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Physics HL · Chapter 7: Thermal Energy Transfers

7.2 Specific Heat Capacity and Calorimetry

Use specific heat capacity as a quantitative slope between energy transfer and temperature change, then solve thermal-equilibrium mixing problems.

Estimated time: 26 minutes

Specific Heat Capacity as an Energy-Slope

Q=mcΔTQ = mc\Delta T

For no phase change, transferred thermal energy is proportional to mass, specific heat capacity, and temperature change.

Specific heat capacity c tells you how much energy per kilogram is needed for a one-kelvin temperature rise. High-c materials need more energy to warm by the same amount, which is why water moderates temperature changes so effectively in climate and engineering contexts.

The equation is linear only inside one phase and over ranges where c can be treated as approximately constant. That approximation is usually valid in introductory problems and IB-style calculations, but the physical interpretation should stay explicit: Q measures transferred thermal energy, not stored 'heat content'.

Calorimetry Equation and Sign Discipline

Qi=0(isolated system)\sum Q_i = 0\quad (\text{isolated system})

In an ideal insulated setup, energy lost by warmer components equals energy gained by cooler components.

A robust method is to write one equation per body with its own mm, cc, and ΔT\Delta T to the common final temperature, then enforce zero net sum. This avoids memorizing special-case formulas and scales well when problems include multiple solids, water, and container corrections.

Sign mistakes usually come from guessing QQ signs too early. Instead, write ΔT\Delta T symbolically as (TfinalTinitial)(T_{\text{final}} - T_{\text{initial}}). Bodies above the final equilibrium temperature automatically get negative ΔT\Delta T, and bodies below it automatically get positive ΔT\Delta T. The algebra then handles direction correctly.

Two-Body Thermal Equilibrium Problems

For two bodies with no phase change and negligible losses, final temperature is a weighted average where weights are thermal capacities m times c. A body with large m c acts as a thermal anchor: its temperature changes little while forcing the other body to adjust more.

This is why dropping hot metal into water often causes a large drop in the metal's temperature but only a modest rise in water temperature. Even if the metal starts much hotter, water's large c and often larger mass can dominate the weighted average.

Simulation: Two-Body Calorimetry Mixer

Set masses, specific heat capacities, and initial temperatures to see equilibrium temperature and energy flow consistency.

Calorimetry + Phase Lab

Two-body mixing scene

Q = 10.9 kJHot body (78.0 deg C)Cold body (20.0 deg C)m = 0.45 kg, c = 450m = 0.65 kg, c = 4186

Final equilibrium temp

24.02 deg C

Hot-side temperature drop

53.98 K

Cold-side temperature rise

4.02 K

Energy mismatch

0.0000 J

Use Mixing mode for conservation in calorimetry, then switch to Heating curve mode to see where energy raises temperature and where it is redirected into latent phase change.

Interpretation guidance: first hold both specific heats fixed and vary only mass ratio to see weighted-average behavior directly. Then hold masses fixed and switch specific heats to compare low-c solids against water-like materials. Watch how the equilibrium point shifts and verify that hot-side energy loss matches cold-side gain.

Test Yourself

A 0.20kg0.20\,\text{kg} aluminum block (c=900Jkg1K1c = 900\,\text{J}\,\text{kg}^{-1}\,\text{K}^{-1}) at 100C100\,^\circ\text{C} is placed in 0.50kg0.50\,\text{kg} water (c=4200Jkg1K1c = 4200\,\text{J}\,\text{kg}^{-1}\,\text{K}^{-1}) at 20C20\,^\circ\text{C}. Ignore losses and enter the final temperature in C^\circ\text{C}.

Hint: Set heat lost by aluminum equal to heat gained by water.