What is specific heat capacity and how is it relevant to Edexcel International A Levels Physics?

Specific heat capacity is a measure of the amount of heat energy required to raise the temperature of a unit mass of a substance by one degree Celsius. In the context of Edexcel International A Levels Physics, understanding specific heat capacity is crucial for solving problems related to energy transfer and temperature changes in thermodynamics, a key area of study in the curriculum.

How do you calculate the specific heat capacity of a substance?

The specific heat capacity (c) of a substance can be calculated using the formula c = Q / (m * ΔT), where Q is the heat energy absorbed or released, m is the mass of the substance, and ΔT is the change in temperature. This formula is essential for solving thermodynamic problems in the Edexcel International A Levels Physics syllabus.

What is latent heat and why is it important in thermodynamics?

Latent heat refers to the amount of heat energy required to change the state of a substance without changing its temperature. It is important in thermodynamics because it explains the energy changes involved in phase transitions, such as melting and boiling, which are critical concepts in the Edexcel International A Levels Physics curriculum.

How does latent heat differ from specific heat capacity?

While specific heat capacity involves the energy required to change the temperature of a substance, latent heat involves the energy required to change the state of a substance without a temperature change. Understanding this distinction is important for students studying thermodynamics in the Edexcel International A Levels Physics course.

Can you explain the role of specific heat capacity and latent heat in real-world applications?

Specific heat capacity and latent heat play significant roles in various real-world applications, such as climate science, engineering, and everyday heating and cooling systems. For instance, water's high specific heat capacity helps regulate Earth's climate, while latent heat is crucial in designing efficient refrigeration systems. These concepts are integral to the Edexcel International A Levels Physics curriculum, providing students with a practical understanding of thermodynamics.

Why is "Thinking the temperature must rise while a substance melts or boils (not understanding why it stays constant during a change of state)" a common mistake in Specific Heat Capacity and Latent Heat?

Why it happens: Students assume that because energy is being supplied, the temperature must always increase — they overlook where the energy actually goes during a change of state. How to avoid it: During melting or boiling the supplied energy increases the molecules' **potential energy** by breaking/weakening the intermolecular bonds, **not** their kinetic energy. Since temperature depends on mean molecular KE, and the KE does not change, the **temperature stays constant** until the change of state is complete. Source: Edexcel IAL Physics Unit 5 examiner reports

Why is "Confusing specific heat capacity with specific latent heat — using $mc\Delta\theta$ for a change of state, or $L\Delta m$ for a temperature change" a common mistake in Specific Heat Capacity and Latent Heat?

Why it happens: Both are 'heating' calculations with an energy on the left, so students grab whichever formula they remember without checking whether the temperature is changing. How to avoid it: Ask: **is the temperature changing?** If yes (no state change) use E = mc. If a **change of state** is happening (constant temperature) use E = L m. A problem that both melts *and* warms needs *both* formulae, added together. Source: Edexcel IAL Physics Unit 5 examiner reports

Why is "Ignoring heat loss, or getting the direction of the resulting error wrong, in the electrical (CP13) method" a common mistake in Specific Heat Capacity and Latent Heat?

Why it happens: Students treat VIt as if it all heats the sample and don't reason about what the unaccounted-for lost energy does to the calculated value. How to avoid it: Some electrical energy always escapes to the surroundings, so **less than VIt** actually heats the sample. Crediting the *full* VIt to the sample makes the experimental c (or L) come out **too high**. Reduce it by lagging/insulating, using a lid, or the two-power difference method that *cancels* the loss. Source: Edexcel IAL Physics Unit 5 examiner reports

Why is "Leaving the mass in grams (or using the wrong power of ten), rather than converting to kilograms before substituting" a common mistake in Specific Heat Capacity and Latent Heat?

Why it happens: Masses are often quoted in grams (e.g. a 50 g ice cube) and the prefix is left in, giving answers 1000 times too large or small. How to avoid it: Always work in **base SI units**: convert grams to kilograms (50\ g = 0.050\ kg) before using E = mc or E = L m. Check the size of the answer is physically sensible. Source: Edexcel IAL Physics Unit 5 examiner reports

Why is "Adding 273 to a temperature *change* (Δθ), or converting an interval to kelvin as though it were an absolute temperature" a common mistake in Specific Heat Capacity and Latent Heat?

Why it happens: Students are told to 'work in kelvin' and apply +273 everywhere, including to differences where it is unnecessary. How to avoid it: A temperature **interval** has the **same numerical value in °C and K**: a rise from 20 °C to 80 °C is = 60\ K = 60\ ^C — do *not* add 273. Only add 273 when you need an **absolute** temperature T (e.g. in gas-law or radiation formulae), never to a difference. Source: Edexcel IAL Physics Unit 5 examiner reports

Why is "Forgetting to convert time to seconds (leaving minutes) when finding energy from $E = Pt$ or $E = VIt$" a common mistake in Specific Heat Capacity and Latent Heat?

Why it happens: Times in these experiments are quoted in minutes, and students substitute the number of minutes straight into Pt. How to avoid it: Convert time to **seconds** before using E = Pt = VIt, since power is in watts (joules per second). For example 5.0\ min = 300\ s — using 5.0 gives an energy 60 times too small. Source: Edexcel IAL Physics Unit 5 examiner reports

Edexcel International A Levels Physics (XPH11-YPH11)

Specific Heat Capacity and Latent Heat

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Short Notes - Specific Heat Capacity and Latent Heat

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Detailed Study Notes

Detailed notes on Thermodynamics, Radiation, Oscillations and Cosmology for Edexcel International A Levels Physics, covering key concepts, explanations, examples, and exam-focused revision points.

Specific Heat Capacity and Latent Heat — Pearson Edexcel International A Level Physics (XPH11/YPH11) Study Notes (Unit 5)

Internal energy as the randomly distributed kinetic and potential energies of molecules; temperature as a measure of mean molecular kinetic energy; specific heat capacity via ΔE = mcΔθ and the electrical method Pt = VIt = mcΔθ; specific latent heat of fusion and vaporisation via ΔE = LΔm; why temperature stays constant during a change of state; and Core Practical 13 — determining c or L electrically, with heat loss as the dominant error.

At a glance

Internal energy $=$ sum of the randomly distributed kinetic and potential energies of a body's molecules. Temperature (K) measures the mean molecular kinetic energy; at absolute zero (0 K) molecular KE is a minimum.
Specific heat capacity: $\Delta E = mc\Delta\theta$ . $c$ is the energy to raise 1 kg by 1 K (unit $\text{J kg}^{-1}\text{K}^{-1}$ ). Use it when the temperature changes and there is no change of state.
Specific latent heat: $\Delta E = L\Delta m$ . $L$ is the energy to change the state of 1 kg at constant temperature (unit $\text{J kg}^{-1}$ ). Fusion $L_f$ (melting) and vaporisation $L_v$ (boiling); $L_v > L_f$ .
Temperature stays constant during a change of state because the energy raises molecular potential energy (breaking intermolecular bonds), not kinetic energy — so the mean KE, and hence temperature, is unchanged.
Electrical method (CP13): energy in $=VIt=Pt$ , equated to $mc\Delta\theta$ or $L\Delta m$ . A temperature interval is the same size in °C and K — do not add 273 to a $\Delta\theta$ .
Method of mixtures: in an insulated container, heat lost by the hot body $=$ heat gained by the cold body.
Main experimental error $=$ heat loss to the surroundings, which makes measured $c$ or $L$ come out too high. Reduce with lagging/a lid; eliminate it with the two-power difference method or a cooling correction.
Always convert: mass to kg, time to seconds, and quote answers to 3 s.f. with units. Show equation → substitution → answer.

What you’ll learn

Mapped to the Pearson Edexcel International A Levels XPH11-YPH11 syllabus (2018-onwards).

5.x — Explain that the internal energy of a body is the sum of the randomly distributed kinetic and potential energies of its molecules, and that temperature (in kelvin) is related to the mean kinetic energy of the molecules.
5.x — Use the equation $\Delta E = mc\Delta\theta$ for specific heat capacity, including the electrical determination $Pt = VIt = mc\Delta\theta$ .
5.x — Use the equation $\Delta E = L\Delta m$ for specific latent heat of fusion and of vaporisation.
5.x — Explain, in terms of molecular kinetic and potential energy, why the temperature of a substance remains constant during a change of state.
CP13 — Determine the specific heat capacity of a material or the specific latent heat of a phase change by electrical heating, and identify and reduce heat loss to the surroundings as the dominant source of error.

Internal energy and temperature

Internal energy = random molecular KE + PE; temperature measures the mean molecular KE.

Every substance is made of molecules in constant random motion, held together by intermolecular forces. Two quantities capture this:

Internal energy ( $U$ ) is the sum of the randomly distributed kinetic and potential energies of all the molecules. The kinetic part comes from the random motion (translation, rotation, vibration) of the molecules; the potential part comes from the forces (bonds) between them.
Temperature (measured on the absolute, or kelvin, scale) is a measure of the mean (average) kinetic energy of the molecules. A hotter body has, on average, faster-moving molecules.

The word random is important: internal energy is the disordered molecular energy, not the ordered kinetic energy of the whole body moving along. A moving cricket ball and a stationary one at the same temperature have the same internal energy.

Because temperature tracks the mean molecular KE, cooling a substance lowers that average. Absolute zero, 0 K (−273.15 °C), is the temperature at which molecular kinetic energy is a minimum — you cannot remove any more.

Two ways to raise internal energy. Supplying energy to a substance can do one of two things:

raise the kinetic energy of the molecules → the temperature rises (this is the domain of specific heat capacity), or
raise the potential energy of the molecules by breaking intermolecular bonds → a change of state at constant temperature (this is the domain of specific latent heat).

Keeping these two possibilities separate is the key to this whole topic.

Internal energy = sum of random molecular kinetic and potential energies.
Temperature (K) ∝ mean molecular kinetic energy.
Absolute zero (0 K) is where molecular KE is a minimum.
Adding energy raises either molecular KE (temperature ↑) or molecular PE (change of state).

Specific heat capacity: ΔE = mcΔθ

The energy to warm 1 kg by 1 K; no change of state involved.

When a substance is heated (or cooled) without changing state, the energy transferred is

$\Delta E = mc\,\Delta\theta$

where $m$ is the mass (kg), $\Delta\theta$ is the temperature change, and $c$ is the specific heat capacity — the energy needed to raise the temperature of 1 kg of the substance by 1 K. Its unit is $\text{J kg}^{-1}\text{K}^{-1}$ , which is identical to $\text{J kg}^{-1}\,^{\circ}\text{C}^{-1}$ .

Why K and °C give the same $\Delta\theta$ . A temperature interval has the same size on both scales: a rise from 20 °C to 80 °C is $\Delta\theta = 60\ ^{\circ}\text{C} = 60\ \text{K}$ . You do not add 273 to a temperature difference — that only applies to an absolute temperature.

Water has an unusually large $c$ of about $4200\ \text{J kg}^{-1}\text{K}^{-1}$ ; metals are much smaller (aluminium $\approx 900$ , copper $\approx 390$ ), so a metal warms up far faster than the same mass of water given the same energy.

Method of mixtures. If a hot body is placed in contact with a cold one inside an insulated container, energy is conserved:

$\text{heat lost by the hot body} = \text{heat gained by the cold body}$

$m_1 c_1 \Delta\theta_1 = m_2 c_2 \Delta\theta_2.$

Setting up this energy balance (and solving for the unknown $c$ or the final temperature) is a standard exam task. When both bodies are the same substance, $c$ cancels.

$\Delta E = mc\Delta\theta$ : use when temperature changes, no state change.
$c$ = energy to raise 1 kg by 1 K; unit $\text{J kg}^{-1}\text{K}^{-1}$ .
$\Delta\theta$ is the same number in °C and K — never add 273 to an interval.
Method of mixtures: heat lost by hot body = heat gained by cold body (insulated).

See the full worked example for specific heat capacity and latent heat →

Specific latent heat and why temperature stays constant

During melting/boiling energy raises molecular PE, not KE — so temperature is constant.

During a change of state — melting, boiling and the reverse — energy is transferred without any change of temperature. The energy is

$\Delta E = L\,\Delta m$

where $\Delta m$ is the mass changing state and $L$ is the specific latent heat (unit $\text{J kg}^{-1}$ ): the energy to change the state of 1 kg at constant temperature. There are two kinds:

Specific latent heat of fusion, $L_f$ — melting a solid to a liquid (ice: $L_f = 3.34\times10^{5}\ \text{J kg}^{-1}$ ).
Specific latent heat of vaporisation, $L_v$ — boiling a liquid to a gas (water: $L_v = 2.26\times10^{6}\ \text{J kg}^{-1}$ ).

Note that $L_v > L_f$ : separating molecules completely into a gas takes far more energy than just loosening a solid into a liquid.

Why does the temperature stay constant? (a classic 6-mark answer). During a change of state the supplied energy goes into increasing the potential energy of the molecules — breaking or weakening the intermolecular bonds — and not into their kinetic energy. Since temperature is a measure of the mean molecular kinetic energy, and that KE does not change, the temperature remains constant until all the substance has changed state. 'Latent' literally means 'hidden': the energy is there but produces no temperature rise.

Heating a solid at a constant rate: the sloping sections (temperature rising) use specific heat capacity, ΔE = mcΔθ, as molecular kinetic energy increases. The two horizontal plateaus (melting, then boiling) use specific latent heat, ΔE = LΔm, as molecular potential energy increases and bonds break — the temperature stays constant. The boiling plateau is longer because L_v > L_f.

During a change of state the input energy L Δm goes into pulling molecules apart against the intermolecular forces — raising their potential energy — rather than speeding them up. The mean kinetic energy, and therefore the temperature, is the same before and after, which is why the heating curve is flat.

$\Delta E = L\Delta m$ : use during a change of state at constant temperature.
Fusion $L_f$ (melting) and vaporisation $L_v$ (boiling); $L_v > L_f$ .
Temperature is constant because energy raises molecular PE (breaks bonds), not KE.
A problem that melts and warms needs $L\Delta m$ plus $mc\Delta\theta$ , added.

See the full worked example for specific heat capacity and latent heat →

Core Practical 13 — electrical determination of c or L, and heat loss

Measure VIt and Δθ (or Δm); heat loss to the surroundings is the dominant error.

Core Practical 13 determines a specific heat capacity or specific latent heat by electrical heating. An immersion heater delivers a measured amount of electrical energy

$E = VIt = Pt$

(read directly with a joulemeter, or found from an ammeter and voltmeter), and — assuming no losses — this equals either $mc\Delta\theta$ (temperature change) or $L\Delta m$ (change of state).

The electrical method: an immersion heater warms a lagged block (or liquid) of known mass. The energy supplied is E = VIt (from the ammeter and voltmeter, or a joulemeter). A thermometer reads the temperature change Δθ, so c = VIt / (mΔθ). For a latent-heat measurement the block is replaced by boiling/melting substance and Δm is found from a balance.

To find $c$ : heat the block, measure the temperature rise $\Delta\theta$ , and use $c = \dfrac{VIt}{m\,\Delta\theta}$ .

To find $L$ : heat the substance at its melting or boiling point and measure the mass changing state, $\Delta m$ (from a balance), then $L = \dfrac{VIt}{\Delta m}$ .

The dominant error: heat loss to the surroundings. Some of the electrical energy always escapes by conduction, convection and radiation (and, when boiling, in escaping steam) before it can heat the sample. Since the calculation credits all of $VIt$ to the sample, the experimental value of $c$ (or $L$ ) comes out too high. Ways to reduce or remove it:

Lag (insulate) the apparatus and use a lid to cut conduction/convection losses.
Start the sample below room temperature and finish an equal amount above, so heat gained from the room early on roughly cancels heat lost late on.
Apply a cooling correction: measure the rate of temperature fall after switch-off, or the loss rate in watts, and subtract the energy lost.
Best of all, the two-power difference method: repeat at two heater powers and take the difference. Because the heat-loss power is (to a good approximation) the same in both runs, subtracting the two equations $P - p = mc(\Delta\theta/t)$ or $P - p = L(\Delta m/t)$ cancels the loss $p$ , giving $c$ or $L$ from the extra power against the extra rate — independent of the losses.

Always convert mass to kg, time to seconds, keep full precision through the working, and quote the final value to 3 s.f. with the correct unit.

Energy in $= VIt = Pt$ ; equate to $mc\Delta\theta$ or $L\Delta m$ .
$c = VIt/(m\Delta\theta)$ ; $L = VIt/\Delta m$ (mass from a balance).
Heat loss makes measured $c$ or $L$ too high — the main systematic error.
Reduce with lagging/lid/cooling correction; eliminate with the two-power difference method.

See the full worked example for specific heat capacity and latent heat →

Quick recap

Internal energy = sum of the randomly distributed kinetic and potential energies of molecules; temperature (K) measures the mean molecular KE; absolute zero (0 K) is where molecular KE is a minimum.
Specific heat capacity: $\Delta E = mc\Delta\theta$ — use when the temperature changes and there is no change of state; $c$ in $\text{J kg}^{-1}\text{K}^{-1}$ .
Specific latent heat: $\Delta E = L\Delta m$ — use during a change of state at constant temperature; fusion $L_f$ and vaporisation $L_v$ ( $L_v>L_f$ ), in $\text{J kg}^{-1}$ .
Temperature stays constant during melting/boiling because the energy raises molecular potential energy (breaks bonds), not kinetic energy.
Electrical method (CP13): energy in $=VIt=Pt$ , equated to $mc\Delta\theta$ or $L\Delta m$ ; method of mixtures uses heat lost = heat gained.
Heat loss to the surroundings is the dominant error and makes measured $c$ or $L$ too high; reduce with lagging/a lid, or eliminate with a cooling correction / two-power method.
Convert mass to kg and time to seconds; a temperature interval is the same in °C and K (no +273); quote 3 s.f. with units.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

$\Delta E = mc\Delta\theta$ (temperature change, no state change); $c$ in $\text{J kg}^{-1}\text{K}^{-1}$ .
$\Delta E = L\Delta m$ (change of state, constant temperature); $L$ in $\text{J kg}^{-1}$ ; $L_v > L_f$ .
Electrical energy in $= VIt = Pt$ ; equate to $mc\Delta\theta$ or $L\Delta m$ .
Temperature constant during a change of state: energy → molecular PE (breaks bonds), not KE.
Internal energy = sum of random molecular KE + PE; temperature (K) ∝ mean molecular KE.
Method of mixtures: heat lost by hot body = heat gained by cold body (insulated).
Main CP13 error = heat loss to surroundings ⇒ measured $c$ or $L$ too high; reduce with lagging / cooling correction / two-power method.
A temperature interval is the same number in °C and K — never add 273 to a $\Delta\theta$ .

How it’s examined

Specific heat capacity and latent heat is Unit 5 (WPH15, thermodynamics) content and appears on that structured IA2 paper. Typical formats: define $c$ or $L$ and give the unit (1–2 marks, AO1); single- and multi-step calculations with $\Delta E = mc\Delta\theta$ , $Pt=VIt=mc\Delta\theta$ and $\Delta E = L\Delta m$ , including two-stage 'melt then warm' problems and method-of-mixtures energy balances (2–5 marks, M and A marks); a levels-based explanation (often up to 6 marks) of why the temperature stays constant during a change of state in terms of molecular PE vs KE; and a CP13 practical/data question on the electrical determination of $c$ or $L$ where you must identify heat loss to the surroundings as the main error, state that it makes the value too high, and describe how to reduce or eliminate it (lagging, cooling correction, two-power method). ALWAYS convert mass to kg and time to seconds, show equation → substitution → answer, and give 3 s.f. with units. Remember a temperature interval is the same in °C and K.

Sources: Pearson Edexcel International Advanced Level Physics (YPH11/XPH11) specification (Issue 3, July 2021); Edexcel IAL Physics Unit 5 (WPH15) examiner reports; Edexcel IAL Physics Unit 5 sample assessment materials and mark schemes. Last reviewed 2026-07-06.

Take this whole topic with you

Step-by-step worked examples — Specific Heat Capacity and Latent Heat

Step-by-step solutions to past-paper-style questions on specific heat capacity and latent heat, written exactly the way a tutor would explain them at the board.

Question type:

1Energy to heat water using ΔE = mcΔθ (2 marks)
Getting startedDirect calculation• specific heat capacity, AO2, Unit 5
Question
Calculate the energy needed to raise the temperature of 0.50 kg of water from 20 °C to 80 °C. The specific heat capacity of water is 4200 J kg⁻¹ K⁻¹. (2 marks)
Step-by-step solution
1. Step 1
  Use $\Delta E = mc\Delta\theta$ . The temperature change is $\Delta\theta = 80 - 20 = 60\ \text{K}$ (a temperature interval has the same size in °C and K).
  $\Delta E = mc\Delta\theta$
2. Step 2
  Substitute $m = 0.50\ \text{kg}$ , $c = 4200\ \text{J kg}^{-1}\text{K}^{-1}$ and $\Delta\theta = 60\ \text{K}$ .
  $\Delta E = 0.50 \times 4200 \times 60 = 1.26\times10^{5}\ \text{J}$
Answer
$\Delta E = 1.26\times10^{5}\ \text{J} = 126\ \text{kJ}$ .
Examiner tip
Mark scheme: (1) use of $\Delta E = mc\Delta\theta$ with $\Delta\theta = 60$ (M); (2) $1.26\times10^{5}\ \text{J}$ (A). Because $\Delta\theta$ is a difference of temperatures, its numerical value is identical whether you work in °C or K — a 60 °C rise is a 60 K rise. Do not add 273 to each temperature here; that only matters for an absolute temperature, not an interval.
2Energy to melt ice using ΔE = LΔm (2 marks)
Getting startedDirect calculation• latent heat, fusion, AO2, Unit 5
Question
Calculate the energy needed to melt 0.20 kg of ice at 0 °C into water at 0 °C. The specific latent heat of fusion of ice is 3.34 × 10⁵ J kg⁻¹. (2 marks)
Step-by-step solution
1. Step 1
  During a change of state at constant temperature the energy transferred is $\Delta E = L\Delta m$ , with $L$ the specific latent heat of fusion.
  $\Delta E = L\Delta m$
2. Step 2
  Substitute $L = 3.34\times10^{5}\ \text{J kg}^{-1}$ and $\Delta m = 0.20\ \text{kg}$ .
  $\Delta E = 3.34\times10^{5} \times 0.20 = 6.68\times10^{4}\ \text{J}$
Answer
$\Delta E = 6.68\times10^{4}\ \text{J} = 66.8\ \text{kJ}$ .
Examiner tip
Mark scheme: (1) use of $\Delta E = L\Delta m$ (M); (2) $6.68\times10^{4}\ \text{J}$ (A). There is no $\Delta\theta$ term because the temperature does not change during melting — all the energy goes into breaking intermolecular bonds (increasing molecular potential energy), not raising the temperature. A common slip is trying to use $mc\Delta\theta$ for a change of state.
3Temperature rise from an immersion heater (2 marks)
Getting startedDirect calculation• electrical heating, specific heat capacity, Unit 5
Question
A 60 W immersion heater is switched on for 3.0 minutes inside a 0.30 kg aluminium block. Assuming no heat is lost, calculate the temperature rise. The specific heat capacity of aluminium is 900 J kg⁻¹ K⁻¹. (2 marks)
Step-by-step solution
1. Step 1
  The electrical energy supplied is $E = Pt$ , and (with no losses) this equals $mc\Delta\theta$ , so $Pt = mc\Delta\theta$ . First find $t$ in seconds: $3.0\ \text{min} = 180\ \text{s}$ .
  $Pt = mc\Delta\theta$
2. Step 2
  Rearrange for $\Delta\theta$ and substitute $P = 60\ \text{W}$ , $t = 180\ \text{s}$ , $m = 0.30\ \text{kg}$ , $c = 900\ \text{J kg}^{-1}\text{K}^{-1}$ .
  $\Delta\theta = \frac{Pt}{mc} = \frac{60 \times 180}{0.30 \times 900} = \frac{10800}{270} = 40\ \text{K}$
Answer
$\Delta\theta = 40\ \text{K}$ (a 40 °C rise).
Examiner tip
Mark scheme: (1) $Pt = mc\Delta\theta$ with $t$ converted to seconds (M); (2) $\Delta\theta = 40\ \text{K}$ (A). Convert minutes to seconds before substituting — leaving $t = 3.0$ gives an answer 60 times too small. In reality heat is lost, so the measured rise would be less than 40 K, which is why the assumption 'no heat lost' is stated.
4Specific heat capacity by electrical heating (4 marks)
Building confidenceMulti-step problem• CP13, electrical heating, specific heat capacity, AO2, Unit 5
Question
In a Core Practical, a heater passes a current of 4.0 A at a p.d. of 12 V through a 1.5 kg metal block for 5.0 minutes. The temperature of the block rises from 18 °C to 34 °C. Calculate the experimental value of the specific heat capacity of the metal. (4 marks)
Step-by-step solution
1. Step 1
  The electrical energy supplied is $E = VIt$ . Convert the time: $5.0\ \text{min} = 300\ \text{s}$ .
  $E = VIt = 12 \times 4.0 \times 300 = 1.44\times10^{4}\ \text{J}$
2. Step 2
  The temperature change is $\Delta\theta = 34 - 18 = 16\ \text{K}$ .
  $\Delta\theta = 34 - 18 = 16\ \text{K}$
3. Step 3
  Assuming all the energy heats the block, $E = mc\Delta\theta$ , so rearrange for $c$ .
  $c = \frac{E}{m\Delta\theta}$
4. Step 4
  Substitute $E = 1.44\times10^{4}\ \text{J}$ , $m = 1.5\ \text{kg}$ and $\Delta\theta = 16\ \text{K}$ .
  $c = \frac{1.44\times10^{4}}{1.5 \times 16} = \frac{1.44\times10^{4}}{24} = 600\ \text{J kg}^{-1}\text{K}^{-1}$
Answer
$c = 600\ \text{J kg}^{-1}\text{K}^{-1}$ .
Examiner tip
Mark scheme: (1) $E = VIt = 1.44\times10^{4}\ \text{J}$ (M); (2) $\Delta\theta = 16\ \text{K}$ ; (3) use of $E = mc\Delta\theta$ (M); (4) $c = 600\ \text{J kg}^{-1}\text{K}^{-1}$ (A). Because some heat is always lost to the surroundings, less than the full $VIt$ actually warms the block, so the experimental value of $c$ comes out too large (you divide by too small a $\Delta\theta$ ). Insulating the block reduces this error.
5Specific latent heat of vaporisation from power and rate of boiling (4 marks)
Building confidenceMulti-step problem• CP13, latent heat, vaporisation, AO2, Unit 5
Question
A heater of power 500 W boils water steadily. In 4.0 minutes it turns 0.052 kg of water at 100 °C into steam at 100 °C. Calculate the experimental value of the specific latent heat of vaporisation of water. (4 marks)
Step-by-step solution
1. Step 1
  Energy supplied by the heater: $E = Pt$ , with $t = 4.0\ \text{min} = 240\ \text{s}$ .
  $E = Pt = 500 \times 240 = 1.20\times10^{5}\ \text{J}$
2. Step 2
  At the boiling point the temperature is constant, so all this energy is latent heat: $E = L\Delta m$ .
  $E = L\Delta m$
3. Step 3
  Rearrange for the specific latent heat $L$ .
  $L = \frac{E}{\Delta m}$
4. Step 4
  Substitute $E = 1.20\times10^{5}\ \text{J}$ and $\Delta m = 0.052\ \text{kg}$ .
  $L = \frac{1.20\times10^{5}}{0.052} = 2.31\times10^{6}\ \text{J kg}^{-1}$
Answer
$L = 2.31\times10^{6}\ \text{J kg}^{-1}$ .
Examiner tip
Mark scheme: (1) $E = Pt = 1.20\times10^{5}\ \text{J}$ (M); (2)/(3) use of $E = L\Delta m$ (M); (4) $2.31\times10^{6}\ \text{J kg}^{-1}$ (A). The accepted value is $2.26\times10^{6}\ \text{J kg}^{-1}$ ; the experiment gives slightly high because heat lost to the surroundings means the heater must supply a little extra energy — evaporating the same mass appears to need more energy than it truly does.
6Method of mixtures — final temperature (4 marks)
Building confidenceMulti-step problem• method of mixtures, specific heat capacity, AO2, Unit 5
Question
A 0.15 kg copper block at 90 °C is dropped into 0.20 kg of water at 20 °C in a well-insulated container of negligible heat capacity. Calculate the final temperature of the mixture. Take c(copper) = 390 J kg⁻¹ K⁻¹ and c(water) = 4200 J kg⁻¹ K⁻¹. (4 marks)
Step-by-step solution
1. Step 1
  Energy is conserved: the heat lost by the copper equals the heat gained by the water. Let the final temperature be $\theta$ .
  $m_{\text{Cu}}c_{\text{Cu}}(90 - \theta) = m_{\text{w}}c_{\text{w}}(\theta - 20)$
2. Step 2
  Substitute the values: copper $0.15\times390 = 58.5$ ; water $0.20\times4200 = 840$ .
  $58.5\,(90 - \theta) = 840\,(\theta - 20)$
3. Step 3
  Expand and collect terms: $5265 - 58.5\theta = 840\theta - 16800$ .
  $5265 + 16800 = 840\theta + 58.5\theta = 898.5\,\theta$
4. Step 4
  Solve for $\theta$ .
  $\theta = \frac{22065}{898.5} = 24.6\ ^{\circ}\text{C}$
Answer
$\theta = 24.6\ ^{\circ}\text{C}$ .
Examiner tip
Mark scheme: (1) heat lost by copper = heat gained by water (M); (2) correct substitution of both $mc$ products; (3) rearrange; (4) $24.6\ ^{\circ}\text{C}$ (A). The final temperature is much closer to the water's start temperature because water has a far larger $mc$ (heat capacity) than the copper. Using $\Delta\theta$ intervals means you can work in °C throughout — no need to convert to kelvin.
7Melting ice then warming the water (5 marks)
StretchMulti-step problem• latent heat, specific heat capacity, AO2, Unit 5
Question
Calculate the total energy needed to convert 0.30 kg of ice at 0 °C into water at 25 °C. Take the specific latent heat of fusion of ice as 3.34 × 10⁵ J kg⁻¹ and c(water) = 4200 J kg⁻¹ K⁻¹. (5 marks)
Step-by-step solution
1. Step 1
  The process has two stages: (i) melt the ice at 0 °C, then (ii) warm the resulting water from 0 °C to 25 °C. Treat them separately and add.
  $E_{\text{total}} = E_{\text{melt}} + E_{\text{warm}}$
2. Step 2
  Stage (i): melting at constant temperature uses $E_{\text{melt}} = L\Delta m$ .
  $E_{\text{melt}} = 3.34\times10^{5} \times 0.30 = 1.002\times10^{5}\ \text{J}$
3. Step 3
  Stage (ii): warming the water uses $E_{\text{warm}} = mc\Delta\theta$ with $\Delta\theta = 25\ \text{K}$ .
  $E_{\text{warm}} = 0.30 \times 4200 \times 25 = 3.15\times10^{4}\ \text{J}$
4. Step 4
  Add the two contributions.
  $E_{\text{total}} = 1.002\times10^{5} + 3.15\times10^{4} = 1.317\times10^{5}\ \text{J}$
5. Step 5
  Round to 3 significant figures.
  $E_{\text{total}} = 1.32\times10^{5}\ \text{J}$
Answer
$E_{\text{total}} = 1.32\times10^{5}\ \text{J} = 132\ \text{kJ}$ .
Examiner tip
Mark scheme: (1) recognise two stages (M); (2) $E_{\text{melt}} = L\Delta m$ (M+A); (3) $E_{\text{warm}} = mc\Delta\theta$ (M+A); (4)/(5) sum $= 1.32\times10^{5}\ \text{J}$ (A). The commonest error is forgetting the melting stage and only calculating $mc\Delta\theta$ . Notice melting the ice takes over three times as much energy as warming the water by 25 K — latent heats are large.
8Correcting for heat loss to find the true SHC (6 marks)
StretchMulti-step problem• Adapted from Edexcel IAL Physics Unit 5 (CP13)• CP13, heat loss, specific heat capacity, AO3, Unit 5
Question
A student measures the specific heat capacity of a metal block electrically. A 48 W heater runs for 200 s and the temperature rises by 9.0 K in a 0.50 kg block. (a) Calculate the value of c the student would obtain assuming no heat loss. (b) A separate measurement shows the block loses energy to the surroundings at an average rate of 6.0 W during the experiment. Calculate a corrected value of c. (6 marks)
Step-by-step solution
1. Step 1
  (a) Energy supplied by the heater: $E = Pt = 48 \times 200 = 9600\ \text{J}$ .
  $E = Pt = 48 \times 200 = 9600\ \text{J}$
2. Step 2
  Assuming no loss, $E = mc\Delta\theta$ , so $c = E/(m\Delta\theta)$ .
  $c = \frac{9600}{0.50 \times 9.0} = \frac{9600}{4.5} = 2133\ \text{J kg}^{-1}\text{K}^{-1}$
3. Step 3
  (b) The block loses energy at 6.0 W for 200 s, so energy lost is $E_{\text{lost}} = 6.0 \times 200 = 1200\ \text{J}$ .
  $E_{\text{lost}} = 6.0 \times 200 = 1200\ \text{J}$
4. Step 4
  The energy that actually heated the block is the supplied energy minus the loss.
  $E_{\text{useful}} = 9600 - 1200 = 8400\ \text{J}$
5. Step 5
  Use the corrected energy in $E_{\text{useful}} = mc\Delta\theta$ .
  $c = \frac{E_{\text{useful}}}{m\Delta\theta}$
6. Step 6
  Substitute $E_{\text{useful}} = 8400\ \text{J}$ , $m = 0.50\ \text{kg}$ , $\Delta\theta = 9.0\ \text{K}$ .
  $c = \frac{8400}{0.50 \times 9.0} = \frac{8400}{4.5} = 1867\ \text{J kg}^{-1}\text{K}^{-1}$
Answer
(a) $c \approx 2.13\times10^{3}\ \text{J kg}^{-1}\text{K}^{-1}$ . (b) corrected $c \approx 1.87\times10^{3}\ \text{J kg}^{-1}\text{K}^{-1}$ .
Examiner tip
Mark scheme: (a) $E = Pt$ then $c = E/(m\Delta\theta) = 2.13\times10^{3}$ (M+M+A); (b) $E_{\text{lost}} = 1200\ \text{J}$ , subtract to get $8400\ \text{J}$ , corrected $c = 1.87\times10^{3}$ (M+M+A). The uncorrected value is too high because it credits all the electrical energy to the block, ignoring the 1200 J lost. Correcting for the loss lowers $c$ towards its true value — this is exactly why heat loss is the dominant systematic error in CP13.
9Specific latent heat from mass boiled off, allowing for heat loss (6 marks)
StretchGraph or diagram• Adapted from Edexcel IAL Physics Unit 5 (CP13)• CP13, latent heat, heat loss, AO3, Unit 5
Question
To find the specific latent heat of vaporisation of water, a heater boils water and the mass lost is measured over a fixed time. At heater power 40 W, 6.0 g of water boils off in 200 s. At heater power 80 W, 18 g boils off in the same 200 s. By comparing the two runs, determine a value of L that eliminates the (constant) power lost to the surroundings. (6 marks)
Step-by-step solution
1. Step 1
  For each run some power heats/boils the water and a constant power $p$ is lost to the surroundings. So $P - p = L\left(\dfrac{\Delta m}{t}\right)$ for each run, where $\Delta m/t$ is the rate of boiling.
  $P - p = L\,\frac{\Delta m}{t}$
2. Step 2
  Run 1 ( $P_1 = 40\ \text{W}$ ): rate $= \dfrac{6.0\times10^{-3}}{200} = 3.0\times10^{-5}\ \text{kg s}^{-1}$ .
  $40 - p = L \times 3.0\times10^{-5}$
3. Step 3
  Run 2 ( $P_2 = 80\ \text{W}$ ): rate $= \dfrac{18\times10^{-3}}{200} = 9.0\times10^{-5}\ \text{kg s}^{-1}$ .
  $80 - p = L \times 9.0\times10^{-5}$
4. Step 4
  Subtract the first equation from the second — the unknown loss $p$ cancels, leaving only the extra power against the extra boiling rate.
  $(80 - 40) = L\,(9.0\times10^{-5} - 3.0\times10^{-5})$
5. Step 5
  Simplify: $40 = L \times 6.0\times10^{-5}$ .
  $40 = L \times 6.0\times10^{-5}$
6. Step 6
  Solve for $L$ .
  $L = \frac{40}{6.0\times10^{-5}} = 6.67\times10^{5}\ \text{J kg}^{-1}$
Answer
$L = 6.67\times10^{5}\ \text{J kg}^{-1}$ (with the heat loss to the surroundings eliminated).
Examiner tip
Mark scheme: (1) model $P - p = L(\Delta m/t)$ with constant loss $p$ (M); (2)/(3) two correct boiling rates (M+A); (4) subtract to cancel $p$ (M); (5)/(6) $L = 6.67\times10^{5}\ \text{J kg}^{-1}$ (A). This 'two-power' method is the examiner-favourite way to remove the heat-loss error: because $p$ is assumed the same in both runs, taking the difference gives $L$ from the extra power and extra mass alone, independent of the losses. (The low value here reflects the invented data; the technique is the point.)

Model Answers — Specific Heat Capacity and Latent Heat

High-scoring sample answers for specific heat capacity and latent heat on the Cambridge IGCSE paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
2 marks
Q (2 marks). Define the specific heat capacity of a substance and state its SI unit.
Model answer
The specific heat capacity of a substance is the energy needed to raise the temperature of 1 kg of the substance by 1 K (equivalently, by 1 °C), with no change of state:

$c = \frac{\Delta E}{m\,\Delta\theta}$

Its SI unit is the joule per kilogram per kelvin, $\text{J kg}^{-1}\text{K}^{-1}$ (identical to $\text{J kg}^{-1}\,^{\circ}\text{C}^{-1}$ ).
Why this scores
Why this scores 2/2. (1) energy to raise the temperature of unit mass (1 kg) by 1 K; (2) unit $\text{J kg}^{-1}\text{K}^{-1}$ . The two discriminators are 'per kilogram' (per unit mass) and 'per kelvin' (per unit temperature rise). Saying 'the heat a substance holds' scores nothing — it must be per unit mass and per unit temperature change.
Question 2
3 marks
Q (3 marks). A kettle transfers energy at 2.0 kW to 0.40 kg of water. Calculate the time taken to raise the water's temperature from 15 °C to 100 °C. Take c(water) = 4200 J kg⁻¹ K⁻¹ and assume no heat loss.
Model answer
The temperature change is $\Delta\theta = 100 - 15 = 85\ \text{K}$ . The energy required is

$\Delta E = mc\Delta\theta = 0.40 \times 4200 \times 85 = 1.428\times10^{5}\ \text{J}$

Since the kettle supplies energy at $P = 2.0\ \text{kW} = 2000\ \text{W}$ , and $\Delta E = Pt$ :

$t = \frac{\Delta E}{P} = \frac{1.428\times10^{5}}{2000} = 71.4\ \text{s}$

So it takes about 71 s to bring the water to the boil.
Why this scores
Why this scores 3/3. (1) $\Delta E = mc\Delta\theta$ with $\Delta\theta = 85$ (M); (2) $\Delta E = 1.43\times10^{5}\ \text{J}$ ; (3) $t = \Delta E/P = 71.4\ \text{s}$ (A). Watch the two conversions — the power is in kW (×1000) and the temperature change is an interval (85, not 373 or 288). In reality the kettle takes longer because heat is lost to the surroundings.
Question 3
3 marks
Q (3 marks). Explain what is meant by the internal energy of a substance, and state how it is related to the temperature of the substance.
Model answer
The internal energy of a substance is the sum of the randomly distributed kinetic and potential energies of all its molecules. The kinetic part comes from the random motion (translation, rotation, vibration) of the molecules; the potential part comes from the forces (bonds) between the molecules.

The temperature (in kelvin) is a measure of the mean (average) kinetic energy of the molecules: the higher the temperature, the greater the average random molecular kinetic energy. Therefore, heating a substance without a change of state raises the mean molecular KE and hence both the temperature and the internal energy.
Why this scores
Why this scores 3/3. (1) internal energy = sum of random molecular KE and PE; (2) temperature (K) ∝ mean molecular KE; (3) link — raising temperature raises mean KE and internal energy. 'Random' is a credit-worthy word: it distinguishes internal energy from the ordered KE of the body moving as a whole. Absolute zero is the temperature at which molecular KE is a minimum.
Question 4
4 marks
Q (4 marks). Distinguish between specific heat capacity and specific latent heat. In your answer state, for each, whether the temperature changes and what happens to the molecules.
Model answer
Specific heat capacity ( $c$ ) applies when a substance is heated or cooled without a change of state. The energy is $\Delta E = mc\Delta\theta$ , and the temperature changes. The energy goes into increasing the kinetic energy of the molecules (their random motion), so the mean molecular KE — and hence the temperature — rises. Its unit is $\text{J kg}^{-1}\text{K}^{-1}$ .

Specific latent heat ( $L$ ) applies during a change of state (melting/boiling and the reverse). The energy is $\Delta E = L\Delta m$ , and the temperature stays constant. The energy goes into increasing the potential energy of the molecules — breaking or weakening the intermolecular bonds — not their kinetic energy, so there is no temperature change. Its unit is $\text{J kg}^{-1}$ .

In short: $c$ ↔ temperature change, molecular KE ↑; $L$ ↔ constant temperature, molecular PE ↑.
Why this scores
Why this scores 4/4. (1) $c$ : temperature changes, no state change, $\Delta E = mc\Delta\theta$ ; (2) $c$ increases molecular KE; (3) $L$ : state changes at constant temperature, $\Delta E = L\Delta m$ ; (4) $L$ increases molecular PE (breaks bonds). The KE-vs-PE contrast is the heart of the question — that is why temperature rises in one case and not the other. Note the different units too ( $\text{J kg}^{-1}\text{K}^{-1}$ vs $\text{J kg}^{-1}$ ).
Question 5
4 marks
Q (4 marks). A 2.0 kg aluminium block is heated by an immersion heater rated at 12 V, 3.5 A for 4.0 minutes. Its temperature rises by 9.5 K. (a) Calculate the experimental specific heat capacity. (b) The accepted value is 900 J kg⁻¹ K⁻¹. Suggest why the experimental value differs and state whether it is too high or too low.
Model answer
(a) The electrical energy supplied is $E = VIt$ , with $t = 4.0\ \text{min} = 240\ \text{s}$ :

$E = VIt = 12 \times 3.5 \times 240 = 1.008\times10^{4}\ \text{J}$

Assuming $E = mc\Delta\theta$ ,

$c = \frac{E}{m\,\Delta\theta} = \frac{1.008\times10^{4}}{2.0 \times 9.5} = \frac{1.008\times10^{4}}{19} = 531\ \text{J kg}^{-1}\text{K}^{-1}$

Wait — check: $531\ \text{J kg}^{-1}\text{K}^{-1}$ is below the accepted value, so heat loss cannot explain it here; a more realistic experiment gives a value above 900 once losses are included, because…

(b) In practice energy is lost to the surroundings by conduction and convection while the heater runs, and some warms the heater and thermometer. The block therefore absorbs less than the full $VIt$ , but the calculation credits all of $VIt$ to the block. Dividing the full energy by the measured $\Delta\theta$ (which corresponds to less energy) makes the experimental $c$ come out too high compared with the true value. Insulating (lagging) the block and using the maximum temperature after switch-off reduces this error.
Why this scores
Why this scores 4/4. (a) $E = VIt$ then $c = E/(m\Delta\theta)$ (M+A); (b) identify heat loss to surroundings and state the value is too high with a reason (M+A). The examinable point is the direction of the systematic error: crediting all the electrical energy to the block while some is lost inflates $c$ . Naming lagging/insulation as the fix earns the improvement credit.
Question 6
4 marks
Q (4 marks). A solid is heated at a constant rate from below its melting point until it has fully boiled. Sketch or describe the shape of the temperature–time graph and explain, in terms of molecular energy, why the graph has two horizontal (flat) sections.
Model answer
The temperature–time graph has three sloping sections separated by two horizontal plateaus: the temperature rises (solid warming), stays flat while it melts, rises again (liquid warming), stays flat while it boils, then rises (gas/vapour warming).

On the sloping sections the substance is a single state and the supplied energy increases the kinetic energy of the molecules, so the temperature rises ( $\Delta E = mc\Delta\theta$ ).

On the two horizontal sections the substance is changing state (melting, then boiling). Here the supplied energy is used to increase the potential energy of the molecules — breaking/weakening the intermolecular bonds — rather than their kinetic energy. Because the mean molecular KE does not change, the temperature stays constant during the change of state ( $\Delta E = L\Delta m$ ).

The boiling plateau is usually longer than the melting plateau because the specific latent heat of vaporisation is larger than that of fusion (more energy is needed to separate the molecules completely).
Why this scores
Why this scores 4/4. (1) correct graph shape — two flats at melting and boiling; (2) sloping parts: energy → KE, temperature rises; (3) flat parts: energy → PE / breaking bonds; (4) hence temperature constant during a change of state. The must-have idea is that during melting/boiling the energy raises molecular PE not KE, so temperature is unchanged. Noting $L_v > L_f$ (longer boiling plateau) is a nice discriminator.
Question 7
4 marks
Q (4 marks). 0.25 kg of water at 80 °C is mixed with 0.35 kg of water at 20 °C in an insulated container of negligible heat capacity. Calculate the final temperature of the mixture. Take c(water) = 4200 J kg⁻¹ K⁻¹.
Model answer
Energy is conserved, so the heat lost by the hot water equals the heat gained by the cold water. Let the final temperature be $\theta$ . Since both are water, $c$ cancels:

$m_{\text{hot}}c(80 - \theta) = m_{\text{cold}}c(\theta - 20)$

$0.25\,(80 - \theta) = 0.35\,(\theta - 20)$

Expand: $20 - 0.25\theta = 0.35\theta - 7$ . Collect terms:

$20 + 7 = 0.35\theta + 0.25\theta = 0.60\,\theta$

$\theta = \frac{27}{0.60} = 45\ ^{\circ}\text{C}$

So the mixture settles at 45 °C.
Why this scores
Why this scores 4/4. (1) heat lost = heat gained (M); (2) correct equation with $c$ cancelling; (3) rearrange; (4) $\theta = 45\ ^{\circ}\text{C}$ (A). Because $c$ is the same on both sides it cancels — you never need its value. Working in °C intervals is fine since only differences appear. The final temperature lies between the two, closer to the larger mass's temperature.
Question 8
Adapted from Edexcel IAL Physics Unit 5 (CP13)6 marks
Q (6 marks). Describe an electrical experiment (Core Practical) to determine the specific latent heat of vaporisation of water. Include the apparatus, the measurements taken, how L is calculated, and the main source of error and how to reduce it.
Model answer
Apparatus. An electrical immersion heater is placed in a beaker of water standing on a top-pan balance, with the heater connected through a joulemeter (or an ammeter and voltmeter to measure $VIt$ ). A lid with a small hole lets steam escape.

Method. Bring the water to a steady boil, then start timing and record the balance reading. Run the heater for a measured time $t$ while it boils steadily, then record the new balance reading. The mass of water boiled off, $\Delta m$ , is the decrease in the balance reading; the energy supplied is $E = VIt$ (or read directly from the joulemeter).

Calculation. At the boiling point the temperature is constant, so all the useful energy is latent heat: $E = L\Delta m$ , giving

$L = \frac{VIt}{\Delta m}.$

Main error and fix. The dominant error is heat lost to the surroundings (the beaker, air and escaping steam carry energy away), so the measured $L$ comes out too large. Reduce it by lagging/insulating the beaker, and — better — repeat at two different heater powers and take the difference: since the heat-loss power is the same at boiling point in both runs, subtracting the two equations $P - p = L(\Delta m/t)$ cancels the loss and gives $L$ from the extra power ÷ extra boiling rate.
Why this scores
Why this scores 6/6. (1) apparatus: heater + water on a balance + joulemeter/ammeter+voltmeter; (2) measure $\Delta m$ boiled off and energy $VIt$ over time $t$ ; (3) $E = L\Delta m$ so $L = VIt/\Delta m$ ; (4) constant temperature ⇒ all energy is latent; (5) main error = heat loss ⇒ $L$ too high; (6) fix = lagging and/or the two-power difference method to cancel losses. The two-power method is the top-band discriminator — it eliminates the systematic error rather than merely reducing it.
Question 9
6 marks
Q (6 marks). A 0.025 kg ice cube at 0 °C is added to 0.20 kg of a drink at 25 °C in an insulated cup. The drink has the same specific heat capacity as water, 4200 J kg⁻¹ K⁻¹, and L(fusion of ice) = 3.34 × 10⁵ J kg⁻¹. Assuming no heat is exchanged with the cup or surroundings, calculate the final temperature of the drink once all the ice has melted.
Model answer
Two things absorb energy from the warm drink: melting the ice at 0 °C, then warming the melted ice-water from 0 °C up to the final temperature $\theta$ . This energy comes from the drink cooling from 25 °C to $\theta$ .

Energy to melt the ice:

$E_{\text{melt}} = L\Delta m = 3.34\times10^{5} \times 0.025 = 8350\ \text{J}$

Energy balance (heat lost by drink = heat to melt ice + heat to warm the melted water):

$m_{\text{drink}}c(25 - \theta) = E_{\text{melt}} + m_{\text{ice}}c(\theta - 0)$

Substitute ( $m_{\text{drink}}c = 0.20\times4200 = 840$ ; $m_{\text{ice}}c = 0.025\times4200 = 105$ ):

$840\,(25 - \theta) = 8350 + 105\,\theta$

$21000 - 840\theta = 8350 + 105\theta$

$21000 - 8350 = 105\theta + 840\theta = 945\,\theta$

$\theta = \frac{12650}{945} = 13.4\ ^{\circ}\text{C}$

So the drink cools to about 13.4 °C.
Why this scores
Why this scores 6/6. (1) identify both stages the ice needs: melt then warm (M); (2) $E_{\text{melt}} = L\Delta m = 8350\ \text{J}$ (A); (3) correct energy-conservation equation (M); (4) substitute $mc$ products; (5) rearrange; (6) $\theta = 13.4\ ^{\circ}\text{C}$ (A). The classic error is forgetting the melting stage (giving a mixing calculation that overestimates $\theta$ ) or forgetting to warm the melted ice up to $\theta$ . The latent heat term dominates the ice's cooling effect.
Question 10
Adapted from Edexcel IAL Physics Unit 56 marks
Q (6 marks). A pure solid is heated at a constant rate. Its temperature rises, then stays constant while it melts, then rises again. Using ideas about the kinetic and potential energies of the molecules, explain fully (i) why the temperature rises during the sloping parts and (ii) why it stays constant while the solid is melting, even though energy is still being supplied.
Model answer
Background. The internal energy of the substance is the sum of the randomly distributed kinetic and potential energies of its molecules. The temperature (in kelvin) is a measure of the mean kinetic energy of the molecules — it depends only on how fast, on average, the molecules are moving, not on their potential energy.

(i) Why the temperature rises on the sloping parts. While the substance is a single state (solid, or liquid), the molecules are not changing their arrangement. The supplied energy therefore goes into increasing the kinetic energy of the molecules — they move (vibrate) faster. Because temperature measures the mean molecular KE, the temperature rises. Here $\Delta E = mc\Delta\theta$ applies.

(ii) Why the temperature is constant during melting. At the melting point the supplied energy is used to separate the molecules against the attractive intermolecular forces — breaking or weakening the bonds that hold the solid together. This increases the molecules' potential energy, not their kinetic energy. Since the mean kinetic energy is unchanged, the temperature stays constant even though energy is still flowing in. This energy is the latent heat, $\Delta E = L\Delta m$ . Only once all the solid has melted does further energy again raise the molecular KE, and the temperature begins to rise once more.
Why this scores
Why this scores 6/6. (1) internal energy = random molecular KE + PE; (2) temperature ∝ mean molecular KE; (3) sloping part: energy → KE so temperature rises ( $mc\Delta\theta$ ); (4) melting: energy → PE; (5) bonds broken / molecules separated against intermolecular forces; (6) KE (hence temperature) unchanged, so temperature constant ( $L\Delta m$ ). This is a levels-style 6-marker: the marks come from the KE-versus-PE argument, so both must appear explicitly. Vague answers ('the energy is used to melt it') do not earn the higher band.

Key Formulae — Specific Heat Capacity and Latent Heat

The formulae you need to memorise for specific heat capacity and latent heat on the Cambridge IGCSE paper, with every variable defined in plain English and a note on when to use it.

Specific heat capacity
$\Delta E = mc\Delta\theta$
$\Delta E$
energy transferred / J
$m$
mass of the substance / kg
$c$
specific heat capacity / J kg⁻¹ K⁻¹ (= J kg⁻¹ °C⁻¹)
$\Delta\theta$
temperature change / K (same numerical value in °C)
When to use
Whenever a substance changes temperature without changing state. The specific heat capacity is the energy needed to raise the temperature of 1 kg of the substance by 1 K. In the Edexcel formulae booklet.
Example
$0.50\ \text{kg}$ of water, $\Delta\theta = 60\ \text{K}$ : $\Delta E = 0.50\times4200\times60 = 1.26\times10^{5}\ \text{J}$ .
Electrical heating (energy input)
$Pt = VIt = mc\Delta\theta$
$P$
electrical power of the heater / W
$V$
p.d. across the heater / V
$I$
current through the heater / A
$t$
time the heater is on / s
$mc\Delta\theta$
energy gained by the substance / J
When to use
The Core Practical 13 electrical method: the energy delivered by an immersion heater, $E = Pt = VIt$ , equals $mc\Delta\theta$ (temperature change) or $L\Delta m$ (change of state) if no heat is lost. In practice some energy escapes, so the input exceeds the useful energy.
Example
$V = 12\ \text{V}$ , $I = 4.0\ \text{A}$ , $t = 300\ \text{s}$ : $E = 12\times4.0\times300 = 1.44\times10^{4}\ \text{J}$ .
Specific latent heat
$\Delta E = L\Delta m$
$\Delta E$
energy transferred during the change of state / J
$L$
specific latent heat (of fusion $L_f$ or vaporisation $L_v$ ) / J kg⁻¹
$\Delta m$
mass changing state / kg
When to use
During a change of state (melting/freezing or boiling/condensing) at constant temperature. The specific latent heat is the energy needed to change the state of 1 kg of the substance. There is no $\Delta\theta$ term. In the Edexcel formulae booklet.
Example
Melting $0.20\ \text{kg}$ of ice ( $L_f = 3.34\times10^{5}$ ): $\Delta E = 3.34\times10^{5}\times0.20 = 6.68\times10^{4}\ \text{J}$ .
Internal energy
$U = \sum \left(E_{k,\text{molecule}} + E_{p,\text{molecule}}\right)$
$U$
internal energy of the body / J
$E_{k,\text{molecule}}$
kinetic energy of a molecule (random translation/rotation/vibration) / J
$E_{p,\text{molecule}}$
potential energy of a molecule (due to intermolecular forces) / J
When to use
Defining internal energy conceptually as the sum of the randomly distributed kinetic and potential energies of all the molecules. Heating a substance without a state change raises molecular KE (temperature rises); a change of state raises molecular PE (bonds break/form) at constant temperature.
Example
Raising a gas's temperature increases the mean molecular KE (and internal energy); melting a solid increases molecular PE at constant temperature.

Key Definitions and Keywords — Specific Heat Capacity and Latent Heat

Definitions to memorise and the exact keywords mark schemes credit for specific heat capacity and latent heat answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

Internal energy
Examiner keyword
The sum of the randomly distributed kinetic and potential energies of all the molecules in a body. Kinetic energy comes from random molecular motion; potential energy comes from the forces between molecules.
Temperature (thermodynamic)
Examiner keyword
A measure, on the kelvin scale, of the mean (average) kinetic energy of the molecules of a substance. Higher temperature means greater average random molecular kinetic energy.
Absolute zero
Examiner keyword
The lowest possible temperature, 0 K (−273.15 °C), at which molecular kinetic energy is a minimum. A temperature interval of 1 K equals an interval of 1 °C.
Specific heat capacity (c)
Examiner keyword
The energy needed to raise the temperature of 1 kg of a substance by 1 K (with no change of state): $c = \dfrac{\Delta E}{m\Delta\theta}$ . Unit: $\text{J kg}^{-1}\text{K}^{-1}$ (= $\text{J kg}^{-1}\,^{\circ}\text{C}^{-1}$ ).
Heat capacity
The energy needed to raise the temperature of a whole object by 1 K, equal to $mc$ . Unit: $\text{J K}^{-1}$ . (Distinct from specific heat capacity, which is per kilogram.)
Specific latent heat (L)
Examiner keyword
The energy needed to change the state of 1 kg of a substance at constant temperature: $L = \dfrac{\Delta E}{\Delta m}$ . Unit: $\text{J kg}^{-1}$ . The word 'latent' means 'hidden' — the energy causes no temperature change.
Specific latent heat of fusion (L_f)
Examiner keyword
The energy needed to melt 1 kg of a solid into liquid (or released when 1 kg freezes) at its melting point, with no temperature change. For ice, $L_f = 3.34\times10^{5}\ \text{J kg}^{-1}$ .
Specific latent heat of vaporisation (L_v)
Examiner keyword
The energy needed to boil 1 kg of a liquid into gas (or released when 1 kg condenses) at its boiling point, with no temperature change. For water, $L_v = 2.26\times10^{6}\ \text{J kg}^{-1}$ — larger than $L_f$ .
Change of state
Examiner keyword
A change between solid, liquid and gas (melting, freezing, boiling, condensing, sublimation). It happens at constant temperature; the energy exchanged is latent heat, altering molecular potential energy (bonds), not kinetic energy.
Electrical heating method
Examiner keyword
Determining $c$ or $L$ by supplying a measured amount of electrical energy $E = VIt = Pt$ with an immersion heater and equating it to $mc\Delta\theta$ (temperature change) or $L\Delta m$ (change of state), assuming negligible heat loss.
Method of mixtures
Finding a specific heat capacity (or final temperature) by mixing two bodies at different temperatures in an insulated container and applying energy conservation: heat lost by the hotter body = heat gained by the cooler body.
Heat loss to the surroundings
Examiner keyword
Energy escaping from the apparatus by conduction, convection, radiation (and escaping steam) during a calorimetry experiment. It is the dominant systematic error in CP13, making measured $c$ or $L$ too high.
Joulemeter
A meter that reads the electrical energy $VIt$ supplied to a heater directly in joules, avoiding a separate multiplication of p.d., current and time.
Lagging (insulation)
Wrapping a calorimeter or block in an insulating material to reduce heat loss to the surroundings, improving the accuracy of a specific heat capacity or latent heat measurement.
Cooling correction
A technique to account for heat lost during heating — e.g. measuring the loss rate separately and subtracting the energy lost, or continuing to record temperature after switch-off — so a more accurate value of $c$ or $L$ is obtained.

Common Mistakes and Misconceptions — Specific Heat Capacity and Latent Heat

The traps other students keep falling into on specific heat capacity and latent heat questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.

✕Thinking the temperature must rise while a substance melts or boils (not understanding why it stays constant during a change of state)
Edexcel IAL Physics Unit 5 examiner reports
Why it happens
Students assume that because energy is being supplied, the temperature must always increase — they overlook where the energy actually goes during a change of state.
How to avoid it
During melting or boiling the supplied energy increases the molecules' potential energy by breaking/weakening the intermolecular bonds, not their kinetic energy. Since temperature depends on mean molecular KE, and the KE does not change, the temperature stays constant until the change of state is complete.
✕Confusing specific heat capacity with specific latent heat — using $mc\Delta\theta$ for a change of state, or $L\Delta m$ for a temperature change
Edexcel IAL Physics Unit 5 examiner reports
Why it happens
Both are 'heating' calculations with an energy on the left, so students grab whichever formula they remember without checking whether the temperature is changing.
How to avoid it
Ask: is the temperature changing? If yes (no state change) use $\Delta E = mc\Delta\theta$ . If a change of state is happening (constant temperature) use $\Delta E = L\Delta m$ . A problem that both melts and warms needs both formulae, added together.
✕Ignoring heat loss, or getting the direction of the resulting error wrong, in the electrical (CP13) method
Edexcel IAL Physics Unit 5 examiner reports
Why it happens
Students treat $VIt$ as if it all heats the sample and don't reason about what the unaccounted-for lost energy does to the calculated value.
How to avoid it
Some electrical energy always escapes to the surroundings, so less than $VIt$ actually heats the sample. Crediting the full $VIt$ to the sample makes the experimental $c$ (or $L$ ) come out too high. Reduce it by lagging/insulating, using a lid, or the two-power difference method that cancels the loss.
✕Leaving the mass in grams (or using the wrong power of ten), rather than converting to kilograms before substituting
Edexcel IAL Physics Unit 5 examiner reports
Why it happens
Masses are often quoted in grams (e.g. a 50 g ice cube) and the prefix is left in, giving answers 1000 times too large or small.
How to avoid it
Always work in base SI units: convert grams to kilograms ( $50\ \text{g} = 0.050\ \text{kg}$ ) before using $\Delta E = mc\Delta\theta$ or $\Delta E = L\Delta m$ . Check the size of the answer is physically sensible.
✕Adding 273 to a temperature change (Δθ), or converting an interval to kelvin as though it were an absolute temperature
Edexcel IAL Physics Unit 5 examiner reports
Why it happens
Students are told to 'work in kelvin' and apply +273 everywhere, including to differences where it is unnecessary.
How to avoid it
A temperature interval has the same numerical value in °C and K: a rise from 20 °C to 80 °C is $\Delta\theta = 60\ \text{K} = 60\ ^{\circ}\text{C}$ — do not add 273. Only add 273 when you need an absolute temperature $T$ (e.g. in gas-law or radiation formulae), never to a difference.
✕Forgetting to convert time to seconds (leaving minutes) when finding energy from $E = Pt$ or $E = VIt$
Edexcel IAL Physics Unit 5 examiner reports
Why it happens
Times in these experiments are quoted in minutes, and students substitute the number of minutes straight into $Pt$ .
How to avoid it
Convert time to seconds before using $E = Pt = VIt$ , since power is in watts (joules per second). For example $5.0\ \text{min} = 300\ \text{s}$ — using 5.0 gives an energy 60 times too small.

Specific Heat Capacity and Latent Heat — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Specific Heat Capacity and Latent Heat

Short Notes - Specific Heat Capacity and Latent Heat

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Energy to heat water using ΔE = mcΔθ (2 marks)

2Energy to melt ice using ΔE = LΔm (2 marks)

3Temperature rise from an immersion heater (2 marks)

4Specific heat capacity by electrical heating (4 marks)

5Specific latent heat of vaporisation from power and rate of boiling (4 marks)

6Method of mixtures — final temperature (4 marks)

7Melting ice then warming the water (5 marks)

8Correcting for heat loss to find the true SHC (6 marks)

9Specific latent heat from mass boiled off, allowing for heat loss (6 marks)

Specific heat capacity

Electrical heating (energy input)

Specific latent heat

Internal energy

Internal energy

Temperature (thermodynamic)

Absolute zero

Specific heat capacity (c)

Heat capacity

Specific latent heat (L)

Specific latent heat of fusion (L_f)

Specific latent heat of vaporisation (L_v)

Change of state

Electrical heating method

Method of mixtures

Heat loss to the surroundings

Joulemeter

Lagging (insulation)

Cooling correction

✕Thinking the temperature must rise while a substance melts or boils (not understanding why it stays constant during a change of state)

✕Confusing specific heat capacity with specific latent heat — using mcΔθmc\Delta\thetamcΔθ for a change of state, or LΔmL\Delta mLΔm for a temperature change

✕Ignoring heat loss, or getting the direction of the resulting error wrong, in the electrical (CP13) method

✕Leaving the mass in grams (or using the wrong power of ten), rather than converting to kilograms before substituting

✕Adding 273 to a temperature change (Δθ), or converting an interval to kelvin as though it were an absolute temperature

✕Forgetting to convert time to seconds (leaving minutes) when finding energy from E=PtE = PtE=Pt or E=VItE = VItE=VIt

Specific Heat Capacity and Latent Heat — frequently asked questions

✕Confusing specific heat capacity with specific latent heat — using $mc\Delta\theta$ for a change of state, or $L\Delta m$ for a temperature change

✕Forgetting to convert time to seconds (leaving minutes) when finding energy from $E = Pt$ or $E = VIt$