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If you heat a saucepan of water on a hob, it takes a noticeable amount of time to get hot. But if you put the same amount of energy into a similar mass of cooking oil, it heats up much faster. Why? The answer lies in a property called specific heat capacity — one of the most important quantities in thermal physics.
The specific heat capacity (c) of a substance is the amount of energy required to raise the temperature of 1 kg of that substance by 1 K (or equivalently, by 1 °C).
The equation linking energy, mass, specific heat capacity, and temperature change is:
Q = mcΔθ
where:
The units of specific heat capacity are J kg⁻¹ K⁻¹ (joules per kilogram per kelvin).
Different materials have very different specific heat capacities. The table below is essential reference data:
| Material | c / J kg⁻¹ K⁻¹ | Notes |
|---|---|---|
| Water | 4200 | Highest of common liquids; excellent coolant |
| Ice | 2100 | About half that of liquid water |
| Steam | 2000 | Lower than liquid water |
| Ethanol | 2440 | Used in some cooling systems |
| Aluminium | 900 | Light metal, heats quickly |
| Copper | 390 | Good conductor, low SHC |
| Iron/Steel | 450 | Structural metal |
| Lead | 130 | Very low — heats up rapidly |
| Glass | 840 | Similar to concrete |
| Cooking oil | ≈ 2000 | Heats faster than water |
| Concrete | ≈ 800 | Used in thermal mass buildings |
| Air (at constant p) | 1010 | Important for atmospheric physics |
| Mercury | 140 | Used in older thermometers |
Water has an unusually high specific heat capacity. This means it takes a lot of energy to heat water up, but it also means water releases a lot of energy when it cools down. This property is why water is used as a coolant in engines and central heating systems, and why coastal areas have more moderate climates than inland areas.
How much energy is needed to heat 2.0 kg of water from 20 °C to 100 °C? (c for water = 4200 J kg⁻¹ K⁻¹)
Q = mcΔθ Q = 2.0 × 4200 × (100 − 20) Q = 2.0 × 4200 × 80 Q = 672 000 J = 672 kJ
A 0.50 kg block of copper at 200 °C is placed in 1.0 kg of water at 20 °C. Assuming no energy is lost to the surroundings, what is the final temperature? (c for copper = 390 J kg⁻¹ K⁻¹, c for water = 4200 J kg⁻¹ K⁻¹)
Energy lost by copper = Energy gained by water
m_copper × c_copper × (200 − T) = m_water × c_water × (T − 20)
0.50 × 390 × (200 − T) = 1.0 × 4200 × (T − 20)
195(200 − T) = 4200(T − 20)
39 000 − 195T = 4200T − 84 000
39 000 + 84 000 = 4200T + 195T
123 000 = 4395T
T = 123 000 ÷ 4395 = 28.0 °C (3 s.f.)
The final temperature is approximately 28 °C. Notice how the large specific heat capacity of water means the water temperature barely rises despite absorbing all the energy from the hot copper.
A 0.25 kg block of an unknown metal at 95 °C is placed in 0.40 kg of water at 18 °C in an insulated container. The final temperature is 22 °C. Calculate the specific heat capacity of the metal. (c_water = 4200 J kg⁻¹ K⁻¹)
Energy lost by metal = Energy gained by water
0.25 × c_metal × (95 − 22) = 0.40 × 4200 × (22 − 18)
0.25 × c_metal × 73 = 0.40 × 4200 × 4
18.25 × c_metal = 6720
c_metal = 6720 / 18.25 = 368 J kg⁻¹ K⁻¹
This is close to the value for copper (390 J kg⁻¹ K⁻¹), suggesting the metal is likely copper. The slightly lower experimental value could be explained by small heat losses to the surroundings.
A 2.5 kW immersion heater is used to heat 8.0 kg of water in a tank from 15 °C to 55 °C. How long does this take? How much does it cost if electricity is 30p per kWh?
Q = mcΔθ = 8.0 × 4200 × (55 − 15) = 8.0 × 4200 × 40 = 1 344 000 J
Time = Q / P = 1 344 000 / 2500 = 537.6 s ≈ 538 s (about 9 minutes)
Energy in kWh = 1 344 000 / (3.6 × 10⁶) = 0.373 kWh
Cost = 0.373 × 30 = 11.2p ≈ 11p
The most common experimental method for determining SHC uses an electrical heater:
For a solid block: drill two holes — one for the heater, one for the thermometer. Add a small amount of oil or water in the thermometer hole to ensure good thermal contact.
For a liquid: place the liquid in an insulated calorimeter (e.g., a polystyrene cup) with the heater immersed and a lid to reduce evaporation.
graph LR
A["Measure mass m"] --> B["Record initial\ntemperature θ₁"]
B --> C["Heat with known\npower P = VI"]
C --> D["Record time t\nand final θ₂"]
D --> E["Calculate:\nE = VIt"]
E --> F["Calculate:\nc = E / (m × Δθ)"]
F --> G["Compare with\naccepted value"]
An alternative method for liquids is the continuous flow method, sometimes called Callendar and Barnes' method:
Since Δθ is the same in both experiments, the heat lost to the surroundings is the same. The heat loss cancels:
V₁I₁ = (m₁/t)cΔθ + heat loss V₂I₂ = (m₂/t)cΔθ + heat loss
Subtracting:
V₁I₁ − V₂I₂ = (m₁/t − m₂/t) × c × Δθ
c = (V₁I₁ − V₂I₂) / [(m₁/t − m₂/t) × Δθ]
The beauty of this method is that it eliminates the need to know or measure the heat lost to the surroundings.
Understanding errors is essential for the practical endorsement and for 6-mark exam questions:
| Source of error | Effect on calculated c | How to reduce it |
|---|---|---|
| Heat loss to surroundings | c too high (Δθ too small → c = Q/mΔθ larger) | Insulate with lagging; use lid on liquids |
| Uneven heating | Thermometer reads inaccurate average | Stir liquids; wait for thermal equilibrium |
| Energy absorbed by container | c too high (some Q heats container, not substance) | Include calorimeter heat capacity in calculation |
| Thermometer precision | Random uncertainty in Δθ | Use digital thermometer with 0.1 °C resolution |
| Evaporation (liquids) | Mass decreases → effective m too large → c too low | Use a lid; minimise heating time |
| Heating past desired temperature | Systematic error in Δθ | Take multiple readings; plot cooling correction curve |
Central heating systems use water as the heat transfer fluid precisely because of its high SHC. A radiator filled with hot water stores a large amount of thermal energy and releases it slowly, keeping rooms warm for extended periods.
Thermal mass in buildings: Concrete floors and walls (c ≈ 800 J kg⁻¹ K⁻¹) absorb solar energy during the day and release it at night, reducing temperature fluctuations. This principle is used extensively in passive solar building design.
Coastal vs continental climates: The sea (water, c = 4200) absorbs vast amounts of solar energy in summer with only a modest temperature rise, and releases this energy in winter. By contrast, land (rock/soil, c ≈ 800) heats and cools much more rapidly. This is why London (coastal influence) has milder winters and cooler summers than Moscow (continental) at a similar latitude.
Car engine cooling: The cooling system circulates water (mixed with antifreeze) around the engine. Water's high SHC means it can absorb large amounts of waste heat from combustion without boiling.
Edexcel 9PH0 specification, Topic 9 — Thermodynamics introduces specific heat capacity as the energy required to raise the temperature of 1 kg of a substance by 1 K, with the defining relationship Q=mcΔθ (refer to the official specification document for exact wording). The topic sits within Paper 2 content but is examined synoptically across Paper 3 (which can draw on any topic) and underpins core practical CP12, the determination of specific heat capacity by an electrical method. Specific heat capacity also underwrites the broader treatment of internal energy, kinetic theory and the first law of thermodynamics later in the topic. The Edexcel formula booklet does provide ΔE=mcΔθ, but the experimental rearrangement c=E/(mΔθ) and the energy-balance reasoning around heat losses must be reconstructed from first principles in the exam.
Question (8 marks):
A student uses an electrical heater to determine the specific heat capacity of an aluminium block of mass 1.00 kg. The heater is rated at 12.0 V and draws a current of 4.00 A. After the heater has been switched on for 5.00 minutes, the temperature of the block has risen from 19.0 °C to 32.5 °C.
(a) Calculate the value of the specific heat capacity of aluminium obtained from this data, assuming no heat is lost to the surroundings. (4)
(b) The accepted value for aluminium is 900 J kg⁻¹ K⁻¹. Explain whether the student's value would increase or decrease if the heat lost to the surroundings during the experiment were taken into account, and estimate the corrected value of c if 10% of the input energy is lost to the surroundings. (4)
Solution with mark scheme:
(a) Step 1 — calculate the electrical energy supplied.
E=VIt=12.0×4.00×(5.00×60)=14400 J
M1 — correct use of E=VIt with time converted to seconds. Common error: leaving t in minutes, giving 240 J — an immediate factor-of-60 slip.
A1 — correct numerical value 14 400 J (or 1.44 × 10⁴ J).
Step 2 — apply Q=mcΔθ assuming all energy heats the block.
Δθ=32.5−19.0=13.5 K
c=mΔθQ=1.00×13.514400=1067 J kg−1 K−1
M1 — correct rearrangement of Q=mcΔθ and substitution of m, Δθ. The temperature change in K and °C is identical, so no conversion of Δθ is required — examiners look for awareness, not unnecessary working.
A1 — final answer to 3 sig fig: c≈1070 J kg⁻¹ K⁻¹ (accept 1066 to 1070 inclusive).
(b) Step 1 — direction of error.
If heat is lost to the surroundings, the energy actually delivered to the block is less than VIt. Using the full VIt in c=VIt/(mΔθ) therefore overestimates the numerator and gives a value of c that is too high.
B1 — correct direction of the systematic error, with reasoning that the calculated c is inflated because not all the electrical input contributed to Δθ.
Step 2 — corrected energy balance.
If 10% of the input energy is lost, the energy actually absorbed by the block is 0.90×14400=12960 J.
M1 — correct identification that 90% of VIt corresponds to the useful heating energy.
Step 3 — recompute c.
ccorrected=1.00×13.512960=960 J kg−1 K−1
A1 — corrected value 960 J kg⁻¹ K⁻¹, much closer to the accepted 900 J kg⁻¹ K⁻¹.
B1 — explanation that the residual difference (≈60 J kg⁻¹ K⁻¹) reflects further losses (energy absorbed by the heater itself, the thermometer, and uneven temperature distribution within the block).
Total: 8 marks (M3 A2 B3 split as shown).
Question (6 marks): A copper calorimeter of mass 0.080 kg contains 0.150 kg of water at 18.0 °C. A 0.050 kg metal sample at 90.0 °C is dropped into the water. The final equilibrium temperature is 22.4 °C. The specific heat capacity of water is 4200 J kg⁻¹ K⁻¹ and of copper is 385 J kg⁻¹ K⁻¹.
(a) State the principle on which the calculation of the specific heat capacity of the metal sample is based. (1)
(b) Calculate the specific heat capacity of the metal sample. (4)
(c) Suggest one reason why the experimental value may differ from the accepted value for the metal. (1)
Mark scheme decomposition by AO:
(a)
(b)
(c)
Total: 6 marks split AO1 = 4, AO2 = 1, AO3 = 1. This is an AO1-dominant question with method-of-mixtures wording — Edexcel uses these to test routine application of Q=mcΔθ alongside conservation of energy and qualitative awareness of systematic error.
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