Specific Heat Capacity

Specific heat capacity is one of the most important quantities in thermal physics. It tells you how much energy is needed to raise the temperature of a material and is central to many AQA exam questions, including both calculation-based and practical-based problems.

Spec mapping (AQA 7408 §3.6.2.2 — Specific heat capacity & Required Practical 8): This lesson addresses the definition of specific heat capacity c, the equation Q = mcΔθ, the determination of c by electrical methods and by the method of mixtures, treatment of heat losses to the surroundings, and the propagation of uncertainty in c. Required Practical 8 explicitly assesses the determination of c for a metal block using an electrical heater, and candidates must be able to evaluate sources of error and propose improvements. (Refer to the official AQA specification document for exact wording.)

Synoptic links: Specific heat capacity sits at the intersection of three other spec areas. It builds directly on internal energy (§3.6.2.1) — the energy supplied raises the kinetic component of U without changing the potential component, which is why c is only defined "without change of state". It feeds into specific latent heat (§3.6.2.2) by contrast: latent heat raises potential energy, c raises kinetic energy. Synoptically it appears in electrical methods (§3.5 — electrical energy E = IVt is the source term in the calorimetric energy balance) and in uncertainty analysis (§3.1.3 — percentage uncertainties combine quadratically for derived quantities), making it the canonical AQA bridge between practical skills and thermal theory. Heat losses to surroundings invoke Newton's law of cooling as a real-system correction.

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Definition and Equation

Key Definition: The specific heat capacity (c) of a substance is the energy required to raise the temperature of 1 kg of the substance by 1 K (or 1 °C), without a change of state.

The fundamental equation is:

Q = mcΔθ

where:

• Q is the energy transferred (J)
• m is the mass of the substance (kg)
• c is the specific heat capacity (J kg⁻¹ K⁻¹)
• Δθ is the temperature change (K or °C)

Since a change of 1 K is equal in magnitude to a change of 1 °C, you can use either unit for the temperature change. However, if you need an absolute temperature (as in the gas laws), you must use kelvin.

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Specific Heat Capacity Values

Substance	Specific Heat Capacity (J kg⁻¹ K⁻¹)	Notes
Water	4200	Unusually high — excellent coolant
Ice	2100	About half that of water
Steam	2000	Lower than liquid water
Aluminium	900	Light metal, moderate c
Copper	390	Used in calorimetry
Iron/Steel	450	Moderate for a metal
Lead	130	Very low — heats up quickly
Glass	840	Similar to aluminium
Ethanol	2400	High for a liquid
Oil	2000	Used in some heating systems

Water has an exceptionally high specific heat capacity compared to most substances. This has important practical consequences:

• Water is used as a coolant in car engines and nuclear power stations.
• Coastal regions have milder climates because the sea absorbs and releases large amounts of energy with only small temperature changes.
• Central heating systems use water to transport thermal energy around buildings efficiently.

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Worked Examples

Worked Example 1 — Basic Calculation

Question: How much energy is required to heat 2.5 kg of water from 18 °C to 100 °C? (c_water = 4200 J kg⁻¹ K⁻¹)

Solution:

Q = mcΔθ = 2.5 × 4200 × (100 − 18)

Q = 2.5 × 4200 × 82

Q = 861 000 J = 861 kJ

Worked Example 2 — Finding Specific Heat Capacity

Question: A 0.80 kg metal block is heated using a 48 W heater for 5 minutes. The temperature rises from 22.0 °C to 42.0 °C. Assuming no energy losses, calculate the specific heat capacity of the metal.

Solution:

Step 1: Calculate the energy supplied by the heater. Q = Pt = 48 × (5 × 60) = 48 × 300 = 14 400 J

Step 2: Calculate the temperature change. Δθ = 42.0 − 22.0 = 20.0 K

Step 3: Rearrange Q = mcΔθ for c. c = Q / (mΔθ) = 14 400 / (0.80 × 20.0) = 14 400 / 16.0 = 900 J kg⁻¹ K⁻¹

This value is consistent with aluminium (900 J kg⁻¹ K⁻¹).

Worked Example 3 — Mixing Hot and Cold Water

Question: 0.30 kg of water at 80 °C is mixed with 0.50 kg of water at 15 °C in an insulated container. Find the final temperature of the mixture. (c_water = 4200 J kg⁻¹ K⁻¹)

Solution:

At thermal equilibrium, energy lost by hot water = energy gained by cold water.

Let T be the final temperature.

m_hot × c × (80 − T) = m_cold × c × (T − 15)

Since c cancels (same substance):

0.30 × (80 − T) = 0.50 × (T − 15)

24 − 0.30T = 0.50T − 7.5

24 + 7.5 = 0.50T + 0.30T

31.5 = 0.80T

T = 31.5 / 0.80 = 39.4 °C

Worked Example 4 — Electrical Heating with Energy Loss

Question: A 50 W immersion heater is used to heat 0.40 kg of water. The temperature rises from 20 °C to 35 °C in 3 minutes 30 seconds. Calculate the rate of energy loss to the surroundings.

Solution:

Step 1: Calculate total energy supplied. Q_supplied = Pt = 50 × 210 = 10 500 J

Step 2: Calculate energy actually used to heat the water. Q_useful = mcΔθ = 0.40 × 4200 × 15 = 25 200 J

Wait — Q_useful > Q_supplied. Let me re-check. The energy supplied is 10 500 J but the energy needed to heat the water by 15 K is 25 200 J. This is impossible if the heater is the only source. The question must intend us to find the actual energy used and compare.

Actually, the water only received 10 500 J from the heater, but some was lost. So:

Q_useful = mcΔθ = 0.40 × 4200 × 15 = 25 200 J

This is incorrect — the water cannot absorb more energy than was supplied. The issue is that the heater supplies 10 500 J total, and some fraction goes to the water and some is lost.

Let me re-read: the temperature rises by 15 K. The energy needed for this rise is: Q_water = 0.40 × 4200 × 15 = 25 200 J

But only 10 500 J was supplied. This means the calculation is inconsistent as stated — the heater cannot produce the observed temperature rise. Let me correct the example with realistic numbers.

Corrected Question: A 150 W immersion heater is used to heat 0.40 kg of water. The temperature rises from 20 °C to 35 °C in 2 minutes. Calculate the rate of energy loss to the surroundings.

Solution:

Step 1: Calculate total energy supplied. Q_supplied = Pt = 150 × 120 = 18 000 J

Step 2: Calculate energy actually used to heat the water. Q_water = mcΔθ = 0.40 × 4200 × 15 = 25 200 J

This is still more than supplied. Let me use different values.

Corrected Question: A 250 W immersion heater is used to heat 0.20 kg of water. The temperature rises from 20 °C to 45 °C in 100 s. Calculate the rate of energy loss to the surroundings.

Solution:

Step 1: Total energy supplied by heater. Q_supplied = Pt = 250 × 100 = 25 000 J

Step 2: Energy used to heat the water. Q_water = mcΔθ = 0.20 × 4200 × 25 = 21 000 J

Step 3: Energy lost to surroundings. Q_lost = Q_supplied − Q_water = 25 000 − 21 000 = 4000 J

Step 4: Rate of energy loss. P_lost = Q_lost / t = 4000 / 100 = 40 W

So the heater supplies 250 W but only 210 W goes into heating the water; 40 W is lost to the surroundings.

Exam Tip: In practical questions, always consider energy losses. If the energy supplied exceeds the energy calculated from mcΔθ, the difference is the energy lost to the surroundings. If the measured specific heat capacity is higher than the known value, this suggests energy losses made the temperature rise smaller than expected.

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Experimental Determination of Specific Heat Capacity

Method 1: Electrical Heating of a Solid Block

Apparatus: A metal block (e.g. aluminium) with two holes drilled in it — one for a thermometer and one for an electric heater. The block sits on an insulating mat and may be wrapped in insulation.

Procedure:

1. Measure the mass of the block (m).
2. Record the initial temperature (θ₁).
3. Switch on the heater and record the ammeter reading (I) and voltmeter reading (V).
4. Heat for a measured time (t).
5. Record the final temperature (θ₂). Continue monitoring after switching off — the temperature may continue to rise briefly as heat spreads through the block.

Calculation: Energy supplied = VIt Q = mcΔθ, so c = VIt / (mΔθ)

Sources of error:

• Energy lost to surroundings (systematic — causes measured c to be too high, since more energy was supplied than used to heat the block, but the temperature rise was smaller than expected, giving c = Q/(mΔθ) a larger value).
• Poor thermal contact between heater and block.
• Temperature measured at one point may not represent the whole block.

Improvements:

• Lag (insulate) the block to reduce heat losses.
• Use a small amount of oil in the heater hole to improve thermal contact.
• Wait for the temperature to stabilise before recording the final reading.

Method 2: Continuous-Flow Calorimetry (for Liquids)

This elegant method eliminates the need to know the heat capacity of the apparatus.

Procedure:

1. Liquid flows at a steady rate past an electrical heater.
2. Measure the mass flow rate (m₁/t), the electrical power (V₁I₁), and the temperature rise (Δθ).
3. Then change the flow rate and adjust the power to give the same temperature rise (Δθ).
4. Take new readings (m₂/t, V₂I₂).

Analysis:

For run 1: V₁I₁t = m₁cΔθ + H (where H is the heat lost to surroundings) For run 2: V₂I₂t = m₂cΔθ + H

Since Δθ is the same in both runs, H is the same (heat losses depend on the temperature difference between the apparatus and surroundings).

Subtracting: (V₁I₁ − V₂I₂)t = (m₁ − m₂)cΔθ

Therefore: c = (V₁I₁ − V₂I₂)t / [(m₁ − m₂)Δθ]

The heat loss term H cancels, giving a more accurate result.

Exam Tip: The continuous-flow method is a favourite exam topic. Be ready to explain why the heat loss cancels (same Δθ, so same rate of heat loss) and to derive the expression for c by subtracting the two energy equations.

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Thermal Capacity vs Specific Heat Capacity

Thermal capacity (or heat capacity) C of an object is the energy required to raise the temperature of the entire object by 1 K:

C = mc (units: J K⁻¹)

Q = CΔθ

This is useful when dealing with a calorimeter or container whose mass and specific heat capacity are combined into a single quantity.

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Common Misconceptions

1. "Specific heat capacity tells you how hot something gets." It actually tells you how much energy is needed per kg per kelvin — a high specific heat capacity means the substance heats up slowly, not that it reaches a higher temperature.

2. "Water boils at 100 °C because it has a high specific heat capacity." The boiling point has nothing to do with specific heat capacity. The boiling point depends on the intermolecular forces and the external pressure.

3. "Temperature change must be in kelvin." For temperature changes (Δθ), either °C or K gives the same numerical value, since 1 K = 1 °C in magnitude. Use kelvin only when an absolute temperature is needed.

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Additional Worked Example — Mixed-Substance Calorimetry

A 100 g block of iron (c_Fe = 450 J kg⁻¹ K⁻¹) at 95 °C is dropped into a copper calorimeter (mass 80 g, c_Cu = 385 J kg⁻¹ K⁻¹) containing 200 g of water (c_water = 4180 J kg⁻¹ K⁻¹). The water and calorimeter are initially at 18 °C. Assuming no heat loss to the surroundings, calculate the final equilibrium temperature θ_f.

Set up the energy-conservation equation. Heat lost by iron = heat gained by water + heat gained by calorimeter: m_Fe c_Fe (95 − θ_f) = m_water c_water (θ_f − 18) + m_Cu c_Cu (θ_f − 18).