Specific heat capacity and latent heat (VCE Physics Unit 1)

What this dot point is asking

VCAA wants you to apply two formulas: $Q = mc\Delta T$ for temperature change within a phase, and $Q = mL$ for energy absorbed or released during a phase change at constant temperature. Both combine in the standard multi-stage heating problem.

Specific heat capacity

• $Q$ = heat energy (J)
• $m$ = mass (kg)
• $c$ = specific heat capacity (J kg$^{-1}$ K$^{-1}$)
• $\Delta T$ = temperature change (K or °C; same size unit)

Typical values to know: $c_{\rm water} = 4186$, $c_{\rm ice} = 2100$, $c_{\rm steam} = 2010$, $c_{\rm aluminium} = 900$, $c_{\rm copper} = 385$.

Water has an unusually high specific heat, which is why coastal climates are mild and why water is used as a coolant.

Latent heat

During a phase change, energy is absorbed or released at constant temperature. The energy goes into breaking or forming intermolecular bonds.

• $L_f$ = specific latent heat of fusion (solid $\leftrightarrow$ liquid). For water, $L_f = 3.34 \times 10^5$ J kg$^{-1}$.
• $L_v$ = specific latent heat of vaporisation (liquid $\leftrightarrow$ gas). For water, $L_v = 2.26 \times 10^6$ J kg$^{-1}$.

Vaporisation is several times more energy-intensive than fusion because all intermolecular bonds must be broken to form a gas.

Conservation of energy in heat exchanges

In an insulated container, heat lost by hot bodies equals heat gained by cold bodies:

Calorimetry questions set up this equation, substitute $mc\Delta T$ on each side, and solve for the final temperature or unknown specific heat.

VCAA exam style

Multi-stage heating problems are the standard. Watch for the phase boundary (the moment $\Delta T$ stops applying and $L$ takes over). Convert grams to kilograms before substituting. Report energy in scientific notation.

Common traps

Forgetting the phase change

A problem like "ice at $-5$°C to water at $20$°C" has three stages. Missing $Q_2 = mL_f$ gives an answer about $4.5$ times too small (it leaves out $78\%$ of the energy).

Using $\Delta T$ across a phase change

Inside a phase change, temperature is constant. $Q = m c \Delta T$ does not apply.

Mixing units

Specific heats are written for SI units. A $200$ g sample is $0.200$ kg.

Treating $L_f$ and $L_v$ as interchangeable

They differ by an order of magnitude. Use $L_f$ for melting/freezing, $L_v$ for boiling/condensing.

In one sentence

Within a single phase, heat is related to temperature change by $Q = mc\Delta T$ (specific heat capacity), and at a phase change heat is absorbed or released at constant temperature according to $Q = mL$ (latent heat of fusion or vaporisation), with energy conserved in insulated heat exchanges.

Examples in context

Example 1. Snowy 2.0 reservoir thermal mass. Tantangara Reservoir, the upper storage for Snowy 2.0, holds $254$ GL ($2.54 \times 10^{11}$ kg) of water. To cool this reservoir from $15^\circ$C to $5^\circ$C overnight in autumn would require removing $Q = mc\Delta T = 2.54 \times 10^{11} \times 4186 \times 10 = 1.06 \times 10^{16}$ J. By comparison, Australia's entire daily electricity demand is approximately $5.6 \times 10^{14}$ J, so the reservoir's thermal mass is enormous. This thermal inertia stabilises water temperature for turbine bearings and reduces ice-formation risk on the spillway gates during cold snaps.

Example 2. Latrobe Valley brown coal moisture and energy penalty. Latrobe Valley brown coal contains about $60\%$ water by mass. Drying $1$ kg of raw coal requires removing $0.6$ kg of water. Energy to heat that water from $20^\circ$C to $100^\circ$C is $Q_1 = 0.6 \times 4186 \times 80 = 2.01 \times 10^5$ J; energy to vaporise it at $100^\circ$C is $Q_2 = mL_v = 0.6 \times 2.26 \times 10^6 = 1.36 \times 10^6$ J. Total moisture penalty per kg of coal is $1.56$ MJ, against a dry-coal heat value of around $26$ MJ. Around $6\%$ of Loy Yang A's energy output is therefore spent evaporating water, which is part of why brown coal generation has lower thermal efficiency than black coal.

Try this

Q1. Define specific heat capacity and state SI units. [2 marks]

• Cue. Energy required to raise the temperature of $1$ kg of a substance by $1^\circ$C (or $1$ K); J kg$^{-1}$ K$^{-1}$.

Q2. A $500$ g aluminium block ($c = 900$ J kg$^{-1}$ K$^{-1}$) absorbs $13.5$ kJ. Calculate the temperature rise. [3 marks]

• Cue. $\Delta T = Q/(mc) = 13500/(0.5 \times 900) = 30^\circ$C.

Q3. A $200$ g ice cube at $-10^\circ$C is added to $400$ g of water at $25^\circ$C in an insulated cup. Take $c_{\rm ice} = 2100$, $c_{\rm water} = 4186$ J kg$^{-1}$ K$^{-1}$, $L_f = 3.34 \times 10^5$ J kg$^{-1}$. (a) Calculate the energy required to warm the ice to $0^\circ$C. (b) Calculate the energy required to melt the ice at $0^\circ$C. (c) Determine the final equilibrium temperature. [2+2+3 marks]

• Cue. (a) $0.2 \times 2100 \times 10 = 4200$ J. (b) $0.2 \times 3.34 \times 10^5 = 66800$ J. (c) The water can release only $0.4 \times 4186 \times 25 = 41\,860$ J cooling to $0^\circ$C, which is less than the $71\,000$ J needed to warm and melt all the ice. So the final temperature is $0^\circ$C. After warming the ice ($4\,200$ J), the remaining $37\,660$ J melts $37\,660 / (3.34 \times 10^5) = 0.113$ kg, leaving about $0.087$ kg of ice.