Edexcel A-Level Physics Thermodynamics

Practise Edexcel A-Level Physics thermodynamics through heating and latent heat, core practicals, internal energy, ideal gases and molecular kinetic theory.

Exam points

calculate heating or latent-heat transfers using E=mcΔθ, E=mL and power-time data
use pV=NkT or pV/T constants with kelvin temperatures to model ideal gases
link molecular kinetic energy, r.m.s. speed and pressure changes to particle motion

Covered syllabus points

5.3.125 - Heating and latent heat equations 5.3.126 - Core Practical 12 - thermistor calibration 5.3.127 - Core Practical 13 - specific latent heat 5.3.128 - Internal energy 5.3.129 - Absolute zero and molecular kinetic energy 5.3.130 - Ideal gas equation 5.3.131 - Core Practical 14 - gas pressure and volume 5.3.132 - Molecular kinetic theory equation

[Maximum number: 3]

The water in a swimming pool at the Olympic Games should be at a temperature of $28.0^{\circ}\mathrm{C}$ .

A pool contains $2750\,\mathrm{m^3}$ of water at $16.5^{\circ}\mathrm{C}$ .

Calculate the increase in internal energy $\Delta E$ of the water when it is heated to $28.0^{\circ}\mathrm{C}$ .

Density of water $=998\,\mathrm{kg\,m^{-3}}$

Specific heat capacity of water $=4190\,\mathrm{J\,kg^{-1}\,K^{-1}}$

Use ho=m/V, so $m= ho V=(998)(2750)=2.74 imes10^6\,\mathrm{kg}$ . Then

$\Delta E=mc\Delta T=(2.74 imes10^6)(4190)(28.0-16.5)=1.32 imes10^{11}\,\mathrm{J}.$

[Maximum number: 6]

The diagram shows some equipment that could be used to calibrate a thermistor.

Figure for Question 5.3 - Thermodynamics question 2 — Edexcel A-Level Physics A2

Describe a method a student could use to obtain a calibration curve for the thermistor from $0^{\circ} \mathrm{C}$ to $100^{\circ} \mathrm{C}$ .

You should identify any extra equipment that would be needed.

[Maximum number: 1]

In an experiment to determine the specific latent heat of vaporisation of water $L_{\mathrm{V}}$ , a student used an electrical heater to boil water in a beaker.

The experimental data gave $L_{\mathrm{V}}=2.20 \mathrm{MJ} \mathrm{kg}^{-1}$ . The textbook value of $L_{\mathrm{V}}$ is $2.26 \mathrm{MJ} \mathrm{kg}^{-1}$ .
Which of the following could be an explanation for this difference in the values of $L_{\mathrm{V}}$ ?

Some energy is transferred to the surroundings.

The heater power was underestimated.

The student did not stir the water.

The heater is inefficient.

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Covered syllabus points

5.3 - Thermodynamics question 1

5.3 - Thermodynamics question 2

5.3 - Thermodynamics question 3