Free Pearson Edexcel International A Level Physics lesson
Measure latent heat and gas behaviour
Separate useful energy transfer from losses and wait for the state you claim.
What you’ll learn
- Plan and evaluate core practical 13.
- Control temperature and dead volume in core practical 14.
- Use slopes and paired trials to diagnose energy losses.
Core practical 13: determine energy per mass of phase change
Specific latent heat describes energy needed per unit mass for a phase change at the stated conditions, without a temperature rise during the change. An electrical method supplies measured power VI and measures the mass changing phase over a timed interval. Decide whether the practical concerns melting or vaporisation, because background heat transfer affects the two arrangements differently.
For a boiling-water method, establish steady boiling before timing a mass decrease on a suitable balance. Record voltage across and current through the heater rather than relying on its label rating. Protect the balance from heat and steam and prevent the heater running dry. Hot liquid, steam and mains electrical equipment require a supervised apparatus-specific safety arrangement.
The simple ratio L = VIt/Δm assumes all measured input produces the measured phase change. In reality heat loss makes a boiling estimate too high if uncorrected. A two-power method can remove an approximately constant loss by comparing steady mass-loss rates at the same boiling temperature. For melting ice, ambient heating can itself melt mass, so an unheated control may be appropriate; the direction of bias must follow the energy balance.
Ephase = LΔm; ideal L = VIt/Δm; steady boiling P = Lṁ + Ploss
Symbols and units
- L is specific latent heat in J kg⁻¹
- Δm is mass undergoing phase change in kg
- V is heater voltage in V
- I is current in A
- t is time in s
- ṁ is mass-loss rate in kg s⁻¹.
How the equation explains the idea
- Electrical input is VIt for steady values. If all of it causes the measured phase change, L = VIt/Δm; heat losses make this simple input-based estimate too high.
- For steady boiling with approximately constant loss power, P = Lṁ + Ploss. Plot input power against mass-loss rate to separate latent heat as gradient from loss as intercept.
Worked example
An uncorrected boiling estimate
- A heater supplies 60 W for 300 s while measured mass decreases by 0.0060 kg.
- Luncorrected = 60 × 300/0.0060 = 3.0 × 10⁶ J kg⁻¹.
- If 15 W is lost to the surroundings, useful phase-change power is 45 W and L = 2.25 × 10⁶ J kg⁻¹.
The uncorrected calculation overestimates latent heat in this boiling model.
Keep this idea. Measure the phase-change mass and electrical input, then justify the energy-loss correction.
Core practical 14: keep the gas sample and temperature fixed
Trap a fixed mass of gas in a syringe or adjustable-volume chamber connected to a pressure sensor. The ideal-gas model predicts pV = NkT, so p is inversely proportional to total volume only while gas amount and absolute temperature remain fixed. Use absolute pressure: a gauge reading relative to atmospheric pressure must be converted before applying this relation.
Adjust volume slowly and wait for pressure to settle as the gas returns to the surroundings’ temperature. Rapid compression heats the gas and gives a pressure above the isothermal prediction. Check seals for leakage, avoid excessive pressure and secure the apparatus. Include the volume inside connecting tubes and the sensor if it contains part of the same trapped gas sample.
Measure a suitable range and repeat reference volumes to check drift. Plot p against 1/V for a straight relation or log p against log V for an exponent expected near −1. A dead-volume correction affects small syringe readings most strongly. Show uncertainty and do not force a desired slope by choosing an unsupported volume correction.
pabsolute Vtotal = NkT; Vtotal = Vsyringe + Vdead; log(p/pref) = constant − log(V/Vref)
Symbols and units
- pabsolute is absolute gas pressure in Pa
- Vtotal is total trapped volume in m³
- N is number of gas particles
- k is Boltzmann’s constant in J K⁻¹
- T is absolute temperature in K.
How the equation explains the idea
- For fixed N and equilibrated T, pabsolute Vtotal = NkT is constant. A gauge pressure must be converted to absolute pressure before this relation is tested.
- Add dead volume to the syringe reading. Doubling the actual total volume halves absolute pressure; doubling only the syringe reading does not necessarily double total volume.
Conditions. Ideal fixed gas sample at constant temperature with pV = 4000 kPa cm³. The volume already includes dead space; pressure is absolute, not gauge pressure.
Total trapped volume / cm³; Absolute pressure / kPa.
At total volume 20 cm³ pressure is 200 kPa; at 40 cm³ it is 100 kPa; at 80 cm³ it is 50 kPa. Equal volume increments do not produce equal pressure drops.
Open the interactive lesson to view the plotted curves and their coordinate tables.
Worked example
Include the connecting tube
- Syringe volume is 20 cm³ and trapped connecting volume is 5 cm³, so total V₁ = 25 cm³.
- Compress the syringe to 10 cm³, giving V₂ = 15 cm³.
- At unchanged temperature, p₂/p₁ = 25/15 = 1.67, not 20/10 = 2.0.
The gas law uses the total gas volume, not only the marked syringe volume.
Keep this idea. Use absolute pressure, total volume and thermal equilibrium for an inverse pressure–volume test.
Use a paired measurement to isolate a constant loss
A model with an unwanted constant term can sometimes be tested using differences. At steady boiling, P = Lṁ + Ploss. Taking two trials at the same boiling temperature and comparable surroundings gives ΔP = LΔṁ if heat loss is approximately unchanged. The gradient of a power-against-mass-loss-rate graph then gives L, while its intercept estimates the loss.
This method replaces an assumption that heat loss is zero with an assumption that it remains approximately constant. The latter can fail if surface area, lid position, draughts or vessel temperature change between trials. Use several power settings rather than one pair where possible, and inspect whether the relation is linear with a physically plausible intercept.
Differences can have large relative uncertainties when the two values are close. Add absolute uncertainties in the two power values and in the two mass rates before converting to percentage contributions for their quotient. Choose a safe power separation large enough to resolve Δṁ without changing the thermal regime or producing uncontrolled splashing.
P = Lṁ + Ploss; L = (P₂ − P₁)/(ṁ₂ − ṁ₁)
Symbols and units
- P₁,P₂ and Ploss are powers in W
- ṁ₁ and ṁ₂ are phase-change mass rates in kg s⁻¹
- L is specific latent heat in J kg⁻¹.
How the equation explains the idea
- Write P₁ = Lṁ₁ + Ploss and P₂ = Lṁ₂ + Ploss. Subtract the equations to eliminate the same loss term.
- Then L = (P₂ − P₁)/(ṁ₂ − ṁ₁). The cancellation relies on comparable thermal-loss conditions; if temperature or exposed area changes, the loss may not cancel.
Conditions. Steady phase-change model with L = 2.26 × 10⁶ J kg⁻¹ and constant heat loss 50 W. Mass rate is plotted in g s⁻¹, so P = 2260 × numerical rate + 50 W. The intercept is a fitted-model extrapolation.
Mass-loss rate / g s⁻¹; Input power / W.
The slope is 2260 W per g s⁻¹, equivalent to 2.26 × 10⁶ J kg⁻¹. Rates 0.10 and 0.20 g s⁻¹ give 276 and 502 W. Subtracting these pairs eliminates the 50 W loss term.
Open the interactive lesson to view the plotted curves and their coordinate tables.
Worked example
Two steady boiling powers
- At 40 W, mass-loss rate is 1.0 × 10⁻⁵ kg s⁻¹; at 60 W it is 2.0 × 10⁻⁵ kg s⁻¹.
- L = (60 − 40)/(2.0 − 1.0) × 10⁵ = 2.0 × 10⁶ J kg⁻¹.
- Ploss = 40 − (2.0 × 10⁶)(1.0 × 10⁻⁵) = 20 W.
The two-point model estimates both latent heat and a common loss.
Keep this idea. A subtraction method needs a justified common-loss condition and uncertainty in both differences.
Open the interactive lesson for free practice and reviewed three-question samples when available. New full question sets and timed unit checks are in preparation. Existing complete Physics revision packs retain paid access; existing free teaching PDFs remain free.
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