Free Pearson Edexcel International A Level Physics lesson

Thermistors and thermal measurements

Build a calibration, then test the energy measurement.

What you’ll learn

  • Calibrate across the operating range and allow thermal equilibrium before each reading.
  • Draw and label the divider before deciding the switching direction.
  • Use a control or a fitted intercept to estimate unwanted energy transfer.

Core practical 12: calibrate the sensor

A negative-temperature-coefficient thermistor has a resistance that decreases as its temperature rises over its intended working range. Its response is strongly nonlinear. Calibration records that response against a reference thermometer, so the thermistor becomes a useful temperature sensor without assuming that equal resistance changes mean equal temperature changes.

Place the thermistor and reference sensor close together in a stirred water bath, electrically insulating exposed connections from the water. Use a suitable low-voltage measuring circuit and wait for both readings to settle at each temperature. Take several points spanning the intended operating range, with extra points near a switching temperature where the local sensitivity matters.

Keep measuring current small to reduce thermistor self-heating. Repeat on cooling to investigate thermal lag and repeatability; do not average away a consistent lag without explaining it. Record thermometer resolution, voltage or resistance resolution, and interpolation uncertainty. A calibration prepared in one circuit can shift if the supply, series resistance or electrical loading changes.

NTC response: increasing T → decreasing R local sensitivity ≈ ΔR/ΔT

Symbols and units

  • R is thermistor resistance in ohms (Ω); 1 kΩ = 1000 Ω.
  • T denotes sensor temperature here; calibration readings may be labelled in °C. ΔT is a temperature difference, with the same numerical size in K and °C.
  • ΔR = R₂ − R₁ and ΔT = T₂ − T₁ compare nearby settled calibration readings. Their ratio is a local slope in Ω K⁻¹ or kΩ °C⁻¹.
  • NTC means negative temperature coefficient: resistance decreases as temperature increases over the operating range.

How the equation explains the idea

  1. Subtract the two calibration readings in the same order: ΔR = 2.8 − 3.2 = −0.4 kΩ and ΔT = 44 − 40 = 4 °C.
  2. The local slope is ΔR/ΔT = −0.4/4 = −0.10 kΩ °C⁻¹. Its negative sign encodes the NTC response; it is not a negative resistance.
  3. Use the short-interval approximation R − 3.2 ≈ (−0.10)(T − 40), with R in kΩ and T in °C. At R = 3.0, T − 40 ≈ (−0.2)/(−0.10) = 2 °C, giving approximately 42 °C.
  4. This interpolation is supported only near the measured 40–44 °C interval and after thermal settling. The same slope need not describe a distant temperature or a self-heated sensor.

Worked example

Interpolate near a threshold

  1. A measured sensor gives 3.2 kΩ at 40 °C and 2.8 kΩ at 44 °C.
  2. Over this short interval the approximate slope is −0.10 kΩ °C⁻¹.
  3. A 3.0 kΩ reading corresponds approximately to 42 °C, not a global straight-line law.

Use local interpolation with a stated calibration range.

Keep this idea. Calibrate across the operating range and allow thermal equilibrium before each reading.

Turn the calibration into a switching signal

A potential divider converts thermistor resistance into an output voltage. For a thermistor below a fixed resistor, with output measured across the thermistor, Vout = Vs RT/(Rfixed + RT). Heating an NTC thermistor reduces RT and therefore reduces this output. If the output is instead taken across the fixed resistor, its voltage increases as the thermistor warms.

Select the divider resistance so that the output changes usefully around the desired threshold. Calibrate the voltage in the actual circuit and compare it with a chosen reference voltage. A comparator or control circuit can then switch an isolated heater driver. The direction of switching must match the sensor arrangement: a low output in this example indicates a high temperature.

A thermostat often uses different turn-on and turn-off thresholds, called hysteresis, to avoid rapid chatter near one boundary. Thermal lag also means that the temperature may continue rising briefly after heating switches off. Investigating a low-voltage demonstration does not authorise building a mains-powered heater controller; actual construction and hot-water work require appropriate supervision.

Vout = Vs RT/(Rfixed + RT) output taken across the thermistor

Symbols and units

  • Vs is supply voltage and Vout is the voltage measured across the thermistor, both in V.
  • RT is thermistor resistance and Rfixed is the fixed series resistance, both in Ω or both in kΩ. The subscript T identifies the thermistor.
  • I is the common series current in A; the divider equation assumes negligible current drawn by the output circuit.

How the equation explains the idea

  1. The same current flows through both resistors: I = Vs/(Rfixed + RT). The thermistor voltage is IRT, giving Vout = VsRT/(Rfixed + RT).
  2. With equal 4.0 kΩ resistances, the thermistor takes half the supply: Vout = 6.0 × 4.0/(4.0 + 4.0) = 3.0 V.
  3. When warming lowers RT to 2.0 kΩ, its fraction becomes 2/(4 + 2), so Vout = 2.0 V. Lower thermistor resistance gives a lower voltage in this particular arrangement.
  4. The fixed-resistor output is the complement Vs − Vout, so it rises from 3.0 to 4.0 V. Choose the comparator’s switching direction using the output actually measured.

Worked example

Choose a 3 V threshold

  1. A 6.0 V supply feeds a 4.0 kΩ fixed resistor in series with an NTC thermistor.
  2. At RT = 4.0 kΩ, Vout = 6.0 × 4.0/(4.0 + 4.0) = 3.0 V.
  3. At RT = 2.0 kΩ, Vout falls to 2.0 V; hotter conditions give a lower output.

The comparator direction depends on which resistor supplies the output.

Keep this idea. Draw and label the divider before deciding the switching direction.

Core practical 13: correct the mass-change rate

To estimate a latent heat of vaporisation, supply measured electrical power to a liquid already boiling steadily and record the mass lost over a timed interval. Measure voltage across the heater and current through it, or use a calibrated energy meter. Use the mass decrease attributable to vaporisation, not the initial mass placed in the vessel.

Not all electrical power necessarily becomes latent energy. A useful model is P = Lr + Ploss, where r is the steady mass-loss rate. With two powers and approximately unchanged surroundings and boiling temperature, subtracting the two equations gives L = (P2 − P1)/(r2 − r1). Several powers permit a graph of P against r whose slope gives L and intercept estimates the loss term.

For melting ice, surrounding warmth can add energy even without the heater. A comparable control measurement estimates this background melt rate and must be subtracted from the heater-on rate. These corrections have opposite physical origins but the same lesson: identify the unwanted transfer. Use stable apparatus, low-voltage heaters, protection from hot water and steam, and supervised handling; record scatter and the uncertainties in both rates.

P = Lr + Ploss L = ΔP/Δr when losses remain approximately constant

Symbols and units

  • P is electrical heating power in W; Ploss is power leaving the boiling system without vaporising the measured liquid, also in W.
  • L is specific latent heat of vaporisation in J kg⁻¹; r is the positive mass-loss rate in kg s⁻¹.
  • ΔP = P₂ − P₁ and Δr = r₂ − r₁ compare two steady operating points. 1 g s⁻¹ = 10⁻³ kg s⁻¹.

How the equation explains the idea

  1. Write the two energy balances P₁ = Lr₁ + Ploss and P₂ = Lr₂ + Ploss. Subtract them only if the loss power is approximately unchanged: ΔP = LΔr.
  2. Rearrange to L = ΔP/Δr. Here ΔP = 60 − 40 = 20 W and Δr = 0.020 − 0.010 = 0.010 g s⁻¹ = 1.0 × 10⁻⁵ kg s⁻¹.
  3. Calculate L = 20/(1.0 × 10⁻⁵) = 2.0 × 10⁶ J kg⁻¹. The slope has the right units because W/(kg s⁻¹) = J kg⁻¹.
  4. Check the intercept with the first point: Ploss = 40 − (2.0 × 10⁶)(1.0 × 10⁻⁵) = 20 W. Agreement with both points supports this balance but does not independently prove constant losses.
A power–rate graph separates latent heat and heat loss

Conditions. Computed boiling model P = Lr + Ploss, with L = 2.0 × 10⁶ J kg⁻¹. The liquid stays at its boiling temperature and losses are constant within each comparison. The zero-rate endpoint is the model intercept.

Vaporisation mass rate / kg s⁻¹; Electrical power / W.

Both lines have gradient L. A constant 20 W loss shifts the line upward, so dividing total P by r overestimates L. Differences in P divided by differences in r recover L if the loss really remains unchanged.

Open the interactive lesson to view the plotted curves and their coordinate tables.

Worked example

Use two boiling rates

  1. At 40 W the steady loss rate is 0.010 g s⁻¹; at 60 W it is 0.020 g s⁻¹.
  2. ΔP = 20 W and Δr = 0.010 g s⁻¹ = 1.0 × 10⁻⁵ kg s⁻¹.
  3. L = 20/(1.0 × 10⁻⁵) = 2.0 × 10⁶ J kg⁻¹. The model implies Ploss = 20 W.

The two-power method separates an approximately constant heat loss.

Keep this idea. Use a control or a fitted intercept to estimate unwanted energy transfer.

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.

Browse Physics Unit 5 lessons and resources