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Thermodynamic Systems overview

Topic 3 of 4

Temperature change and calorimetry

Specific heat capacity relates a sample's heating to its temperature change. An apparatus input may also warm the vessel or supply losses, so isolate the sample's share before calculating c.

Define the property and its conditions

Specific heat capacity c is the energy required by heating per unit mass per unit temperature rise, without a change of state and under the stated process conditions.

Q = mcΔT
c = Q/(mΔT)

Use c in J/(kg K), m in kg and the temperature change in K. A Celsius temperature interval has the same numerical size, so J/(kg °C) gives the same numerical c. An absolute temperature is not interchangeable with a temperature change.

Use a supplied approximately constant c over the interval. Its value can depend on temperature and process. Heating a gas at fixed pressure and at fixed volume are different conditions; do not treat one gas value as process-independent.

Choose the calorimeter boundary

For a system containing the sample, heater, vessel and temperature sensor, electrical input through the heater is work on that assembly. Energy lost to the surroundings by heating is a separate outward transfer. If the sample alone is selected instead, energy passing from the heater into the sample can be heating across its boundary.

Both descriptions can be consistent, but do not count the same input as both electrical work and heating in one account. A calibrated input-energy reading lets the thermal calculation proceed without assuming how the electrical instrument obtained it.

Choose the assembly boundary before allocating input energy

The dashed boundary includes the sample, heater, vessel and temperature sensor. A calibrated 6000 J electrical input is work into this assembly. Energy lost to the surroundings crosses it by heating.

Electrical work warms both the sample and apparatus while some energy leaves by heatingA dashed assembly boundary encloses a 0.250-kilogram sample, its vessel, an immersed heater and a temperature sensor. Two wires connect the heater to an external calibrated input device. An orange downward energy-transfer arrow crosses the assembly boundary beside the wires and denotes 6000 joules of electrical work, not an additional heating input. A separate outward orange arrow denotes 680 joules lost by heating. The apparatus has independently supplied heat capacity forty joules per kelvin, and both apparatus and sample have an eight-kelvin endpoint temperature rise. Of the 5320 joules retained by the assembly, 320 warm the apparatus and 5000 warm the sample. The sample's specific heat capacity is therefore 2500 joules per kilogram per kelvin. The liquid-like sample drawing is schematic and does not identify the sample as water or prescribe an instrument specification.Calibratedwork input6000 JSensorSample0.250 kgHeaterLoss680 JSettled rise: 8.0 KAssembly boundary

For this assembly, 6000 - 680 = 5320 J remains. The independently supplied apparatus contribution is 40 × 8.0 = 320 J, leaving 5000 J for the sample. The apparatus is assumed to share the sample's stated endpoint temperature change.

A boundary around the sample alone uses the energy transferred into the sample by heating. Do not count the same electrical input once as assembly work and again as a second assembly heating input.

The marked assembly includes the sample, heater, vessel and sensor. Electrical input is work into that assembly, while heat loss leaves it. Sample mass and apparatus heat capacity are distinct; a sample-only account uses the net energy reaching that sample.

Decide whether corrections matter

Roughly 0.3 kg warming by about 10 K with a few kilojoules input suggests c of order 103 J/(kg K). Apparatus warming of a few hundred joules plus a loss of several hundred joules is of order 1 kJ, which is not automatically negligible beside a 6 kJ input.

Worked corrected calorimetry

Separate sample warming, apparatus warming and loss

A 0.250 kg condensed sample rises by 8.0 K during a measured 6000 J input. There is no phase change and volume work is negligible. Independently supplied apparatus heat capacity is 40 J/K, and the apparatus has the same 8.0 K change between endpoints. Estimated energy lost to the surroundings over that run is 680 J.

Apparatus energy increase = 40 × 8.0 = 320 J
Net energy retained by sample
= 6000 - 320 - 680 = 5000 J
c = 5000/(0.250 × 8.0)
= 2500 J/(kg K)

The assembly's internal-energy increase is 6000 - 680 = 5320 J, shared as 5000 J in the sample and 320 J in the apparatus. For the sample-only boundary with negligible work, its net heating is 5000 J.

Allocating all the input to the sample would give c = 6000/(0.250 × 8.0) = 3000 J/(kg K), an overestimate. With positive apparatus and loss terms and no extra source, the sample receives less than the full input, so the corrected c must be below that uncorrected value.

The 1000 J correction is one sixth of the input. The uncorrected c is 20% above the corrected value, because (3000 - 2500)/2500 = 0.20. These percentages use different denominators.

Optional check A 0.250 kg sample warms by 8.0 K during a 6000 J input. Independently supplied corrections are 320 J for apparatus warming and 680 J lost to the surroundings. With no phase change or significant volume work, what is the sample's specific heat capacity?
A 0.250 kg sample warms by 8.0 K during a 6000 J input. Independently supplied corrections are 320 J for apparatus warming and 680 J lost to the surroundings. With no phase change or significant volume work, what is the sample's specific heat capacity?

Make the measurement represent the sample

  • Mass: measure the sample separately or subtract the empty container mass. Record the balance range, resolution and zero or tare.
  • Temperature: establish sensor contact, record its range and resolution, and use stable readings. Stir a suitable liquid so one sensor reading represents it; avoid treating a local heater hot spot as the whole sample's temperature.
  • Input: measure the energy delivered over a defined interval. State whether the heater and vessel belong to the chosen system.
  • Apparatus response: use independently determined heat capacity and justify the component temperature changes. Equal endpoint changes do not require every component to share one temperature at every instant.
  • Loss: reduce avoidable transfer and estimate the remaining loss under comparable conditions. Record what evidence supports that correction.

Keep the vessel stable and the heater and sensor securely supported so a pulled lead cannot tip or spill the sample. Switch off before adjusting the heater, and let hot parts cool before handling them.

A cooling record or matched reference can help estimate background transfer only if relevant temperatures, exposed areas and surroundings are comparable. A no-heater control at a different temperature does not automatically reproduce the loss during heating. Correction values must come from calibration or other evidence, not be chosen to produce an expected c.

Apply energy conservation to thermal contact

Worked water mixture

Unequal masses need not meet at the arithmetic mean

Mix 0.200 kg water at 50.0°C with 0.300 kg water at 20.0°C. Use c = 4200 J/(kg K) for both. Neglect vessel heat capacity, losses, other work and changes of state.

Taking both water samples as an isolated combined system, energy lost by the hotter sample equals energy gained by the colder one. If the common final temperature is θf:

0.200c(50.0 - θf) = 0.300c(θf - 20.0)
θf = (0.200 × 50.0 + 0.300 × 20.0)/0.500
= 32.0°C

The energy transferred is 0.200(4200)(50.0 - 32.0) = 15120 J, equal to 0.300(4200)(32.0 - 20.0). The transfer is internal to the combined system; it is not an extra energy input to it.

The final temperature is closer to the initial temperature of the larger heat capacity. Equal final temperature establishes thermal equilibrium, not equal total internal energies.