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4.2 Calorimetry and phase changes

4.2 Calorimetry and phase changes

Written by the Fiveable Content Team • Last updated August 2025
Written by the Fiveable Content Team • Last updated August 2025
🥵Thermodynamics
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Calorimetry measures heat transfer between systems, giving you a way to quantify how energy moves during reactions and phase transitions. It's the experimental backbone for finding specific heat capacities and enthalpy changes, and it connects directly to the first law of thermodynamics.

Calorimetry and Heat Transfer

Principles of calorimetry

Calorimetry rests on conservation of energy: in an isolated system, heat lost by one substance equals heat gained by another. This simple constraint is what makes calorimetry calculations work.

  • Enables determination of specific heat capacity, the amount of heat required to raise the temperature of 1 gram of a substance by 1°C
  • Investigates enthalpy changes during chemical reactions (like combustion) or phase transitions (melting, boiling)
  • Used to evaluate efficiency of thermal systems such as heat engines and heat exchangers
Principles of calorimetry, Calorimetry | Chemistry for Majors

First law in heat transfer

The first law of thermodynamics connects heat transfer to changes in internal energy and work:

ΔU=QW\Delta U = Q - W

where ΔU\Delta U is the change in internal energy, QQ is heat transferred into the system, and WW is work done by the system. Internal energy depends on the system's temperature, pressure, and volume.

For calorimetry problems, you'll rely on two key equations:

When temperature changes (no phase change occurring):

Q=mcΔTQ = mc\Delta T

  • mm = mass of the substance (g or kg)
  • cc = specific heat capacity (J/g·°C or J/kg·K)
  • ΔT\Delta T = temperature change (°C or K)

When a phase change occurs (temperature stays constant):

Q=mLQ = mL

  • LL = latent heat of fusion or vaporization (J/g or J/kg)

The critical distinction: Q=mcΔTQ = mc\Delta T applies while temperature is changing, and Q=mLQ = mL applies during a phase change when temperature is not changing. Many problems require you to use both equations in sequence.

Principles of calorimetry, 5.2 Calorimetry – Chemistry

Phase Changes and Energy

Energy changes during phase transitions

During a phase transition, a substance absorbs or releases latent heat while its temperature remains constant. All the energy goes into breaking or forming intermolecular bonds rather than increasing kinetic energy.

  • Latent heat of fusion (LfL_f): Energy to convert between solid and liquid at the melting point
    • For water: 334 J/g at 0°C
  • Latent heat of vaporization (LvL_v): Energy to convert between liquid and gas at the boiling point
    • For water: 2260 J/g at 100°C

Notice that vaporization requires roughly 6.8 times more energy than fusion for water. That's because separating molecules entirely (liquid → gas) takes far more energy than just loosening them from a rigid structure (solid → liquid).

Example: Melting ice in water

Suppose you drop 10 g of ice at 0°C into 100 g of water at 20°C. Here's how to think through it:

  1. The ice must first absorb latent heat to melt: Qmelt=mice×Lf=10 g×334 J/g=3340 JQ_{\text{melt}} = m_{\text{ice}} \times L_f = 10 \text{ g} \times 334 \text{ J/g} = 3340 \text{ J}
  2. That energy comes from the warm water, which cools down.
  3. Once the ice is fully melted, the resulting 0°C meltwater and the now-cooler liquid water exchange heat until they reach the same final temperature.
  4. Setting heat lost by the warm water equal to heat gained (melting + warming the meltwater) and solving gives a final temperature of about 11.7°C.

Interpretation of calorimetry data

To find specific heat capacity, measure the temperature change when a known amount of heat is added or removed, then rearrange:

c=QmΔTc = \frac{Q}{m\Delta T}

Example: Heating 50 g of aluminum from 20°C to 70°C requires 1130 J. c=1130 J50 g×50°C=0.452 J/g\cdotp°Cc = \frac{1130 \text{ J}}{50 \text{ g} \times 50\text{°C}} = 0.452 \text{ J/g·°C}

To find latent heat, measure the total heat transferred during a constant-temperature phase change:

L=QmL = \frac{Q}{m}

Example: Condensing 5 g of steam at 100°C releases 11,300 J. L=11,300 J5 g=2260 J/gL = \frac{11{,}300 \text{ J}}{5 \text{ g}} = 2260 \text{ J/g}

When working through calorimetry data, follow these steps:

  1. Identify the initial and final temperatures of each substance.
  2. Note the mass of each substance involved.
  3. Determine whether a phase change occurs (look for a flat region on a heating/cooling curve where temperature holds steady).
  4. Apply Q=mcΔTQ = mc\Delta T for temperature-change segments and Q=mLQ = mL for phase-change segments.
  5. Set heat lost equal to heat gained and solve for the unknown.
  6. Compare your calculated values to accepted literature values to check the validity of your results.
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