4.3 - Specific Latent Heat

Latent Heat and Specific Latent Heat

• Matter exists as solid, liquid and gas. Molecular bonds are stronger in solids than liquids; gas molecules move freely and have the weakest bonds.
• Latent heat is absorbed or released during a change of phase without a temperature change.
• Melting (solid to liquid) and boiling (liquid to gas) absorb heat. Freezing (liquid to solid) and condensation (gas to liquid) release heat.
• The heat required for a phase change depends on the mass and type of substance.
• Specific latent heat, $l$: heat absorbed or released when 1 kg of a substance changes phase without changing temperature.

• $Q$ is heat in J; $m$ is mass in kg. S.I. unit of $l$: $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$.
• Specific latent heat of fusion, $l_f$: heat absorbed in melting, or released in freezing, 1 kg at constant temperature.
• Specific latent heat of vaporisation, $l_v$: heat absorbed in boiling, or released in condensation, 1 kg at constant temperature.
• Temperature depends on average molecular kinetic energy. During phase change, latent heat changes molecular bonds rather than increasing or decreasing this average kinetic energy.

Heating and Cooling Curves

Stage	Molecular explanation
Heating a solid, liquid or gas	Heat absorption increases molecular kinetic energy and temperature.
Melting plateau	Solid and liquid coexist. Absorbed latent heat weakens bonds and frees molecules from fixed positions; temperature remains constant until all solid melts.
Boiling plateau	Liquid and gas coexist. Absorbed latent heat breaks bonds and separates molecules far apart; temperature remains constant until all liquid boils.
Cooling a gas, liquid or solid	Heat release decreases molecular kinetic energy and temperature.
Condensation plateau	Gas and liquid coexist. Latent heat is released as molecular bonds form; temperature remains constant until all gas condenses.
Freezing plateau	Liquid and solid coexist. Latent heat is released as bonds strengthen and molecules vibrate around fixed positions; temperature remains constant until all liquid freezes.

Comparing Substances

• Each substance has its own specific latent heat. Ice requires more heat per kilogram to melt than wax; water requires more heat per kilogram to vaporise than petrol.

Substance	Melting point / ${}^{\circ }\mathrm{C}$	$l_f$ / $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$
Wax	46–68	$1.45\times 10^5$–$2.10\times 10^5$
Lead	327	$0.25\times 10^5$
Copper	1 083	$2.07\times 10^5$
Ice	0	$3.34\times 10^5$
Olive oil	6	$2.67\times 10^5$
Ethanol	$-114$	$1.04\times 10^5$
Oxygen	$-219$	$0.14\times 10^5$
Nitrogen	$-210$	$0.26\times 10^5$

Substance	Boiling point / ${}^{\circ }\mathrm{C}$	$l_v$ / $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$
Lead	1 750	$8.59\times 10^5$
Copper	2 566	$47.3\times 10^5$
Water	100	$2.26\times 10^6$
Petrol	35–200	$3.49\times 10^5$
Diesel	180–360	$2.56\times 10^5$
Ethanol	78	$8.55\times 10^5$
Oxygen	$-183$	$2.13\times 10^5$
Nitrogen	$-196$	$2.00\times 10^5$

• Petrol and diesel are different hydrocarbons with different boiling ranges.
• Vaporisation requires more heat per kilogram than fusion of the same substance: boiling breaks molecular bonds and separates molecules far apart; melting weakens bonds while molecules remain close.

Experiment 4.3A - Specific Latent Heat of Fusion of Ice

• Aim: determine $l_f$ of ice.
• Apparatus: immersion heater, filter funnels, beakers, electronic balances, power supply, stopwatch and retort stands. Material: crushed ice.

1. Zero each balance with its collecting beaker in place.
2. Prepare an experiment set with a heater in crushed ice in a funnel, and a control set with a switched-off heater in crushed ice in a funnel. Initially keep collecting beakers and balances away from the funnels.
3. Switch on the experiment-set heater only. Once water drips at a steady rate, put the beakers and balances under their funnels and start timing.
4. After 10 minutes, record collected water masses: experiment set $m_1$ and control set $m_2$.
5. Switch off the heater and record its power, $P$.

• Record $m_1$, $m_2$ (kg), $P$ (W) and $t$ (s). Use the mass difference in the calculation:

Experiment 4.3B - Specific Latent Heat of Vaporisation of Water

• Aim: determine $l_v$ of water.
• Apparatus: 500 W immersion heater, power supply, beaker, electronic balance and stopwatch. Materials: water and tissue paper.

1. Set the tissue-wrapped beaker of water and immersion heater on the balance.
2. Switch on the heater and wait until the water boils.
3. At boiling, start timing and record initial balance reading, $m_1$.
4. After five minutes, record final reading, $m_2$.
5. Switch off the heater and record its power, $P$.

• Record $m_1$, $m_2$ (kg), $t$ (s) and $P$ (W). The decrease in mass is the water vaporised:

Cooling and Heating Applications

• Evaporation causes cooling: liquid molecules absorb heat from the surroundings to break bonds and become vapour; the surroundings lose heat.
• Melting ice cools drinks and food. Boiling liquid nitrogen provides cooling.
• Condensation releases latent heat for heating, including steaming food and heating espresso coffee.
• Sweat evaporation: sweat absorbs heat from the body as it evaporates. Air circulation increases the evaporation rate.
• Refrigerators and air conditioners use cooling from evaporation. In a refrigerator, the cooling agent circulates through these stages:
  1. The compressor compresses the gas, increasing its pressure and temperature.
  2. In the condenser, the gas condenses and releases latent heat outside.
  3. The liquid flows through the expansion valve.
  4. In the evaporator, it evaporates and absorbs latent heat from inside the refrigerator.
  5. The gas returns to the compressor.

Worked Calculations

• Melting ice: a 480 W heater operates for 120 s; the balance reading decreases by 0.172 kg as melted water drains away. The ice mass melted is 0.172 kg.

• Assumptions: only the heater melts the ice; all meltwater leaves the container; all heater energy is absorbed by the ice; no heat enters from the surroundings.
• Water to steam: change 0.75 kg water at $30{}^{\circ }\mathrm{C}$ to steam at $100{}^{\circ }\mathrm{C}$ in two stages: heat water to its boiling point, then vaporise it without further temperature rise.

• Assumptions: all heater energy is absorbed by the water; no heat is lost during heating or phase change.