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Physics · Thermal Physics, Waves and Oscillations

Heat Capacity and Specific Heat Capacity

Heat capacity as a property of an object and specific heat capacity as a property of a material, the relation Q equals m c delta T, and the electrical and method-of-mixtures experiments with their sources of error.

Heat Capacity and Specific Heat Capacity

The previous chapter established that heat is energy transferred because of a temperature difference. This chapter answers the quantitative question that immediately follows: how much heat does it take to change a body's temperature by a given amount? The answer depends on how much material there is, on what the material is, and — for gases — on whether the material is allowed to expand while it is heated. Getting a numerical handle on all three is the business of the two quantities in the title.

These are among the most heavily examined ideas in the whole block, partly because the arithmetic is straightforward and partly because the experiments to measure them — the electrical method and the method of mixtures — are a rich source of "explain the sources of error" questions. Both experiments are worked through in detail here.

Reading note. Keep the distinction between CC (a property of an object) and cc (a property of a material) in view from the start. Almost every mistake in this topic is a confusion between the two, or a slip in which mass or temperature difference has been left in the wrong units.

1. Heat Capacity of a Body

The heat capacity of a particular object is the heat needed to raise its temperature by one kelvin.

Definition. The heat capacity CC of a body is

C=QΔT[JK]C = \frac{Q}{\Delta T} \qquad \left[\dfrac{\text{J}}{\text{K}}\right]

where QQ is the heat supplied and ΔT\Delta T the resulting temperature change.

Heat capacity is a property of that object — this kettle, this block of aluminium, this calorimeter — and it bundles together the object's mass and the material it is made of. A large copper tank and a small copper coin are the same material but have very different heat capacities. Because a temperature difference is the same number in kelvin and in degrees Celsius, ΔT\Delta T may be measured on either scale without conversion.

A negative QQ (heat removed) produces a negative ΔT\Delta T (cooling), and CC itself is always positive.

2. Specific Heat Capacity of a Material

To separate out the amount of substance, divide by the mass. The specific heat capacity is the heat needed to raise the temperature of one kilogram of a material by one kelvin.

Definition. The specific heat capacity cc of a material is

c=Qm ΔT[Jkg⋅K]c = \frac{Q}{m\,\Delta T} \qquad \left[\dfrac{\text{J}}{\text{kg}\cdot\text{K}}\right]

so that the heat exchanged when a mass mm changes temperature by ΔT\Delta T is

Q=mc ΔTQ = mc\,\Delta T

Specific heat capacity is now a property of the material alone, independent of how much of it there is. The heat capacity of an object and the specific heat capacity of its material are linked by

C=mcC = mc

Some representative values, worth a rough memory: water, 4180 Jkg⋅K4180\ \dfrac{\text{J}}{\text{kg}\cdot\text{K}}; aluminium, 900900; copper, 385385; lead, 130130. Water's value is strikingly large — larger than almost any common substance — which is why it is used as a coolant in car engines and power stations, why coastal climates are milder than inland ones, and why a hot-water bottle stays warm for hours.

Worked example. A 2.5 kW2.5\ \text{kW} kettle contains 0.80 kg0.80\ \text{kg} of water at 18 ∘C18\ ^{\circ}\text{C}. Assuming all the electrical energy heats the water, how long does it take to reach 100 ∘C100\ ^{\circ}\text{C}?

Q=mc ΔT=0.80 (4180)(100−18)=2.74×105 JQ = mc\,\Delta T = 0.80\,(4180)(100 - 18) = 2.74\times 10^{5}\ \text{J} t=QP=2.74×1052500=1.1×102 st = \frac{Q}{P} = \frac{2.74\times 10^{5}}{2500} = 1.1\times 10^{2}\ \text{s}

about two minutes — an underestimate in practice, because some heat is lost to the kettle body and the surroundings.

3. Molar Heat Capacity and the Special Case of Gases

For chemists and for gas problems it is often more natural to count in moles than in kilograms. The molar heat capacity Cm=McC_m = Mc (where MM is the molar mass) is the heat per mole per kelvin. For a monatomic ideal gas heated at constant volume, the previous chapter's result U=32nRTU = \tfrac{3}{2}nRT gives immediately

CV,m=1nQΔT=1nΔUΔT=32R≈12.5 Jmol⋅KC_{V,m} = \frac{1}{n}\frac{Q}{\Delta T} = \frac{1}{n}\frac{\Delta U}{\Delta T} = \tfrac{3}{2}R \approx 12.5\ \dfrac{\text{J}}{\text{mol}\cdot\text{K}}

Key distinction. A gas has two principal specific heat capacities, not one. Heated at constant volume, all the energy raises the internal energy. Heated at constant pressure, the gas also expands and does work pushing back its surroundings, so more heat is needed for the same temperature rise: cp>cVc_p > c_V. For solids and liquids the expansion is tiny and the distinction is usually ignored.

4. Measuring Specific Heat Capacity: The Electrical Method

An electrical heater of known power is embedded in the sample, with a thermometer alongside. Energy conservation gives

VIt=mc ΔT+(energy lost to surroundings)VIt = mc\,\Delta T + (\text{energy lost to surroundings})

If losses could be eliminated, c=VItm ΔTc = \dfrac{VIt}{m\,\Delta T}.

Worked example. A 1.00 kg1.00\ \text{kg} aluminium block is heated by a 12.0 V12.0\ \text{V}, 4.00 A4.00\ \text{A} heater for 5.0 minutes5.0\ \text{minutes}. The temperature rises by 16.0 K16.0\ \text{K}. Estimate the specific heat capacity of aluminium.

VIt=12.0 (4.00)(300)=1.44×104 JVIt = 12.0\,(4.00)(300) = 1.44\times 10^{4}\ \text{J} c=VItm ΔT=1.44×1041.00 (16.0)=9.0×102 Jkg⋅Kc = \frac{VIt}{m\,\Delta T} = \frac{1.44\times 10^{4}}{1.00\,(16.0)} = 9.0\times 10^{2}\ \dfrac{\text{J}}{\text{kg}\cdot\text{K}}

This happens to match the accepted value well, but in a real run the answer comes out too high: some of the VItVIt is lost to the room rather than going into the block, so the measured ΔT\Delta T is smaller than it should be for that energy input, and VItm ΔT\dfrac{VIt}{m\,\Delta T} is inflated.

Reducing the error. Lag the block with insulation; start the experiment a few kelvin below room temperature and finish an equal amount above it, so that heat gained from the room in the first half roughly cancels heat lost in the second; use a large mass so the loss is a smaller fraction of the total; and correct for the heat capacity of the heater and thermometer themselves.

5. Measuring Specific Heat Capacity: The Method of Mixtures

A hot body is dropped into a cooler liquid in an insulated container (a calorimeter), and the mixture is stirred until it reaches a common final temperature TfT_f. If the container is well insulated, no energy leaves the system, so the heat lost by the hot body equals the heat gained by everything else:

m1c1(T1−Tf)⏟lost by hot body  =  m2c2(Tf−T2)⏟gained by liquid  +  Ccal(Tf−T2)⏟gained by calorimeter\underbrace{m_1 c_1 (T_1 - T_f)}_{\text{lost by hot body}} \;=\; \underbrace{m_2 c_2 (T_f - T_2)}_{\text{gained by liquid}} \;+\; \underbrace{C_{\text{cal}} (T_f - T_2)}_{\text{gained by calorimeter}}

Equivalently, writing every term as "heat gained" with its natural sign, ∑mc ΔT=0\sum m c\,\Delta T = 0.

Worked example. A 0.150 kg0.150\ \text{kg} copper block (c=385 Jkg⋅Kc = 385\ \dfrac{\text{J}}{\text{kg}\cdot\text{K}}) at 95.0 ∘C95.0\ ^{\circ}\text{C} is dropped into 0.200 kg0.200\ \text{kg} of water (c=4180c = 4180) at 18.0 ∘C18.0\ ^{\circ}\text{C}, held in a calorimeter of heat capacity 40 JK40\ \dfrac{\text{J}}{\text{K}} also at 18.0 ∘C18.0\ ^{\circ}\text{C}. Find the final temperature.

Balancing heat lost against heat gained:

0.150 (385)(95.0−Tf)=0.200 (4180)(Tf−18.0)+40 (Tf−18.0)0.150\,(385)(95.0 - T_f) = 0.200\,(4180)(T_f - 18.0) + 40\,(T_f - 18.0) 57.75 (95.0−Tf)=876 (Tf−18.0)57.75\,(95.0 - T_f) = 876\,(T_f - 18.0) 5486−57.75 Tf=876 Tf−157685486 - 57.75\,T_f = 876\,T_f - 15768 Tf=5486+15768876+57.75=21254933.75=22.8 ∘CT_f = \frac{5486 + 15768}{876 + 57.75} = \frac{21254}{933.75} = 22.8\ ^{\circ}\text{C}

Neglecting the calorimeter's 40 JK40\ \dfrac{\text{J}}{\text{K}} would give 23.0 ∘C23.0\ ^{\circ}\text{C} — a small but real systematic shift, and exactly the kind of term examiners check you have included.

6. Sources of Error Common to Both Methods

  • Heat exchange with the surroundings — the dominant error. Minimised by lagging, lids, and the "start below, end above room temperature" trick.
  • The heat capacity of the container, heater, thermometer and stirrer. Each absorbs some energy; a good experiment measures or estimates these and includes them.
  • Incomplete or non-uniform mixing, so the thermometer does not read the true mean temperature. Stirring helps.
  • Evaporation of the liquid, which removes latent heat (next chapter) and is easy to overlook.
  • Transfer losses in the method of mixtures: the hot body cools slightly while being moved from the heater to the calorimeter.
  • Thermometer resolution and reading lag: the temperature may still be changing when it is read.

Common Exam Pitfalls

  • Confusing heat capacity CC with specific heat capacity cc. CC belongs to an object and has units JK\dfrac{\text{J}}{\text{K}}; cc belongs to a material and has units Jkg⋅K\dfrac{\text{J}}{\text{kg}\cdot\text{K}}. They are related by C=mcC = mc.
  • Leaving mass in grams. Q=mc ΔTQ = mc\,\Delta T needs mass in kilograms when cc is in Jkg⋅K\dfrac{\text{J}}{\text{kg}\cdot\text{K}}.
  • Forgetting the calorimeter in a method-of-mixtures calculation. Its heat capacity term is small but systematic.
  • Sign confusion in the mixtures equation. Decide once whether each term is "heat lost" or "heat gained" and keep it consistent; the hot body's ΔT\Delta T and the cold body's ΔT\Delta T have opposite signs.
  • Assuming no heat is lost and then being surprised the measured cc is too high in the electrical method — heat loss inflates VItm ΔT\dfrac{VIt}{m\,\Delta T}.
  • Using Q=mc ΔTQ = mc\,\Delta T across a change of state. During melting or boiling the temperature does not change; that is latent heat, covered next.
  • Treating a gas as having a single specific heat capacity. Specify constant volume or constant pressure.

Summary

  • Heat capacity of a body: C=QΔTC = \dfrac{Q}{\Delta T}, unit JK\dfrac{\text{J}}{\text{K}} — a property of the object.
  • Specific heat capacity of a material: c=Qm ΔTc = \dfrac{Q}{m\,\Delta T}, unit Jkg⋅K\dfrac{\text{J}}{\text{kg}\cdot\text{K}} — a property of the material, with Q=mc ΔTQ = mc\,\Delta T and C=mcC = mc.
  • Water has an unusually large cc (4180 Jkg⋅K4180\ \dfrac{\text{J}}{\text{kg}\cdot\text{K}}), which underlies its use as a coolant and the mildness of maritime climates.
  • Molar heat capacity Cm=McC_m = Mc; a monatomic ideal gas at constant volume has CV,m=32RC_{V,m} = \tfrac{3}{2}R. A gas has cp>cVc_p > c_V because heating at constant pressure also does expansion work.
  • Electrical method: VIt=mc ΔT+lossesVIt = mc\,\Delta T + \text{losses}; heat loss makes the measured cc come out too large.
  • Method of mixtures: heat lost by the hot body = heat gained by the liquid + calorimeter; ∑mc ΔT=0\sum mc\,\Delta T = 0 for an insulated system.
  • The dominant experimental error in both methods is heat exchange with the surroundings; the heat capacity of the apparatus is the standard systematic correction.