Practical skills in core practicals 12 to 14

Core practical 12: calibrating a thermistor

• Thermometer close to the thermistor in a water bath; stir; heat slowly so they are at the same temperature.
• Take readings while heating and while cooling, and average, to cancel lag.
• The calibration curve (output p.d. against temperature) is non-linear. Its sensitivity at a temperature is the gradient of the tangent there; choose the fixed resistor so the circuit is most sensitive near the switching temperature.
• Safety: hot water and hotplate: heat-resistant gloves or tongs; stable beaker.

Core practical 13: specific latent heat

• Fusion: heater in a funnel of crushed ice plus an identical control funnel without heating; L = Pt / (m_heater - m_control). The control removes the systematic effect of energy from the room.
• Vaporisation: two-power method. Losses are about the same at both powers, so L = (P_1 - P_2) / (rate_1 - rate_2), where rate is mass evaporated per second. This removes a systematic error.
• Use a joulemeter or measure V, I and t; time with a stopwatch.

Core practical 14: pressure and volume

• Wait after each compression for the gas to return to room temperature (T constant).
• Read the Bourdon gauge and the column length at eye level; for a uniform tube, length is proportional to volume.
• Plot p against 1/V: straight line through the origin, gradient = NkT. N = gradient / kT.

Worked example

An ice-melting run: P = 36 W for 300 s. Meltwater: 48.6 g from the heated funnel, 16.2 g from the control. Find L.

1. Energy = Pt = 36 x 300 = 10 800 J
2. Mass melted by the heater = 48.6 - 16.2 = 32.4 g = 0.0324 kg
3. L = 10 800 / 0.0324 = 333333 J/kg

Answer: L = 3.33 x 10⁵ J/kg

Check yourself

1. A thermistor circuit's output rises from 2.10 V at 30 °C to 2.58 V at 40 °C. Find the average sensitivity in V/K.
   Show answer

   0.048 V/K. 0.48 V / 10 K = 0.048 V/K.

2. A 50 W heater runs for 240 s. The heated funnel yields 52.0 g of meltwater and the control 15.4 g. Find L, in units of 10⁵ J/kg, to 3 s.f.
   Show answer

   3.28 x 10⁵ J/kg. L = 12 000 / 0.0366 = 328000 J/kg.

3. In the two-power method, P_1 = 60 W gives 2.40 x 10^-5 kg/s evaporated and P_2 = 40 W gives 1.52 x 10^-5 kg/s. Find L, in units of 10⁶ J/kg, to 3 s.f.
   Show answer

   2.27 x 10⁶ J/kg. L = 20 / 0.88 x 10^-5 = 2.27 x 10⁶ J/kg.

4. Using L = 2.27 x 10⁶ J/kg, find the power lost to the surroundings in the 60 W run (rate 2.40 x 10^-5 kg/s), in W, to 2 s.f.
   Show answer

   5.5 W. Useful power = 2.27 x 10⁶ x 2.40 x 10^-5 = 54.5 W; loss = 60 - 54.5 = 5.5 W.

5. A gas column has volume 25.0 cm³. What value of 1/V should be recorded, in cm^-3, to 3 s.f.?
   Show answer

   0.0400 cm^-3. 1 / 25.0 = 0.0400 cm^-3.

6. At 293 K, a gas has p = 1.0 x 10⁵ Pa when V = 25 x 10^-6 m³. Find N, in units of 10²⁰, to 3 s.f. (k = 1.38 x 10^-23 J/K)
   Show answer

   6.18 x 10²⁰. N = pV / kT = 2.5 / (1.38 x 10^-23 x 293) = 6.18 x 10²⁰.

7. In core practical 14, why must the student wait after each change of pressure?
   • Compressing the gas warms it; waiting lets it return to room temperature
   • The gauge needs time to reset
   • The oil must drain away
   • To let the gas leak out
   Show answer

   Compressing the gas warms it; waiting lets it return to room temperature. Boyle's law needs constant temperature.

8. What is the purpose of the control funnel in the ice experiment?
   • To keep the heater cool
   • To measure the ice melted by energy from the surroundings, so it can be subtracted
   • To check the heater's power
   • To measure the temperature of the ice
   Show answer

   To measure the ice melted by energy from the surroundings, so it can be subtracted. It removes a systematic error from heat gained from the room.

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Sources

• Pearson Edexcel International Advanced Subsidiary/Advanced Level in Physics specification (XPH11/YPH11), Issue 3
• Pearson qualifications: International Advanced Levels Physics (2018)
• Pearson examiners' report: IAL Physics WPH16/01, October 2024

Last updated 8 November 2026. Spotted a mistake? Tell us.