth power; using 227 °C instead of 500 K is the most common error in this chapter.

Formula sheet

How these chapters are examined

Expect one numerical worth the long-answer marks from calorimetry or conduction, one derivation (the coefficient relation, the method of mixtures, or Newton's law graph), and short conceptual questions on anomalous expansion, black bodies and the three modes of heat transfer. Work the solved board questions on each chapter page in that order, and use the Class 11 Physics important questions view to see which of the three chapters has been asked most in recent years. All chapters of the subject are on the Class 11 Physics page.