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

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.

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

Examples in context

Example 1. A Bundaberg sugar mill heats $1500 \text{ kg}$ of raw juice from $25^\circ \text{C}$ to $90^\circ \text{C}$ before evaporation. With $c = 3850 \text{ J kg}^{-1} \text{ K}^{-1}$ (sugar solution), $Q = mc\Delta T = 1500 \times 3850 \times 65 = 3.75 \times 10^8 \text{ J}$ per batch, supplied by bagasse-fired steam. The mill then vaporises $1000 \text{ kg}$ of water in evaporators using $Q = mL_v = 1000 \times 2.26 \times 10^6 = 2.26 \times 10^9 \text{ J}$, an order of magnitude larger because $L_v$ exceeds $c\Delta T$. The QCAA Unit 1 EA Paper 1 industrial stem is a direct lift from this kind of process calculation.

Example 2. A Sunshine Coast cool-store handles $5000 \text{ kg}$ of strawberries (effective $c \approx 3900 \text{ J kg}^{-1} \text{ K}^{-1}$) at intake at $28^\circ \text{C}$, cooling them to $4^\circ \text{C}$. The sensible-heat load is $Q = 5000 \times 3900 \times 24 = 4.68 \times 10^8 \text{ J}$. If $200 \text{ kg}$ of surface dew freezes en route ($L_f = 3.34 \times 10^5 \text{ J kg}^{-1}$), an extra $Q = 6.68 \times 10^7 \text{ J}$ is removed. Refrigeration sizing in Queensland packing houses combines both terms, as the QCAA dot point demands.

Try this

Q1. State the specific heat capacity equation and the specific latent heat equation, defining each symbol. [2 marks]

• Cue. $Q = mc\Delta T$ (no phase change); $Q = mL$ (at phase change).

Q2. Calculate the energy needed to heat $0.50 \text{ kg}$ of water from $20^\circ \text{C}$ to $100^\circ \text{C}$ ($c = 4186 \text{ J kg}^{-1} \text{ K}^{-1}$) and then boil it all to steam ($L_v = 2.26 \times 10^6 \text{ J kg}^{-1}$). [3 marks]

• Cue. Heating $1.67 \times 10^5 \text{ J}$; vaporising $1.13 \times 10^6 \text{ J}$; total $1.30 \times 10^6 \text{ J}$.

Q3. A Bundaberg mill needs to raise $2000 \text{ kg}$ of juice ($c = 3850 \text{ J kg}^{-1} \text{ K}^{-1}$) from $30^\circ \text{C}$ to $95^\circ \text{C}$. (a) Calculate the heat required. (b) The boiler delivers $750 \text{ kW}$. Determine the heating time. (c) Explain why the boiler may be derated in tropical summer conditions. [3+2+2 marks; ISMG: Analysis and interpretation, Evaluation]

• Cue. (a) $5.01 \times 10^8 \text{ J}$; (b) $668 \text{ s}$ or $11.1 \text{ min}$; (c) higher ambient reduces feedwater $\Delta T$ recovery, lowering throughput.