Thermal Practicals

Supporting practical enrichment

This page develops experimental-transfer skills using methods in the older anchor notes. The 2026 syllabus requires the thermal quantities and energy reasoning, but it does not prescribe this complete apparatus list. Learn the core definitions first.

Overview

This page focuses on experimental determination of thermal quantities in H2 Physics.

Main practical themes:

• specific heat capacity of solids
• specific heat capacity of liquids
• continuous-flow calorimetry
• specific latent heat of fusion
• specific latent heat of vaporisation
• electrical heating methods
• heat-loss correction methods
• apparatus interpretation
• common sources of error

Most practicals use electrical energy:

$$E=IVt$$

Where:

• $I$ = current
• $V$ = potential difference
• $t$ = heating time

This page supports:

• Thermal Physics A

Definition

Thermal practicals are experiments where electrical energy is converted into thermal energy and related to temperature rise or change of state.

Why It Matters

This topic develops practical skills in setting up apparatus safely, identifying heat losses, interpreting apparatus, choosing the right thermal model, and explaining how correction methods improve accuracy.

Key Representations

Core Ideas

• Electrical heating methods usually start from $E=IVt$.
• The thermal model then depends on whether temperature changes or state changes.
• Heat losses, thermometer lag, poor insulation, and non-uniform temperature are common practical limitations.
• Good answers explain apparatus roles, assumptions, and correction methods, not only formulas.
• Practical questions often test whether the measured energy input is equal to the useful thermal energy gained.

Exam Relevance

Use this branch for practical questions involving electrical heating, specific heat capacity, latent heat, calorimetry, continuous-flow methods, heat-loss corrections, apparatus roles, and sources of uncertainty.

Core Experimental Principle

Electrical energy supplied is converted into thermal energy.

In an idealised energy boundary, if losses and the heat capacities of the heater/container are negligible:

$$IVt=Q$$

Then combine with:

$$Q=mc\Delta T$$

or

$$Q=ml$$

depending on the process.

Determination of Specific Heat Capacity of a Solid

Standard Heated Metal Block Method

A metal block contains:

• electric heater
• thermometer or temperature probe
• insulation around block

Figure: In an electrical specific-heat-capacity experiment, electrical energy supplied by the heater is compared with the thermal energy gained by the sample.

Measure:

• mass $m$
• current $I$
• voltage $V$
• heating time $t$
• temperature rise $\Delta T$

Formula

Assuming negligible heat loss:

$$IVt=mc\Delta T$$

Hence:

$$c=\frac{IVt}{m\Delta T}$$

Why Use a Metal Block?

Metals:

• conduct heat well
• become nearly uniform in temperature
• easy to machine holes for heater and probe

Experimental Precautions

• insulate block well
• ensure heater fits tightly
• thermometer inserted deeply
• record stable initial temperature
• avoid drafts
• measure mass accurately

Worked Example 1

Given:

• $I=2.0\mathrm{A}$
• $V=12\mathrm{V}$
• $t=300\mathrm{s}$
• $m=1.5\mathrm{kg}$
• $\Delta T=12\mathrm{K}$

Find $c$.

$$c=\frac{IVt}{m\Delta T}$$ $$c=\frac{(2.0)(12)(300)}{(1.5)(12)}$$ $$c=400\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$$

Determination of Specific Heat Capacity of a Liquid

Calorimeter Method

Liquid placed in insulated calorimeter with:

• immersion heater
• thermometer
• stirrer

Measure:

• mass of liquid
• electrical input
• temperature rise

Formula (Ignoring Container)

$$IVt=mc\Delta T$$

More Accurate Formula

If calorimeter heat capacity $C$:

$$IVt=mc\Delta T+C\Delta T$$

So:

$$IVt=(mc+C)\Delta T$$

Importance of Stirring

Stirring helps:

• uniform temperature
• more accurate thermometer reading
• faster equilibrium

Continuous-Flow Method for Liquids

Principle

Liquid flows continuously through a heated tube.

Measure:

• mass flow rate $\dot{m}$
• inlet temperature
• outlet temperature
• electrical power

Formula

If heat loss is negligible:

$$IV=\dot{m}c\Delta T$$

Hence:

$$c=\frac{IV}{\dot{m}\Delta T}$$

Advantages

Compared with static calorimeter:

• less heat stored in apparatus
• easier steady-state measurement
• more suitable for liquids

Two-Trial Heat-Loss Correction

Real Situation

Some power is lost to surroundings:

$$IV=\dot{m}c\Delta T+h$$

Where:

• $\dot{m}$ is the mass-flow rate in $\mathrm{kg}{\mathrm{s}}^{-1}$
• $h$ is the approximately steady heat-loss power in watts

Repeat with a different flow rate while reproducing the inlet temperature, outlet temperature, mean apparatus temperature and steady operating conditions as closely as possible.

Then:

$$I_1{V}_1={\dot{m}}_{1}c\Delta T+h$$ $$I_2{V}_2={\dot{m}}_{2}c\Delta T+h$$

Subtract equations to eliminate $h$.

This gives more accurate $c$.

Why Keep Same Temperature Difference?

Because heat loss depends strongly on temperature difference between apparatus and surroundings.

Keeping the same $\Delta T$ is useful but is not sufficient by itself; the apparatus must have a similar temperature distribution relative to the surroundings so that $h$ is approximately the same in both trials.

Determination of Specific Latent Heat of Fusion

Typical Ice Method

Crushed melting ice in funnel.

Heater inserted into ice.

Water produced is collected.

Use second identical setup without heater to estimate environmental melting.

Why Must Ice Be Melting?

At the stated pressure, ice already coexisting with water is at its melting temperature (approximately $0^{\circ }\mathrm{C}$ at standard atmospheric pressure).

$$0^{\circ }\mathrm{C}$$

Therefore no initial warming term is needed in that specified condition.

This is only true if the ice is already melting at $0^{\circ }\mathrm{C}$. If the ice starts below $0^{\circ }\mathrm{C}$, an additional warming term $mc\Delta T$ is needed before melting.

Formula

Let:

• $m_1$ = water collected with heater
• $m_2$ = water collected without heater

Mass melted by heater only:

$$m=m_1-m_2$$

Then:

$$IVt=m{l}_f$$

So:

$$l_f=\frac{IVt}{m_1-m_2}$$

Worked Example 2

Given:

• $I=3.0\mathrm{A}$
• $V=10\mathrm{V}$
• $t=600\mathrm{s}$
• $m_1=0.070\mathrm{kg}$
• $m_2=0.016\mathrm{kg}$

$$m=0.054\mathrm{kg}$$ $$l_f=\frac{(3)(10)(600)}{0.054}=3.33\times 10^5\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$$

Determination of Specific Latent Heat of Vaporisation

Boiling Method

Liquid is boiled electrically.

Steam or vapour produced is condensed and collected.

Measure:

• electrical input
• mass vaporised
• time

Formula (Idealised)

$$IVt=m{l}_v$$

Real Case With Heat Loss

$$IV=\dot{m}{l}_v+h$$

Repeat with second power setting:

$$I_1{V}_1={\dot{m}}_1{l}_v+h$$ $$I_2{V}_2={\dot{m}}_2{l}_v+h$$

Subtract to remove $h$.

Apparatus Interpretation Skills

You may be asked:

• why insulation is used
• why stirring is needed
• why thermometer is placed centrally
• why crushed ice is used
• why second control setup is needed
• why steady state is required

These are common exam questions.

Common Sources of Error

Heat Loss to Surroundings

Causes measured values to be too high or too low depending on setup.

Poor Thermal Contact

Heater not fitted tightly into metal block.

Temperature Lag

Thermometer responds slowly.

Incomplete Stirring

Liquid temperature non-uniform.

Evaporation or Splashing

Mass measurement inaccurate.

Reading Errors

Stopwatch, ammeter, voltmeter, thermometer.

Improving accuracy and managing trade-offs

• use insulation
• use digital sensors
• repeat and average to reduce random scatter; this does not remove a systematic heat-loss bias
• wait for steady readings
• stir continuously
• reduce drafts
• use two-trial correction methods

A larger temperature rise reduces the fractional uncertainty in a temperature measurement, but it also increases the temperature difference from the surroundings and usually increases heat loss. A good design chooses a measurable rise without making this loss excessive; it does not treat “larger $\Delta T$” as an unconditional improvement.

Worked Example 3

Continuous flow heater:

• $I=2.5\mathrm{A}$
• $V=12\mathrm{V}$
• flow rate $=0.020\mathrm{kg}{\mathrm{s}}^{-1}$
• temperature rise $=0.36\mathrm{K}$

Find $c$.

Power:

$$IV=30\mathrm{W}$$

Then:

$$30=(0.020)c(0.36)$$ $$c=\frac{30}{0.0072}=4.17\times 10^3\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$$

(close to water)

Data Analysis Tips

Graph Method

If experiment is repeated for different powers:

Plot:

• power $IV$ vs $\dot{m}\Delta T$

Gradient may give $c$.

Intercept may represent heat loss.

Summary

Thermal practicals are mainly about tracking where electrical energy goes, identifying losses, and justifying how the method improves accuracy.

For a solid or liquid being heated with negligible losses:

$$IVt=mc\Delta T$$

so:

$$c=\frac{IVt}{m\Delta T}$$

For a liquid, if the calorimeter also absorbs energy:

$$IVt=mc\Delta T+C\Delta T$$

For mixing calorimetry:

$$\text{heat lost}=\text{heat gained}$$

For latent heat of fusion or vaporisation:

$$IVt=ml$$

This applies to the mass that changes state at the phase-change temperature. If the substance starts below the phase-change temperature, warming terms must be included.

For a continuous flow calorimeter:

$$P=\dot{m}c\Delta T$$

where $\dot{m}$ is mass flow rate.

Useful plots include $\Delta T$ against $t$, $Q$ against $\Delta T$, and temperature against time. With continued heating of a pure substance at fixed pressure, a temperature plateau can indicate a phase change while two phases coexist.

Real systems lose energy by conduction, convection, radiation, and evaporation. Reduce unwanted transfer using insulation, a lid, reduced draughts and an appropriate short heating duration. Repeat-and-average reduces random scatter, not systematic heat loss. A larger $\Delta T$ reduces fractional reading uncertainty but usually increases heat loss, so it is a design trade-off rather than an unconditional improvement.

Common Exam Traps

Do not assume $IVt=mc\Delta T$ is exact unless heat losses are negligible.

Include container heat capacity if given.

Stir liquids for uniform temperature.

Avoid measuring temperature too late after heating stops because cooling begins immediately.

For latent heat of fusion, ice should be dry before measuring mass; surface water causes error.

Convert units carefully: g to kg, min to s, and kJ to J.

Check zero error and calibration for thermometers and balances.

Links

• Thermal Physics A
• Temperature and Heat
• Thermal Properties of Matter
• First Law of Thermodynamics
• Thermal Measurement and Scales
• Heat Capacity and Latent Heat
• Thermal Physics A Common Exam Traps
• Measurement
• Current Electricity Fundamentals