Latent Heat & Heating/Cooling Curves – Chapter 17 Study Notes

Concept of Latent Heat

• Latent (Hidden) Heat

  • Energy required for a phase change rather than a temperature change.

  • Two principal types for water (but concept is universal):

  • Heat of fusion (melting/freezing)

  • Heat of vaporization (boiling/condensing)

  • Units commonly used in class:

  • $\text{J\,g}^{-1}$ (joules per gram)

  • Values are supplied on tests; no memorization required.

• Why “latent”?

  • During a phase change temperature remains constant even though energy is absorbed/released.

  • Energy goes into breaking (or forming) intermolecular forces instead of raising kinetic energy.

Heating–Cooling Curve for Water (1 atm)

• Graph segments (left → right, solid → gas):

  • A — Solid warming

  • Phase: Ice only

  • Temperature: T<0\,^\circ!\text{C} up to $0\,^\circ!\text{C}$

  • Equation: $q = m c_{\text{ice}} \Delta T$

  • B — Melting (fusion)

  • Phases present: Ice + Liquid water

  • Temperature plateau: $0\,^\circ!\text{C}$

  • Equation: $W = m H_{\text{fus}}$

  • C — Liquid warming

  • Phase: Liquid only

  • Temperature: 0 < T < 100\,^\circ!\text{C}

  • Equation: $q = m c_{\text{liq}} \Delta T$

  • D — Boiling (vaporization)

  • Phases present: Liquid + Vapor

  • Temperature plateau: $100\,^\circ!\text{C}$

  • Equation: $W = m H_{\text{vap}}$

  • E — Gas warming

  • Phase: Steam only

  • Temperature: T>100\,^\circ!\text{C}

  • Equation: $q = m c_{\text{gas}} \Delta T$

• Core observations

  • Sloped regions (A, C, E) → Temperature changes ⇒ use specific-heat formula.

  • Flat regions (B, D) → Phase changes ⇒ use latent-heat (work) formula.

Equations & Variables

• Specific-heat relationship (temperature change) $q = m c \Delta T$

  • $m$ = mass of sample

  • $c$ = specific heat capacity for the given phase

  • $\Delta T = T<em>{\text{final}} - T</em>{\text{initial}}$

• Latent-heat (work) relationship (phase change)
  $W = m H$ where $H$ is either $H<em>{\text{fus}}$ or $H</em>{\text{vap}}$.

• Analogy: same mathematical structure as mechanical work $W = F d$ (force × distance).

Typical Constants for Water (provided on exams)

• $c_{\text{ice}} = 2.06\; \text{J\,g}^{-1}\,^\circ!\text{C}^{-1}$

• $c_{\text{liq}} = 4.184\; \text{J\,g}^{-1}\,^\circ!\text{C}^{-1}$

• $c_{\text{gas}} = 2.02\; \text{J\,g}^{-1}\,^\circ!\text{C}^{-1}$

• $H_{\text{fus}} = 334\; \text{J\,g}^{-1}$

• $H_{\text{vap}} = 2260\; \text{J\,g}^{-1}$

  • Note: Fusion and vaporization energies differ greatly; vaporization requires far more energy.

How to Solve “Total Energy” Problems

1. Locate initial and final temperatures on the heating-cooling curve.

2. Identify every segment crossed (A, B, C, D, E).

3. Write separate expressions (either $q$ or $W$) for each segment.

4. Plug numerical values, perform calculations individually.

5. Sum all energies algebraically to obtain $E_{\text{total}}$.

Worked Example 1

"How much energy is required to raise 2.5 g of water from $-3\,^\circ!\text{C}$ to $108\,^\circ!\text{C}$?"

• Segments encountered: A, B, C, D, E (all five!)

A. Ice warming ($-3\rightarrow0\,^\circ!\text{C}$)
$q_A = (2.5\,\text{g})(2.06\,\text{J\,g}^{-1}\,^\circ!\text{C}^{-1})(0-(-3)) = 15.45\,\text{J}$

B. Melting at $0\,^\circ!\text{C}$
$W_B = (2.5\,\text{g})(334\,\text{J\,g}^{-1}) = 835\,\text{J}$

C. Liquid warming ($0\rightarrow100\,^\circ!\text{C}$)
$q_C = (2.5)(4.184)(100-0) = 1.05\times10^3\,\text{J}$

D. Boiling at $100\,^\circ!\text{C}$
$W_D = (2.5)(2260) = 5650\,\text{J}$

E. Steam warming ($100\rightarrow108\,^\circ!\text{C}$)
$q_E = (2.5)(2.02)(108-100) = 40.4\,\text{J}$

Total energy
$E<em>{\text{total}} = q</em>A + W<em>B + q</em>C + W<em>D + q</em>E \approx 7590\,\text{J}$

• Rounding based on least-precise “tens” place reported by instructor.

Worked Example 2

"50 g of water from $33.6\,^\circ!\text{C}$ to $117.1\,^\circ!\text{C}$."

• Segments: C (liquid warming), D (boiling), E (steam warming)

C. Liquid warming ($33.6\rightarrow100\,^\circ!\text{C}$)
$q_C = (50)(4.184)(100-33.6)=1.389\times10^4\,\text{J}$
(Rounds to $1.39\times10^4\,\text{J}$)

D. Boiling at $100\,^\circ!\text{C}$
$W_D = (50)(2260) = 1.13\times10^5\,\text{J}$

E. Steam warming ($100\rightarrow117.1\,^\circ!\text{C}$)
$q_E = (50)(2.02)(17.1) = 1.73\times10^3\,\text{J}$

Total energy (hundreds place common)
$E_{\text{total}} \approx 1.29\times10^5\,\text{J}\;(=128{,}600\,\text{J})$

Practical & Conceptual Takeaways

• Temperature change ⇒ use specific heat; Phase change ⇒ use latent heat.

• Different phases = different $c$ values → always check whether the substance is solid, liquid, or gas.

• Phase-change energies (fusion, vaporization) are often orders of magnitude larger than sensible-heat changes for small $\Delta T$.

• In calorimetry, heat lost by one part of a system often equals heat gained by another (preview for next lecture: $q<em>{\text{lost}} = - q</em>{\text{gained}}$).

• Always track units and significant figures; identify the common column (place value) before final rounding.

Problem-Solving Checklist

• [ ] Sketch or mentally reference the heating-cooling curve.

• [ ] Mark initial and final temperatures.

• [ ] List segments crossed and phase(s) in each.

• [ ] Write the correct formula for every segment.

• [ ] Insert proper constants ($c, H<em>{\text{fus}}, H</em>{\text{vap}}$).

• [ ] Compute, then sum with correct significant figures.

End of Chapter 17 latent-heat & heating-curve summary.