Skip to content

Home
Learn
Practice(sign in to use)
Mock exams
Review(sign in to use)

Plan(sign in to use)
Progress(sign in to use)
Tutor(sign in to use)
Studio(sign in to use)

Your plan

Save your progressA free account keeps your answers, predicts your score and builds a daily plan.Create a free account

Loading your exam home

Home
Learn
Practice, sign in to use
Mocks
Review, sign in to use

Loading lesson

Internal energy, specific heat capacity and latent heat · A-Level Physics · Courselo

A-Level Physics
Learn
Thermal physics

Internal energy, specific heat capacity and latent heat

Thermal physics · Lesson 1 of 3

Lesson 1 of 3

Unit 7

Thermal physics

3 lessons · 4 h 25 min of study

Internal energy, specific heat capacity and latent heatReading · 0%
Ideal gases and kinetic theory1 h 50 min of study
First law of thermodynamics, p–V diagrams and heat engines1 h 5 min of study

This unit is ≈6% of the A-Level Physics, across 3 lessons. Full syllabus

Lesson 1 of 3 · Thermal physics

Internal energy, specific heat capacity and latent heat

0% read · 8 min left

8 min read · about 1 h 30 min with practice

3 quick checks

≈2% of the test

Foundational: Foundational: the groundwork the rest of the unit builds on

Reading is free. Sign in to tick off lessons, keep your place and track your mastery.

What the exam asks
Core ideas
Worked examples
Exam technique
Common mistakes
Quick recap
Check your understanding
Flashcards

This is the most dependable scoring topic in thermal physics. The equations are simple ( $Q = mc\Delta\theta$ and $Q = ml$ ), so the marks go to learners who explain internal energy precisely, set up energy balances correctly (mixtures, continuous flow and phase changes in one problem) and can evaluate a heating experiment. Expect 1-mark MCQs on definitions and units, 3–5 mark calculations, graph analysis and a practical method or 6-mark extended response.

By the end you’ll be able to

Define internal energy and explain changes in kinetic and potential energy during heating and phase changes
Convert between Celsius and kelvin and explain absolute zero
Use Q = mcΔθ and Q = mL, including mixing and continuous-flow problems
Describe experiments to measure specific heat capacity and specific latent heat, including heat-loss corrections
Interpret temperature–time graphs for heating and cooling through phase changes

What the exam asks

Internal energy: its definition, and what happens to the kinetic and potential energies of particles during heating and during a change of state.
Temperature: Celsius–kelvin conversion, absolute zero, and the fact that a temperature change is the same in K and in °C.
Calculations: $Q = mc\Delta\theta$ , $Q = ml$ , electrical heating ( $Q = VIt = Pt$ ), mixtures, continuous flow (), and heating then melting then heating again.

Check your understanding

3 questions on internal energy, specific heat capacity and latent heat. Every option is explained once you answer.

Sign in to try the quick check

Answers are checked on our side, every option is explained, and your result feeds your mastery for this topic. It’s free.

Sign in Create a free account

Flashcards

The first 3 of 11 cards for this topic. Sign in and finish the lesson to review them with spaced repetition.

PromptCard 1 of 3

Define internal energy.

PreviousDamping, forced oscillations and resonance Next lessonIdeal gases and kinetic theory

On this page

What the exam asks
Core ideas
Worked examples
Exam technique
Common mistakes
Quick recap
Check your understanding
Flashcards

Key formulas

$\displaystyle T/\text{K} = \theta/^\circ\text{C} + 273$
$\displaystyle Q = mc\Delta\theta$
$\displaystyle Q = ml$
$\displaystyle P = \dfrac{m}{t}c\Delta\theta$

Practice this skill

16 questions on internal energy, specific heat capacity and latent heat, served weakest-first.

Practise this topic

Q

=

V I t =

P t

P = \frac{m}{t}c\Delta\theta

Graphs: temperature–time curves through a phase change. Gradients give

c

and plateau lengths give

l

.

Practical: electrical methods for

c

and

l

, the method of mixtures, heat-loss corrections, and sources of error.

Core ideas

Internal energy

Internal energy is the sum of the randomly distributed kinetic and potential energies of the particles in a body.

Kinetic energy comes from the random motion of the particles (translation, rotation, vibration). Temperature is a measure of the mean random kinetic energy.
Potential energy comes from the forces (bonds) between particles, and it depends on their separation.
For an ideal gas there are no intermolecular forces, so the internal energy is kinetic only.

Internal energy increases when energy is transferred by heating or when work is done on the system, and it decreases in the reverse cases. This is the first law of thermodynamics in words.

Board differences: first law and symbols

Cambridge writes $\Delta U = q + W$ , with $W$ the work done on the system. AQA (Engineering option), Eduqas and WJEC write $Q = \Delta U + W$ , with $W$ the work done by the gas. Always say which convention you are using. Symbols vary too. AQA gives $Q = mc\Delta\theta$ and $Q = ml$ on its data sheet. OCR A gives $E = mc\Delta\theta$ and $E = mL$ . Edexcel/IAL use $\Delta E = mc\Delta\theta$ and $\Delta E = L\Delta m$ . Cambridge 9702 does not print these on the Paper 4 data sheet, so learn them, and use $T/\text{K} = \theta/^\circ\text{C} + 273.15$ . OCR B calls $c$ the specific thermal capacity. It lists $\Delta E = mc\Delta\theta$ with an electrical method, but does not list specific latent heat as a learning outcome.

Temperature and absolute zero

T/\text{K} = \theta/^\circ\text{C} + 273

Absolute zero (0 K, about −273 °C) is the temperature at which the particles have the minimum possible internal energy. The thermodynamic (kelvin) scale does not depend on the properties of any particular substance. A temperature difference of 1 K equals 1 °C, so $T/\text{K} = \theta/^\circ\text{C} + 273$ in $Q = mc\Delta\theta$ never needs converting. Two bodies in thermal equilibrium are at the same temperature, and there is no net energy transfer between them.

Specific heat capacity

Q = mc\Delta\theta

Specific heat capacity $c$ is the energy needed to raise the temperature of 1 kg of a substance by 1 K, with no change of state. Unit: $\text{J kg}^{-1}\text{ K}^{-1}$ . Water has a very high value (about $4200\ \text{J kg}^{-1}\text{ K}^{-1}$ ), which is why it is used for cooling and heating systems.

Specific latent heat

Q = ml

Specific latent heat $l$ is the energy needed to change the state of 1 kg of a substance without changing its temperature. Fusion means solid to liquid; vaporisation means liquid to gas. Unit: $\text{J kg}^{-1}$ .

During a change of state the energy supplied increases the potential energy of the particles (bonds are broken or weakened and separation increases). The mean kinetic energy, and so the temperature, stays constant.

$l_{\text{vap}}$ is much larger than $l_{\text{fus}}$ (for water, $2.26 \times 10^6$ against $3.34 \times 10^5\ \text{J kg}^{-1}$ ) because:

Vaporising separates the particles completely, breaking essentially all the bonds. Melting only loosens the structure, and the particles stay at almost the same separation.
The gas expands enormously, so work is done pushing back the atmosphere.

Heating curves

At constant power $P$ , the gradient of a temperature–time graph is $\frac{\Delta\theta}{\Delta t} = \frac{P}{mc}$ . So a steeper section means a smaller specific heat capacity. The length of a flat section is $t = \frac{ml}{P}$ . Comparing sections gives ratios such as $\frac{c_{\text{liquid}}}{c_{\text{solid}}}$ or $\frac{l}{c}$ without knowing $P$ or $m$ .

Energy balances

Mixtures (no losses): energy lost by the hot substance = energy gained by the cold substance. If ice is involved, first check whether there is enough energy to melt all of it.
Continuous flow (showers, cooling systems, boilers): mass flows through at a rate $\frac{m}{t}$ , so $P = \frac{m}{t}c\Delta\theta$ In steady state the heater and pipes no longer change temperature, so (ignoring losses) all the power goes into the water flowing through.

Measuring c and l

Electrical method for c: put a heater and a thermometer (oil in the holes for good thermal contact) into a lagged metal block of known mass. Record $P = \frac{m}{t}c\Delta\theta$ , $I$ and the temperature every 30–60 s. Energy $= VIt$ . Plot $\theta$ against $t$ : the initial gradient is $\frac{VI}{mc}$ , so $c = \frac{VI}{m \times \text{gradient}}$ . Energy lost to the surroundings makes the measured temperature rise too small, so $c$ comes out too large. The initial gradient is used because heat losses grow as the block gets hotter, and the graph curves.

Reducing the effect of heat loss:

Insulate the block and put a lid on liquids.
Start below room temperature and finish the same amount above it, so the energy gained and lost roughly cancel.
For latent heat of fusion of ice, run a control funnel of ice without a heater for the same time and subtract its melted mass.
Two-power (differential) method: repeat at a second power. The rate of heat loss is the same at the same temperature, so it cancels when you subtract the two energy equations.

Practical endorsement links

Edexcel/IAL Core Practical 13 determines a specific latent heat. OCR A requires electrical methods for both $c$ and $l$ , and the method of mixtures. AQA suggests investigating temperature change by an electrical method or the method of mixtures, and logging temperature through a phase change.

Worked examples

Continuous-flow shower (3 marks)

A 9.5 kW electric shower heats water flowing at $0.10\ \text{kg s}^{-1}$ from 12 °C. Find the outlet temperature. ( $c = 4200\ \text{J kg}^{-1}\text{ K}^{-1}$ )

$c = 4200\ \text{J kg}^{-1}\text{ K}^{-1}$ , so $\Delta\theta = \frac{9500}{0.10 \times 4200} = 22.6$ K.

Outlet temperature $= 12 + 22.6 = 34.6$ °C. Halving the flow rate would double $\Delta\theta$ , which is how a shower’s temperature control works.

Ice in a drink (4 marks)

0.050 kg of ice at 0 °C is added to 0.25 kg of water at 30 °C in an insulated cup. Find the final temperature. ( $l_f = 3.34 \times 10^5\ \text{J kg}^{-1}$ , $c = 4200\ \text{J kg}^{-1}\text{ K}^{-1}$ )

Check first: melting all the ice needs $0.050 \times 3.34 \times 10^5 = 16\,700$ J. Cooling the water to 0 °C releases $0.25 \times 4200 \times 30 = 31\,500$ J, which is enough, so all the ice melts.

Energy lost by the warm water = energy to melt the ice + energy to warm the meltwater: $0.25(4200)(30 - T) = 16\,700 + 0.050(4200)T$ $31\,500 - 1050T = 16\,700 + 210T$ , so °C.

Two-power method for latent heat (4 marks)

Water is kept boiling by a heater. At 50.0 W, 5.2 g boils away in 300 s. At 80.0 W, 9.2 g boils away in 300 s. Find $l_v$ and the rate of heat loss.

$P_1t = m_1l + Ht$ and $P_2t = m_2l + Ht$ . Subtracting: $(80.0 - 50.0)(300) = (9.2 - 5.2) \times 10^{-3}\,l$ , so $l = \frac{9000}{4.0 \times 10^{-3}} = 2.25 \times 10^6\ \text{J kg}^{-1}$ .

Heat loss: $H = \frac{15\,000 - 0.0052 \times 2.25 \times 10^6}{300} = 11$ W. It is the same in both runs because the water is at 100 °C both times.

Exam technique

Write the energy balance in words first (“energy lost by … = energy gained by …”). Then substitute.
Every mass gets its own term. Meltwater from ice must then be warmed from 0 °C, and condensed steam must then cool from 100 °C.
Kelvin vs Celsius: $\Delta\theta$ is the same in both. Only use kelvin for absolute temperatures (gas laws, $\frac{3}{2}kT$ ).
Graph questions: the gradient of $\theta$ against $t$ is $\frac{P}{mc}$ . A flat section means a change of state, with potential energy increasing and kinetic energy constant.
Evaluation: heat loss always makes the temperature rise too small, so the calculated $c$ or $l$ is too big. Say the direction of the error.

Common mistakes

“Temperature is constant, so internal energy is constant”

During melting or boiling the temperature is constant but the internal energy increases, because potential energy increases.

Converting a temperature change to kelvin

A rise from 20 °C to 50 °C is 30 K, not 303 K or 323 K.

Forgetting the second stage

Steam at 100 °C first condenses (releasing $ml$ ) and then the condensed water cools (releasing $mc\Delta\theta$ ). Leaving out either stage loses the accuracy mark.

Grams in $Q = mc\Delta\theta$

$c$ is per kilogram. 250 g is 0.250 kg.

Using a chord instead of the initial gradient

With heat losses the heating curve bends. A chord from the start to the end underestimates the initial gradient and overestimates $c$ .

Quick recap

Internal energy = random kinetic + potential energies of the particles; ideal gas = kinetic only.
$T/\text{K} = \theta/^\circ\text{C} + 273$ ; absolute zero is minimum internal energy; $\Delta\theta$ is the same in K and °C.
$Q = mc\Delta\theta$ (temperature change) and $Q = ml$ (change of state at constant temperature).
During a phase change, potential energy increases and mean kinetic energy stays constant.
Continuous flow: $P = \frac{m}{t}c\Delta\theta$ . Mixtures: energy lost = energy gained; check whether all the ice melts.
Heating-curve gradient $= \frac{P}{mc}$ ; plateau time $= \frac{ml}{P}$ .
Heat losses make measured $c$ and $l$ too large. Reduce them with insulation, a control, starting below room temperature or two powers.

Key idea

Write the energy balance in words, give every mass and every stage its own term, and remember that a change of state changes potential energy, not temperature.

\frac{9500}{0.10 \times 4200} =

22.6

4200

×

30 =

31500

31

500

−

22.6

16700 +

210 T

T = \frac{14\,800}{1260} = 11.7

2.25 \times

1 0^{6} J kg^{- 1}

=

11

P

m l