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VICPhysicsSyllabus dot point

How do bodies exchange heat?

Investigate and apply theoretically and practically the relationships $Q = mc\Delta T$ (specific heat capacity) and $Q = mL$ (latent heat of fusion and vaporisation), including multi-stage heating problems

A focused answer to the VCE Physics Unit 1 dot point on specific heat capacity and latent heat. Applies $Q = mc\Delta T$ and $Q = mL$, identifies typical values for water, ice, aluminium, and works the VCAA SAC-style multi-stage problem (ice to steam).

Generated by Claude OpusReviewed by Better Tuition Academy5 min answer

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What this dot point is asking

VCAA wants you to apply two formulas: Q=mcΔTQ = mc\Delta T for temperature change within a phase, and Q=mLQ = 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=mcΔTQ = m c \Delta T

  • IMATH_5 = heat energy (J)
  • IMATH_6 = mass (kg)
  • IMATH_7 = specific heat capacity (J kg−1^{-1} K−1^{-1})
  • IMATH_10 = temperature change (K or °C; same size unit)

Typical values to know: cwater=4186c_{\rm water} = 4186, cice=2100c_{\rm ice} = 2100, csteam=2010c_{\rm steam} = 2010, caluminium=900c_{\rm aluminium} = 900, ccopper=385c_{\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.

Q=mLQ = m L

  • IMATH_16 = specific latent heat of fusion (solid ↔\leftrightarrow liquid). For water, Lf=3.34×105L_f = 3.34 \times 10^5 J kg−1^{-1}.
  • IMATH_20 = specific latent heat of vaporisation (liquid ↔\leftrightarrow gas). For water, Lv=2.26×106L_v = 2.26 \times 10^6 J kg−1^{-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:

Qlost=QgainedQ_{\rm lost} = Q_{\rm gained}

Calorimetry questions set up this equation, substitute mcΔTmc\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 ΔT\Delta T stops applying and LL takes over). Convert grams to kilograms before substituting. Report energy in scientific notation.

Common traps

Forgetting the phase change. A problem like "ice at −5-5°C to water at 2020°C" has three stages. Missing Q2=mLfQ_2 = mL_f gives an answer about 2525 times too small.

Using ΔT\Delta T across a phase change. Inside a phase change, temperature is constant. Q=mcΔTQ = m c \Delta T does not apply.

Mixing units. Specific heats are written for SI units. A 200200 g sample is 0.2000.200 kg.

Treating LfL_f and LvL_v as interchangeable. They differ by an order of magnitude. Use LfL_f for melting/freezing, LvL_v for boiling/condensing.

In one sentence

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

Past exam questions, worked

Real questions from past VCAA papers on this dot point, with our answer explainer.

Year 11 SAC5 marksCalculate the total energy required to heat $300$ g of ice from $-5$°C to water at $20$°C, using $c_{\rm ice}=2100$ J kg$^{-1}$ K$^{-1}$, $c_{\rm water}=4186$ J kg$^{-1}$ K$^{-1}$ and $L_f = 3.34 \times 10^5$ J kg$^{-1}$.
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Three stages.

Stage 1 (ice −5-5°C →\to ice 00°C). Q1=(0.300)(2100)(5)=3150Q_1 = (0.300)(2100)(5) = 3150 J.

Stage 2 (melt at 00°C). Q2=mLf=(0.300)(3.34×105)=100 200Q_2 = m L_f = (0.300)(3.34 \times 10^5) = 100\,200 J.

Stage 3 (water 00°C →\to water 2020°C). Q3=(0.300)(4186)(20)=25 116Q_3 = (0.300)(4186)(20) = 25\,116 J.

Total: Q=3150+100 200+25 116=1.28×105Q = 3150 + 100\,200 + 25\,116 = 1.28 \times 10^5 J.

Markers reward splitting at the phase boundary, mass in kg, and a final answer in scientific notation.

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