Specific Heat Capacity

Different substances respond very differently to being heated. Pour 1 kJ of energy into 1 kg of water and its temperature rises by about a quarter of a degree. Pour the same 1 kJ into 1 kg of aluminium, and the temperature rises by more than one full degree. Pour it into 1 kg of mercury, and you will get a rise of about 7 degrees. The quantity that captures this difference is the specific heat capacity, introduced in Module 5.1.2 of the OCR A-Level Physics A specification (H556).

This lesson develops the equation E = mcΔθ, works through a large number of numerical examples, and describes in detail how specific heat capacity is measured in the laboratory.

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Defining Specific Heat Capacity

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 equivalently 1 °C), without a change of state.

The symbol is lowercase c (do not confuse it with the speed of light, which in thermal physics context never arises). The SI units are

[c] = J kg⁻¹ K⁻¹  (equivalently J kg⁻¹ °C⁻¹)

Water, for example, has c = 4200 J kg⁻¹ K⁻¹ — a very high value, which is why the oceans act as thermal reservoirs that moderate the climate. Aluminium has c ≈ 900 J kg⁻¹ K⁻¹, copper 385, iron 440, lead 128, mercury 140.

The word "specific" here is a historical technical term meaning "per unit mass". "Specific heat capacity" therefore literally means "heat capacity per unit mass".

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The Defining Equation

From the definition, if m kilograms of a substance of specific heat capacity c receives an energy E and rises in temperature by Δθ, then

E = m c Δθ

Let us parse this equation carefully:

• E is the energy (in joules) supplied to the substance.
• m is the mass of the substance (in kg).
• c is the specific heat capacity (in J kg⁻¹ K⁻¹).
• Δθ is the change in temperature, Δθ = θ_final - θ_initial.

Because we are dealing with a difference in temperature, we can use Celsius or kelvin interchangeably here — the size of one degree Celsius equals the size of one kelvin, so Δθ is the same number in both.

Exam Tip: Remember that in E = mcΔθ, the symbol is Δθ (theta), not ΔT. On the OCR data sheet it is written with theta for a reason: historically θ denotes Celsius temperature, and the equation uses differences which are identical in °C and K. You may write ΔT if you prefer, provided you are consistent.

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A Table of Common Values

Substance	Specific heat capacity c (J kg⁻¹ K⁻¹)
Water (liquid)	4200
Ice	2100
Steam	2000
Ethanol	2400
Aluminium	900
Glass	840
Iron / Steel	440
Copper	385
Brass	380
Mercury	140
Lead	128

Note how high the value for water is — higher than any common liquid or solid except ammonia. This high specific heat capacity explains why water is such an effective coolant (car engines, power-station condensers), why the sea warms and cools slowly with the seasons, and why coastal climates are more equable than inland ones.

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Worked Example 1: Heating Water

How much energy is required to heat 0.50 kg of water from 20 °C to 100 °C?

E = m c Δθ
  = (0.50)(4200)(100 - 20)
  = (0.50)(4200)(80)
  = 168 000 J
  = 168 kJ

So 168 kJ is needed just to warm half a litre of water from room temperature to boiling, quite apart from the much larger energy needed actually to boil it (Lesson 4). Practical electric kettles have a power of 2–3 kW; this explains why a half-litre of water takes around a minute to boil.

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Worked Example 2: Finding the Temperature Rise

A 2.0 kW heater is used to warm a 1.5 kg block of aluminium for 30 seconds. Assuming no heat loss, find the temperature rise of the block. Take c = 900 J kg⁻¹ K⁻¹.

First find the energy supplied:

E = Pt = (2.0 × 10³)(30) = 60 000 J

Then rearrange E = mcΔθ:

Δθ = E / (mc)
   = 60000 / ((1.5)(900))
   = 60000 / 1350
   ≈ 44.4 K

So the block heats up by about 44 K (equivalently 44 °C).

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Worked Example 3: Finding the Mass

A quantity of water in a mug is heated by a microwave oven delivering 800 W. Over 90 seconds, the water temperature rises from 20 °C to 80 °C. Assuming all the microwave energy goes into the water, what is the mass of water in the mug?

E = Pt = (800)(90) = 72 000 J
Δθ = 80 - 20 = 60 K

m = E / (cΔθ)
  = 72000 / ((4200)(60))
  = 72000 / 252000
  ≈ 0.286 kg

So about 0.29 kg (roughly a large mug's worth) of water.

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Worked Example 4: Mixing Hot and Cold

You mix 0.30 kg of water at 80 °C with 0.20 kg of water at 20 °C in an insulated container. Find the final temperature once thermal equilibrium is reached.

Let the final temperature be θ_f. Heat lost by the hot water equals heat gained by the cold water (no loss to surroundings):

m₁ c (θ₁ - θ_f) = m₂ c (θ_f - θ₂)
(0.30)(4200)(80 - θ_f) = (0.20)(4200)(θ_f - 20)
(0.30)(80 - θ_f) = (0.20)(θ_f - 20)
24 - 0.30 θ_f = 0.20 θ_f - 4.0
28 = 0.50 θ_f
θ_f = 56 °C

Notice that the 4200 (specific heat capacity of water) cancelled because we were mixing two bodies of the same substance. The final temperature is not the simple average 50 °C, because the hot mass (0.30 kg) was larger than the cold mass (0.20 kg). The weighted average gives 56 °C — closer to 80 than to 20, as you would expect.

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Worked Example 5: Mixing Different Substances

A 0.10 kg lump of copper at 200 °C is dropped into 0.50 kg of water at 15 °C in an insulated container. Find the final temperature. (c_water = 4200 J kg⁻¹ K⁻¹, c_copper = 385 J kg⁻¹ K⁻¹.)

Again, heat lost = heat gained:

m_Cu c_Cu (200 - θ_f) = m_w c_w (θ_f - 15)
(0.10)(385)(200 - θ_f) = (0.50)(4200)(θ_f - 15)
38.5 (200 - θ_f) = 2100 (θ_f - 15)
7700 - 38.5 θ_f = 2100 θ_f - 31500
39200 = 2138.5 θ_f
θ_f ≈ 18.3 °C

The copper, though originally 185 K hotter than the water, raised the water temperature by only about 3 K. The reason is twofold: the copper mass is much smaller than the water mass, and the specific heat capacity of copper is about eleven times smaller than that of water. Both factors mean that copper carries relatively little thermal energy for its size.

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Experimental Determination of c

The OCR specification requires you to know how specific heat capacity is measured in the laboratory. There are several standard arrangements. The central idea is to supply a known amount of electrical energy to the substance and measure the resulting temperature rise.

Apparatus for a solid