Solve problems involving specific heat capacity ($Q = mc\Delta T$) and specific latent heat ($Q = mL$) of fusion and vaporisation, including state changes

What this dot point is asking

QCAA wants you to apply $Q = mc\Delta T$ to temperature changes within a single phase, and $Q = mL$ to the energy absorbed or released when a substance changes phase at constant temperature. Both equations together solve the standard multi-stage heating problem.

Specific heat capacity

The specific heat capacity ($c$) of a substance is the energy required to raise the temperature of $1$ kg by $1$ K:

where $Q$ is heat (J), $m$ is mass (kg), $c$ is specific heat capacity (J kg$^{-1}$ K$^{-1}$) and $\Delta T$ is temperature change (K or °C; the size of the unit is identical).

Typical values to know: $c_{\text{water}} = 4186$ J kg$^{-1}$ K$^{-1}$, $c_{\text{ice}} = 2100$, $c_{\text{aluminium}} = 900$, $c_{\text{copper}} = 385$. Water has an unusually high specific heat, which is why coastal climates are mild.

A positive $Q$ means heat absorbed and temperature rises. A negative $Q$ (or negative $\Delta T$) means heat released and temperature falls.

Latent heat

When a substance changes phase, energy is absorbed or released without a temperature change. Particles are gaining or losing the potential energy needed to break or form intermolecular bonds.

• IMATH_23 = specific latent heat of fusion (solid to liquid). For water, $L_f = 3.34 \times 10^5$ J kg$^{-1}$.
• IMATH_26 = specific latent heat of vaporisation (liquid to gas). For water, $L_v = 2.26 \times 10^6$ J kg$^{-1}$.

Vaporisation is much more energy-intensive than fusion because all intermolecular bonds must be broken, not just rearranged.

Conservation of energy in heat exchanges

If two bodies exchange heat in an insulated container, energy is conserved:

This is the principle behind calorimetry. Set up the equation, substitute $m c \Delta T$ on each side, and solve for the unknown (final temperature or unknown specific heat).

Worked example

A $250$ g aluminium block at $90°$C is placed in $300$ g of water at $20°$C in an insulated container. Find the final temperature.

Let final temperature $= T_f$. Energy lost by aluminium = energy gained by water.

$(0.250)(900)(90 - T_f) = (0.300)(4186)(T_f - 20)$

$225 (90 - T_f) = 1255.8 (T_f - 20)$

$20\,250 - 225 T_f = 1255.8 T_f - 25\,116$

$45\,366 = 1480.8 T_f$

$T_f = 30.6°$C.

The system is mostly water (high $c$), so the equilibrium is closer to the water's starting temperature.

Common traps

Using grams instead of kilograms. Both formulas are written for SI units. A $200$ g sample is $0.200$ kg, not $200$ kg.

Treating temperature change in celsius differently from kelvin. $\Delta T = 30°$C $= 30$ K. Both are correct because the units of temperature interval are identical in size.

Forgetting to include the phase change. A common QCAA trap is "ice at $-5°$C to water at $20°$C". You need three stages: warm ice, melt ice, warm water. Skipping the latent heat gives an answer about $25$ times too small.

Using $L_v$ for boiling water at $100°$C all the way to steam at a higher temperature. Latent heat covers the phase change at $100°$C only. Heating the steam further uses $c_{\text{steam}} \Delta T$ on top.

In one sentence

Within a single phase, heat is related to temperature change by $Q = m c \Delta T$ (specific heat capacity), and at a phase change heat is absorbed or released at constant temperature according to $Q = m L$ (latent heat of fusion or vaporisation), with energy conserved in insulated heat exchanges so the heat lost by hotter bodies equals the heat gained by colder bodies.