Specific Heat Capacity

Spec mapping: OCR H556 Module 5.1 — Thermal physics (specific heat capacity defined as the energy per kilogram per kelvin required to raise the temperature of a substance without change of state; the equation $Q = mc\Delta\theta$Q=mcΔθ; method-of-mixtures and electrical methods for measuring $c$c). PAG 7 anchors the practical work. Refer to the official OCR H556 specification document for exact wording.

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, the central quantitative concept of Module 5.1 of the OCR A-Level Physics A specification (H556).

This lesson develops the equation $Q = mc\Delta\theta$Q=mcΔθ, works through a large number of numerical examples, and describes in detail how specific heat capacity is measured in the laboratory. The practical work is anchored in PAG 7 — measuring $c$c of water (or a metal block) using the electrical method — and you should be ready to discuss its sources of error in detail.

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Defining Specific Heat Capacity

Specific heat capacity $c$c of a substance is the energy required to raise the temperature of 1 kg of the substance by 1 K (or equivalently 1 $^{\circ}$∘C), without a change of state.

The symbol is lowercase $c$c (do not confuse it with the speed of light, which in thermal-physics context never arises). The SI units are J kg$^{-1}$−1 K$^{-1}$−1 (equivalently J kg$^{-1}$−1 $^{\circ}$∘C$^{-1}$−1 — both are correct because 1 K and 1 $^{\circ}$∘C are the same size of interval).

Water, for example, has $c = 4200$c=4200 J kg$^{-1}$−1 K$^{-1}$−1 — a very high value, which is why the oceans act as thermal reservoirs that moderate the climate. Aluminium has $c \approx 900$c≈900 J kg$^{-1}$−1 K$^{-1}$−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 $m$m kilograms of a substance of specific heat capacity $c$c receives an energy $Q$Q and rises in temperature by $\Delta\theta$Δθ, then

$Q = m c \Delta \theta.$Q=mcΔθ.

Let us parse this equation carefully:

• $Q$Q is the energy (in joules) supplied to the substance (the OCR data sheet uses $E$E; either is acceptable).
• $m$m is the mass of the substance (in kg).
• $c$c is the specific heat capacity (in J kg$^{-1}$−1 K$^{-1}$−1).
• $\Delta\theta$Δθ is the change in temperature, $\Delta\theta = \theta_{\text{final}} - \theta_{\text{initial}}$Δθ=θfinal​−θinitial​.

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

Exam Tip: Remember that in $Q = mc\Delta\theta$Q=mcΔθ, the symbol is $\Delta\theta$Δθ (theta), not $\Delta T$ΔT. On the OCR data sheet it is written with theta for a reason: historically $\theta$θ denotes Celsius temperature, and the equation uses differences which are identical in $^{\circ}$∘C and K. You may write $\Delta T$ΔT if you prefer, provided you are consistent.

Single-phase only. $Q = mc\Delta\theta$Q=mcΔθ applies only inside a single phase (solid, liquid, or gas). During a phase change the temperature does not change and this formula gives the meaningless answer $Q = 0$Q=0 for any finite mass undergoing melting or boiling. Phase changes need $Q = mL$Q=mL (lesson 4).

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A Table of Common Values

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

Decision tree — which equation do I use?

flowchart TD
    Q["Energy supplied: how much?"] --> A{"Is the substance changing phase?"}
    A -- "No (single phase)" --> B["Use Q = m c Delta-theta"]
    A -- "Yes (melting / boiling)" --> C["Use Q = m L"]
    B --> D["Delta-theta non-zero; temperature rises"]
    C --> E["Delta-theta = 0; latent heat only"]

Multi-stage problems (e.g. ice → water → steam) need both equations, stage by stage, with $Q$Q summed across stages. See worked example 7 below and lesson 4 for the full treatment.

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Worked Example 1 — Heating Water

How much energy is required to heat 0.50 kg of water from $20\,^{\circ}$20∘C to $100\,^{\circ}$100∘C?

$Q = mc\Delta\theta = (0.50)(4200)(100 - 20) = (0.50)(4200)(80) = 168\,000 \text{ J} = 168 \text{ kJ}.$Q=mcΔθ=(0.50)(4200)(100−20)=(0.50)(4200)(80)=168000 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 $c = 900$c=900 J kg$^{-1}$−1 K$^{-1}$−1.

First find the energy supplied: $Q = Pt = (2.0 \times 10^{3})(30) = 60\,000$Q=Pt=(2.0×103)(30)=60000 J.

Then rearrange $Q = mc\Delta\theta$Q=mcΔθ:

$\Delta\theta = \frac{Q}{mc} = \frac{60\,000}{(1.5)(900)} = \frac{60\,000}{1350} \approx 44.4 \text{ K}.$Δθ=mcQ​=(1.5)(900)60000​=135060000​≈44.4 K.

So the block heats up by about 44 K (equivalently 44 $^{\circ}$∘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 s, the water temperature rises from $20\,^{\circ}$20∘C to $80\,^{\circ}$80∘C. Assuming all the microwave energy goes into the water, what is the mass of water in the mug?

$Q = Pt = (800)(90) = 72\,000 \text{ J}, \qquad \Delta\theta = 80 - 20 = 60 \text{ K}.$Q=Pt=(800)(90)=72000 J,Δθ=80−20=60 K.

$m = \frac{Q}{c\Delta\theta} = \frac{72\,000}{(4200)(60)} = \frac{72\,000}{252\,000} \approx 0.286 \text{ kg}.$m=cΔθQ​=(4200)(60)72000​=25200072000​≈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\,^{\circ}$80∘C with 0.20 kg of water at $20\,^{\circ}$20∘C in an insulated container. Find the final temperature once thermal equilibrium is reached.

Let the final temperature be $\theta_f$θf​. Heat lost by the hot water equals heat gained by the cold water (no loss to surroundings):

$m_1 c (\theta_1 - \theta_f) = m_2 c (\theta_f - \theta_2).$m1​c(θ1​−θf​)=m2​c(θf​−θ2​).

$(0.30)(4200)(80 - \theta_f) = (0.20)(4200)(\theta_f - 20).$(0.30)(4200)(80−θf​)=(0.20)(4200)(θf​−20).

The $4200$4200 cancels. Solving:

$(0.30)(80 - \theta_f) = (0.20)(\theta_f - 20) \Rightarrow 24 - 0.30\theta_f = 0.20\theta_f - 4.0$(0.30)(80−θf​)=(0.20)(θf​−20)⇒24−0.30θf​=0.20θf​−4.0

$28 = 0.50\theta_f \Rightarrow \theta_f = 56\,^{\circ}\text{C}.$28=0.50θf​⇒θf​=56∘C.

Notice that the $4200$4200 cancelled because we were mixing two bodies of the same substance. The final temperature is not the simple average $50\,^{\circ}$50∘C, because the hot mass (0.30 kg) was larger than the cold mass (0.20 kg). The weighted average gives $56\,^{\circ}$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\,^{\circ}$200∘C is dropped into 0.50 kg of water at $15\,^{\circ}$15∘C in an insulated container. Find the final temperature. ($c_{\text{water}} = 4200$cwater​=4200 J kg$^{-1}$−1 K$^{-1}$−1, $c_{\text{Cu}} = 385$cCu​=385 J kg$^{-1}$−1 K$^{-1}$−1.)

Heat lost by copper $=$= heat gained by water:

$m_{\text{Cu}} c_{\text{Cu}} (200 - \theta_f) = m_w c_w (\theta_f - 15).$mCu​cCu​(200−θf​)=mw​cw​(θf​−15).

$(0.10)(385)(200 - \theta_f) = (0.50)(4200)(\theta_f - 15).$(0.10)(385)(200−θf​)=(0.50)(4200)(θf​−15).

$38.5(200 - \theta_f) = 2100(\theta_f - 15).$38.5(200−θf​)=2100(θf​−15).

$7700 - 38.5\theta_f = 2100\theta_f - 31\,500 \Rightarrow 39\,200 = 2138.5\theta_f \Rightarrow \theta_f \approx 18.3\,^{\circ}\text{C}.$7700−38.5θf​=2100θf​−31500⇒39200=2138.5θf​⇒θf​≈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 — the product $mc$mc is what determines how much energy a body can hold for a given $\Delta\theta$Δθ.

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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 (PAG 7)

flowchart TD
    PSU["Low-voltage power supply"] --> A["Ammeter (measure I)"]
    A --> H["Immersion heater (inside block)"]
    H --> V["Voltmeter (measure V across heater)"]
    V --> PSU
    H -. inside .-> BLK["Metal block with drilled holes"]
    BLK --> T["Thermometer (in second hole, with thermal paste)"]
    BLK -.- LAG["Insulating lagging"]

A cylindrical metal block is drilled with two holes: one for an electric immersion heater and one for a thermometer. A known electrical power is supplied via a low-voltage power supply, an ammeter and a voltmeter. The block is lagged with insulating material to reduce heat loss to the surroundings. A small drop of thermal paste or oil in the thermometer hole improves heat conduction between the block and the thermometer bulb.

Procedure

1. Record the initial temperature $\theta_1$θ1​.
2. Switch on the heater. Note the current $I$I and the voltage $V$V.
3. Heat for a measured time $t$t.
4. Record the final temperature $\theta_2$θ2​. (Often you continue reading for a little while after switching off to capture the maximum, because heat continues to flow from the heater into the block.)
5. Determine the energy supplied: $Q = IVt$Q=IVt.
6. Calculate $c$c using

$c = \frac{Q}{m\Delta\theta} = \frac{IVt}{m(\theta_2 - \theta_1)}.$c=mΔθQ​=m(θ2​−θ1​)IVt​.

Sources of error and how to reduce them

1. Heat loss to the surroundings. Even with lagging, some thermal energy will leak out through the lagging or from the top of the block. This causes the measured temperature rise to be smaller than it would otherwise have been, so the calculated specific heat capacity will be larger than the true value. To reduce: use good lagging, start the experiment a little below room temperature and finish a little above room temperature so heat gained and lost approximately cancel.
2. Heat capacity of the heater itself. A small fraction of the electrical energy is absorbed by the heater and the wires, not by the block. Using a low-mass heater minimises this; you can also estimate the heater's heat capacity and subtract.
3. Non-uniform temperature. If the heater has only just switched off and the thermometer is at the other end of the block, the reading may not reflect the true average temperature. Stirring a liquid, or waiting for the block to equilibrate, helps.
4. Thermometer reading errors. A digital thermometer reading to 0.1 K is much better than an analogue one reading to 1 K, especially when $\Delta\theta$Δθ is small. Use a larger $\Delta\theta$Δθ to reduce the percentage uncertainty.

Apparatus for a liquid

For a liquid such as water, the apparatus is similar but the immersion heater and thermometer are placed directly in a beaker of the liquid, which is stirred continuously to keep the temperature uniform. The beaker sits on an insulating mat to reduce heat loss to the bench, and is surrounded by lagging or a vacuum flask.

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Worked Example 6 — An Experiment on Aluminium (PAG 7)

An aluminium block of mass 1.00 kg is heated by a 12 V, 4.0 A immersion heater for 120 s. The temperature rises from $19.0\,^{\circ}$19.0∘C to $25.4\,^{\circ}$25.4∘C. What value of $c$c does this experiment suggest?

$Q = IVt = (4.0)(12)(120) = 5760 \text{ J}.$Q=IVt=(4.0)(12)(120)=5760 J. $\Delta\theta = 25.4 - 19.0 = 6.4 \text{ K}.$Δθ=25.4−19.0=6.4 K. $c = \frac{Q}{m\Delta\theta} = \frac{5760}{(1.00)(6.4)} = 900 \text{ J kg}^{-1}\text{ K}^{-1}.$c=mΔθQ​=(1.00)(6.4)5760​=900 J kg−1 K−1.

Comparing with the accepted value ($\approx 897$≈897 J kg$^{-1}$−1 K$^{-1}$−1 for pure aluminium), the experiment is clearly successful to within a few percent. Any small discrepancy can be explained by heat loss to the surroundings and by the heat capacity of the heater itself.

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Worked Example 7 — Method of Mixtures with $\Delta\theta$Δθ both ways