Thermal Properties of Matter · Lesson 3 of 8
Heat Capacity and Calorimetry
“Relate energy input to temperature change, distinguish specific and molar heat capacities, and determine unknown properties by complete calorimeter energy balances.”
• Define heat capacity, specific heat and molar heat capacity with their units and process conditions. • Interpret controlled heating experiments and compare solids, liquids and gases. • Include the vessel and water equivalent in calorimetry and assess heat-loss bias. • Distinguish energy transferred as heat from mechanical work converted to internal energy.
How much energy produces a temperature rise?
Put equal masses of water and oil on identical controlled heaters. Even if each absorbs the same energy, their temperatures need not rise equally. A material property is involved. Now heat twice as much water through the same temperature interval: approximately twice the energy is required. These observations motivate a model in which transferred energy depends on mass, material and temperature change, provided no change of state occurs.
Heat capacity belongs to the complete body: double its mass and its heat capacity doubles. Specific heat belongs to a unit mass under specified conditions and is approximately independent of sample size. A temperature rise in kelvins equals the same numerical rise in Celsius degrees. Energy in joules divided by mass in kilograms and a temperature change in kelvins therefore gives J kg⁻¹ K⁻¹. A value in J g⁻¹ K⁻¹ must be multiplied by 1000 before combining it with masses in kilograms.
For warming, ΔT and heat gained are positive; for cooling they are negative if heat into the body is the positive convention. In elementary calorimetry it is often clearer to equate positive magnitudes of energy lost and gained. The formula assumes a single state and an approximately constant specific heat over the interval. At a phase transition, heat can flow with no temperature change, so dividing that finite energy by zero is not a useful specific-heat model.
A controlled heating investigation
Use a measured mass of water in an insulated vessel, a thermometer and a heater whose absorbed power is known or approximately constant. Stir gently so the measured temperature represents the sample. Record the time for a chosen rise, such as from 20 °C to 40 °C. Repeat with the original water mass over twice the temperature rise, then with twice the water mass over the original rise, and finally with the original mass of oil under comparable conditions. Account for the vessel and allow for environmental losses rather than assuming the electrical rating is exactly the power absorbed by the liquid.
At constant absorbed power, input energy is power multiplied by time. Twice the temperature rise should require about twice the energy for the same mass, while twice the mass should take about twice the time for the same rise if other heat capacities and losses are negligible. Different times for equal masses of water and oil reveal different specific heats. A graph of temperature against absorbed energy has slope 1/(ms); against time its slope is P/(ms). Larger heat capacity means a shallower rising line. Deviations can reveal heat losses, an unaccounted vessel or changing specific heat.
| Substance near ordinary temperatures | Specific heat / J kg⁻¹ K⁻¹ |
|---|---|
| Aluminium | 900 |
| Carbon | 506.5 |
| Copper | 386.4 |
| Lead | 127.7 |
| Silver | 236.1 |
| Tungsten | 134.4 |
| Water | 4186 |
| Ice | 2060 |
| Glass | 840 |
| Iron | 450 |
| Kerosene | 2118 |
| Edible oil | 1965 |
| Mercury | 140 |
Water’s large specific heat means that it gains or loses much energy for a modest temperature change. That helps explain its usefulness in cooling and heating systems and its role in moderating coastal temperature variations. Ice and water have different values because state matters. The table is representative, not a set of exact constants valid at every temperature. Use explicitly supplied problem data when they differ slightly from these rounded values.
Molar heat capacity and why the process matters
A gas warmed at fixed volume cannot push its boundary outward. At fixed pressure it normally expands and does work on its surroundings. Equal temperature rises can therefore require different heat inputs. Denote the molar capacities by CV for constant volume and CP for constant pressure. The heat supplied at constant pressure must provide both the internal-energy increase and the expansion work, explaining why CP is larger in the usual ideal-gas comparison. Do not use a gas heat capacity without specifying its process.
| Gas | CP / J mol⁻¹ K⁻¹ | CV / J mol⁻¹ K⁻¹ |
|---|---|---|
| Helium | 20.8 | 12.5 |
| Hydrogen | 28.8 | 20.4 |
| Nitrogen | 29.1 | 20.8 |
| Oxygen | 29.4 | 21.1 |
| Carbon dioxide | 37.0 | 28.5 |
The table also shows that molecular structure matters. Translational motion has three independent directions. A diatomic molecule can additionally store energy in rotation, so the same energy per mole produces a smaller temperature rise than in a monatomic gas. In the simple ordinary-temperature idealization with two active rotational contributions, CV ≈ (5/2)R ≈ 20.8 J mol⁻¹ K⁻¹, or 4.97 cal mol⁻¹ K⁻¹. The monatomic comparison is (3/2)R ≈ 2.98 cal mol⁻¹ K⁻¹. This is a brief interpretation of the data; detailed molecular statistics belongs to kinetic theory.
Problem
Measured constant-volume molar capacities in cal mol⁻¹ K⁻¹ are H₂ 4.87, N₂ 4.97, O₂ 5.02, NO 4.99, CO 5.01 and Cl₂ 6.17. Why are most similar, and why can chlorine be higher?
- 1.Most values cluster near the diatomic translation-plus-rotation estimate (5/2)R ≈ 4.97 cal mol⁻¹ K⁻¹.
- 2.The additional rotational energy storage makes their capacities larger than a monatomic translation-only value.
- 3.Molecular vibration can contribute when thermally active. Its different excitation scale can give chlorine an additional contribution in the stated range.
- 4.Do not infer that every diatomic molecule has exactly the same capacity at every temperature. The data are approximate and depend on which internal motions can absorb energy.
Calorimetry: make the energy balance complete
A calorimeter measures thermal energy exchanges using known masses, temperatures and capacities. Its vessel, lid, thermometer and stirrer can absorb energy too. For an ideally insulated mixing experiment with no mass loss, no phase change and negligible other work, conservation of energy says that the energy lost by the hotter bodies equals the energy gained by the colder bodies. Insulation alone means approximately adiabatic; the full balance also requires controlling work and matter transfer.
Choose the final common temperature as unknown or use it to infer an unknown specific heat. Each temperature difference must refer to the body whose energy it describes. The hot metal does not cool through the same interval by which the water warms. Check that a passive single-state mixture finishes between its initial temperatures; a result outside that range signals an incorrect sign, missing phase change or an external energy source.
Problem
A 0.047 kg aluminium sphere at 100 °C is placed in 0.25 kg of water in a 0.14 kg copper calorimeter, both initially at 20 °C. The final temperature is 23 °C. Use cwater = 4180 and ccopper = 386 J kg⁻¹ K⁻¹.
- 1.The water and copper together have heat capacity 0.25 × 4180 + 0.14 × 386 = 1099.04 J K⁻¹.
- 2.Their 3 K rise requires 3297.12 J. The aluminium cools through 77 K.
- 3.Set 0.047 × cAl × 77 = 3297.12, obtaining cAl ≈ 911 J kg⁻¹ K⁻¹.
- 4.Neglecting the vessel would omit a real energy gain and give a smaller inferred specific heat. The near-900 result is consistent with the representative material table.
Problem
A 0.200 kg metal sample at 150 °C is put into 150 cm³ of water at 27 °C in a calorimeter of water equivalent 0.025 kg. The final temperature is 40 °C. Take water density as 1.00 g cm⁻³ and cwater = 4186 J kg⁻¹ K⁻¹. What if some energy escapes?
- 1.The water mass is 0.150 kg; the combined effective water mass is 0.150 + 0.025 = 0.175 kg.
- 2.Water and vessel gain 0.175 × 4186 × (40 − 27) = 9523.15 J.
- 3.The sample loses 0.200 × c × (150 − 40), giving c ≈ 432.87 J kg⁻¹ K⁻¹ = 0.433 J g⁻¹ K⁻¹.
- 4.If the initially hot system loses some energy to the external environment, the sample’s actual loss also supplied that escaped energy. Ignoring it attributes too little energy to the metal’s cooling and underestimates its true specific heat.
Mechanical input can also warm a body
The relation mcΔT measures an internal-energy change in this simple model. That change need not have arrived entirely by heat transfer. Friction, drilling or electrical work can convert organized input energy into internal energy. Identify the input and its efficiency before assigning the fraction that warms the sample.
Problem
A 10 kW drill operates for 2.5 min on an 8.0 kg aluminium block. Half the input energy increases the block’s internal energy. Use c = 0.91 J g⁻¹ K⁻¹ and neglect losses from the block during this estimate.
- 1.Time is 150 s, so input energy is 10000 × 150 = 1.50 × 10⁶ J.
- 2.The block receives 0.50 × 1.50 × 10⁶ = 7.50 × 10⁵ J.
- 3.Convert the capacity to 910 J kg⁻¹ K⁻¹, then ΔT = 750000/(8 × 910) ≈ 103 K.
- 4.The temperature rise is also about 103 °C in interval size. This does not specify a final temperature without an initial temperature.
Quick Quiz
Quiz
Two samples of the same material and state have masses m and 2m. With approximately constant specific heat, their heat capacities are in ratio
Heat capacity is mass times specific heat under the stated conditions.
Why does a gas generally require more heat for the same rise at constant pressure than at constant volume?
Energy must supply both internal-energy increase and boundary work.
A calorimeter has water equivalent 25 g. What does that mean?
Water equivalent expresses vessel heat capacity as an equivalent water mass.
At the same constant absorbed power, a sample’s temperature-versus-time slope becomes smaller when
The single-state slope is P/(mc), so larger heat capacity gives a smaller slope.
In the described hot-metal calorimetry experiment, ignoring outward environmental heat loss tends to infer a metal specific heat that is
The metal supplied both the measured gain and escaped energy; omitting the latter undercounts its loss.
Practice Problems
- A 0.500 kg water sample and a 0.500 kg oil sample each absorb 6000 J without changing state. Use cwater = 4186 and coil = 1965 J kg⁻¹ K⁻¹ to calculate and compare their temperature rises.
- An insulated vessel and its contents have total heat capacity 850 J K⁻¹. A heater delivers an absorbed power of 50.0 W. Find the time for a 12.0 K rise, and explain how steady environmental losses would change that time.
- A 0.100 kg copper sample at 100 °C is mixed with 0.200 kg water at 20 °C in a vessel initially at 20 °C whose heat capacity is 40.0 J K⁻¹. Use ccopper = 386 and cwater = 4186 J kg⁻¹ K⁻¹. Find the common temperature, neglecting losses.
- A calorimeter has heat capacity 104.65 J K⁻¹. Find its water equivalent using cwater = 4186 J kg⁻¹ K⁻¹. Explain why adding this equivalent mass to the metal sample instead of the cold water would give a wrong balance.
- A 2.0 kg metal block with c = 450 J kg⁻¹ K⁻¹ receives 60% of the energy supplied by a 200 W motor during 3.0 min. Find its temperature rise. Identify the input as work and state the energy-loss assumption.
Key Takeaways
• Specific heat is per unit mass; molar heat capacity is per mole; both depend on conditions and state. • A gas’s constant-pressure and constant-volume heat capacities differ because the energy accounting differs. • Calorimetry must include the vessel and use each body’s own temperature change. • Mechanical work can raise internal energy; distinguish input energy from energy transferred by a temperature difference.