Pietro MeloniPhD

28 ideas · B.1 · 2025 syllabus

IB Physics IA ideas: thermal energy transfers

Thermal investigations measure how energy moves: cooling curves, specific heat capacity, latent heat and insulation. Temperature is easy to log with a probe, so data is rarely the problem; heat losses to the room almost always are.

By Pietro Meloni, PhD · Updated on 26 September 2026

28 of 28 ideas

B.1 Thermal energy transfers: 28 ideas

Specific heat capacity of salt solutions by electrical heating

Research question. How does the specific heat capacity of sodium chloride solution change as its mass concentration rises from 0 to 250 g per kg of water, in 6 steps, measured by electrical heating in a lagged calorimeter?

B.1 Thermal energy transfers
SL and HL
Needs care
Common: on 3 sites

My take. Sensible but risky, because the effect is small and a sloppy setup will hide it. Pick it if you like careful calorimetry and are ready to state honestly whether a difference was detected.

Method, physics and where marks are lost+

Independent variable: Salt mass per 100 g of water: 0, 5, 10, 15, 20, 25, 30 g (7 values), each run 3 times with a fresh solution.
What you measure: Temperature rise measured with a digital thermometer (0.1 °C) in a lagged polystyrene cup while a 50 W immersion heater runs for a fixed time. Energy comes from a joulemeter or from V, I and t. c is calculated from E = mcΔT with heat capacity of the cup included.
Controlled variables: Mass of solution weighed on a 0.01 g balance. Heater power checked by ammeter and voltmeter. Starting temperature about 20 °C every time. Lid, lagging and stirring rate kept the same. Heating time fixed so ΔT stays near 10 °C.
Physics and graph: E = mcΔT, or VIt = (m c + C)ΔT. Plot ΔT against t for each concentration and use the gradient to find c. Then plot c against concentration; the expected fall is small, about a few per cent, so the linear fit and its uncertainty matter. Compare with published data.
SL and HL: SL students find c at each concentration and a trend. To reach the top band, correct for heat lost to the surroundings by plotting a full heating and cooling curve and discuss whether the change is bigger than the uncertainty. HL adds nothing specific.
Where marks are lost: Research design: changes in c are small compared with heat loss, so a weak method gives no visible trend. Data analysis: uncertainty bars larger than the effect. Conclusion: claiming a trend that the data cannot support. Evaluation: not quantifying the heat loss.
Data: Needs a joulemeter or power supply with meters, and a lagged calorimeter; heat loss and the small size of the effect are the main uncertainty.

Heat flow through insulation slabs at steady state

Research question. How does the thickness of foam or fibreboard insulation (5 to 30 mm, 6 values) affect the electrical power needed to keep a heated block at 60 °C?

B.1 Thermal energy transfers
SL and HL
Needs care
Common: on 2 sites

My take. Worth choosing if you are patient, because the steady state design is sound. Twist: use the cooling curve of a wrapped can as a cheaper check and compare the two k values.

Method, physics and where marks are lost+

Independent variable: Insulation thickness, 5 to 30 mm, six values made by stacking identical sheets, each measured with a calliper. Repeat each thickness twice or three times.
What you measure: Electrical power P = VI, from a heater and a power supply with a meter, once the temperature of the hot side has stayed constant for 5 minutes. Temperature by a probe on each face.
Controlled variables: Hot side temperature, held at 60 °C by adjusting the supply. Cold side or room temperature, recorded. Same area of contact and same heater. Same material, with edges sealed to reduce loss.
Physics and graph: Conduction: P = kAΔT/d. Plot P (y) against 1/d (x). The gradient is kAΔT, so k can be found and compared with tables. Note that edge losses give a positive intercept.
SL and HL: SL students can produce the plot and estimate k. Excellent work adds a correction for edge loss, and a second material for comparison. HL does not change much here.
Where marks are lost: Research design: not waiting for steady state, so the reading is still changing. Data analysis: uncertainty in ΔT not propagated. Conclusion: k value not compared with a source. Evaluation: heat lost from the sides ignored, poor contact between layers and heater.
Data: Needs a heater, two thermometers and a power supply; the main issue is reaching steady state, which can take 20 minutes per run.

Linear expansion coefficient of a heated metal rod

Research question. How does the temperature rise (from 20 °C to 95 °C, 6 or more values) of a 0.50 m aluminium rod affect its extension, measured in mm?

B.1 Thermal energy transfers
SL and HL
Hard data
Common: on 2 sites

My take. Worthwhile only with a dial gauge. A ruler cannot see the change. The twist is comparing aluminium, brass and steel rods.

Method, physics and where marks are lost+

Independent variable: Rod temperature, from 20 °C to 95 °C using steam or a hot water jacket, read at about 10 °C steps.
What you measure: Extension measured with a dial gauge or a micrometer-style lever arm to 0.01 mm, with temperature from a thermometer or thermocouple in contact with the rod. Repeat on cooling.
Controlled variables: Same rod, with initial length measured at room temperature. The rod insulated so the temperature is uniform along it. One end fixed rigidly. Same gauge position each run.
Physics and graph: ΔL = αL₀ΔT. Plot ΔL against ΔT; the gradient is αL₀, so α = gradient/L₀. Compare with the tabulated value (about 23 × 10⁻⁶ /K for aluminium).
SL and HL: SL work extracts α and compares it with the literature. Top band handles thermal lag between water and rod, the expansion of the support frame and a hysteresis check on cooling.
Where marks are lost: Research design: extensions of about 1 mm are too small for a ruler. Data analysis: ignoring temperature uncertainty along the rod. Evaluation: heating the support and gauge as well as the rod.
Data: Needs a dial gauge and a steam or hot water jacket; the extension is tiny, so the gauge resolution and frame expansion dominate.

Painted cans cooling: emissivity and convection compared

Research question. How does the surface finish of identical metal cans (matt black, gloss white, bare shiny, foil wrapped) change the initial cooling rate of water from 80 °C in a 20 min period?

B.1 Thermal energy transfers
SL and HL
Needs care
Common: on 2 sites

My take. Good if you tackle the convection problem directly, since otherwise differences vanish. Twist: use a Leslie cube style set up or a lid to cut evaporation.

Method, physics and where marks are lost+

Independent variable: Surface finish of the can, four to five types, each can filled with the same mass of water and tested three times.
What you measure: Water temperature with a probe every 30 s for 20 min. Calculate initial cooling rate dT/dt from the slope of the first few minutes, plus the total energy lost using Q = mcΔT.
Controlled variables: Same starting temperature of 80 °C. Same mass of water measured on a balance. Same can size and lid with insulated top. Same location out of draughts, room temperature logged, can raised on a cork mat.
Physics and graph: Power radiated P = eσA(T⁴ − Ts⁴), plus convection. Plot ln(T − Ts) against t; the gradient gives the cooling constant which should be larger for higher emissivity. Convection is best reduced by a shield.
SL and HL: SL compares finishes qualitatively and ranks them. Higher marks come from separating radiation from convection, for example by repeating in a closed box, or by using an infrared thermometer to check surface temperature. HL can fit the T⁴ law.
Where marks are lost: Research design: temperature of the surroundings and the draughts not controlled, so differences are noise. Data analysis: comparing raw curves without a cooling constant or uncertainty. Evaluation: not admitting that convection and evaporation from the lid dominate over radiation at these temperatures.
Data: Needs identical cans, paint, a data logger or probe thermometer and a balance; main uncertainty is convection and evaporation swamping the radiation difference.

Spring stiffness of a steel spring in warm water

Research question. How does the spring constant k of a steel spring change as its temperature is raised from 20 °C to 80 °C in steps of 10 °C?

B.1 Thermal energy transfers
SL and HL
Hard data
Common: on 2 sites

My take. Risky because the real effect is very small and may vanish in your uncertainty. It can still be a good result if you plan for a high-resolution measurement and honestly state a null outcome. Otherwise choose something with a clear signal.

Method, physics and where marks are lost+

Independent variable: Temperature of the water bath around the spring, 20 to 80 °C, 7 values, three loading runs per temperature.
What you measure: Extension of the spring with a fixed 200 g load, read by a rule or a phone photo against a scale, and a digital thermometer for temperature. k = F/x is calculated, and better from the gradient of force against extension for 4 masses.
Controlled variables: Same spring throughout. Same masses used and the load applied for a fixed time. Spring fully in the water bath so the temperature is even. Reading taken when the temperature is stable, within 1 °C. Extension below the elastic limit.
Physics and graph: Hooke's law F = kx. Plot k (y) against T (x). The temperature dependence is small, so expect a change of only a few percent over the range, well within noise. The gradient would give a temperature coefficient.
SL and HL: SL students can carry out the measurements and judge whether any trend beats the uncertainty. Reaching the top band needs uncertainty propagation on k and a proper conclusion, including 'no measurable change'. HL students can relate it to the temperature dependence of Young's modulus.
Where marks are lost: Research design: thermal expansion of the spring and the water buoyancy on the mass ignored. Data analysis: no error bars, so any trend is claimed without support. Conclusion: overclaiming a trend. Evaluation: rule resolution compared with the tiny effect not discussed.
Data: Needs a water bath, thermometer and a fine measurement of extension, and the effect is small compared with reading error.

Air gap width in a double glazed model window

Research question. How does the air gap between two glass panes, varied from 2 mm to 20 mm, affect the rate of cooling of 200 g of water at 60 °C in a box with a double glazed window?

B.1 Thermal energy transfers
SL and HL
Needs care
Rarely listed

My take. A good idea if you can make the window the dominant loss path. Test a control run with the window blocked with foam to prove it, which makes the project yours.

Method, physics and where marks are lost+

Independent variable: Air gap width between two microscope slides or acrylic sheets, 2, 5, 8, 12, 16, 20 mm (6 values), 3 repeats each, set with spacers of measured thickness.
What you measure: Temperature of hot water in an insulated box read with a thermometer or data logger every 30 s for 10 min. Initial cooling rate in °C per s is taken from the gradient, and power lost is found using P = mcΔθ/Δt.
Controlled variables: Start temperature and volume of water identical. Window area the same, with other box walls thickly insulated. Room temperature and draughts monitored and kept constant. Gap sealed with tape to stop air exchange.
Physics and graph: Conduction through a layer, P = kAΔT/d, and convection in a gap that becomes important once the gap is wide. Plot power lost against gap width and look for a minimum or levelling off, and 1/gap width if conduction alone is expected.
SL and HL: SL students can measure cooling rate and describe the trend using conduction and convection. Top band work identifies the gap at which convection takes over and uses U-values or thermal resistance to compare with the data.
Where marks are lost: Research design: heat escaping through the box walls swamps the window effect. Data analysis: comparing raw cooling curves rather than initial rates with uncertainties. Conclusion: claiming that larger gaps always insulate better. Evaluation: not quantifying how much heat leaks around the edges.
Data: Needs a well insulated box and a logger; the main uncertainty is heat loss through everything except the window.

Boiling water at fixed power: latent heat from mass loss

Research question. How does the electrical power of an immersion heater, varied from 50 W to 250 W, affect the rate of mass loss of boiling water, and what value of specific latent heat of vaporisation does this give?

B.1 Thermal energy transfers
SL and HL
Needs care
Rarely listed

My take. A strong experiment because the intercept gives you a physical result about heat loss. Take care with electrical safety and keep the water level constant.

Method, physics and where marks are lost+

Independent variable: Heater power: 6 settings from about 50 W to 250 W, set with a variable DC supply and measured as V × I. Each held for 5 minutes of steady boiling, repeated 3 times.
What you measure: Mass of the beaker on a balance (±0.01 g) read every 30 s, giving the mass loss rate dm/dt from the gradient of a mass against time graph. Power from a voltmeter and ammeter.
Controlled variables: Starting water mass: same 300 g each run. Container: same insulated beaker with the same partial lid. Heater position: clamped, fully immersed, not touching the base. Water: start every run already boiling, to ignore warm-up.
Physics and graph: P = L (dm/dt) + P_loss. Plot P (y) against dm/dt (x): gradient is L, and the y-intercept is the power lost to surroundings. Compare with 2.26 MJ/kg.
SL and HL: SL: plot P against dm/dt, read L, and comment on the intercept. Top band or HL depth: consider whether P_loss changes with power, use error bars from the repeats, and discuss steam condensing on the lid and the heater lead.
Where marks are lost: Research design: including the warm-up period in the mass loss. Data analysis: no uncertainty on the gradient, or forcing the line through the origin. Conclusion: not explaining why L comes out too low or too high. Evaluation: ignoring condensed steam or convection forces on the balance.
Data: Needs a mains-safe immersion heater or a low-voltage coil, and a balance able to handle a hot beaker; the main uncertainty is steam condensing and buoyancy forces from the rising steam.

Comparing thermal conductivity of metal rods by steady state

Research question. How does the thermal conductivity of copper, aluminium, brass and steel rods of equal length 20 cm and diameter 1 cm affect the temperature gradient along each rod at steady state with one end in boiling water?

B.1 Thermal energy transfers
SL and HL
Needs care
Rarely listed

My take. Sound if you drop the heat exchanger language and measure k properly. Twist: add a metal from something you own, such as a spoon, and predict its gradient first.

Method, physics and where marks are lost+

Independent variable: Metal type, four to five metals (copper, aluminium, brass, mild steel, and stainless steel if available), with 6 thermometer positions along each rod as second variable.
What you measure: Temperature at fixed points along each rod from thermocouples or a multi-channel logger, reading to 0.1 °C. Temperature gradient dT/dx found from a plot, then relative conductivity k ∝ 1/gradient.
Controlled variables: Rod length and diameter measured with vernier callipers. Hot end held at 100 °C in a kettle or steam. Cold end in an ice bath or at fixed room temperature. Rods lagged with the same insulation and steady state judged when the readings stop changing for two minutes.
Physics and graph: Fourier conduction P = kA(ΔT/Δx). Plot temperature against distance along the rod, gradient is proportional to 1/k for a constant heat flow. Compare with the ratio of literature k values.
SL and HL: SL students can rank the metals and compare gradients. Top band work estimates the heat flow from the cold end water (mcΔT/t) to get an absolute k. HL not needed.
Where marks are lost: Research design: 'heat exchanger efficiency' is not measured, so the RQ should be about conductivity. Data analysis: not reaching steady state before reading. Evaluation: side losses from unlagged rods.
Data: Needs rods of equal size and several thermometers or thermocouples; lateral heat loss and thermal contact at the hot end are the main uncertainties.

Cooling constant of hot water and open surface area

Research question. How does the open surface area of hot water, varied from 20 cm² to 80 cm² using cylindrical containers of different diameter, affect the cooling constant k in Newton's law of cooling?

B.1 Thermal energy transfers
SL and HL
Needs care
Rarely listed

My take. A worthwhile investigation if you keep it about the cooling constant and not a general talk on cooling. Insulating the sides is the twist that isolates the top surface.

Method, physics and where marks are lost+

Independent variable: Exposed surface area: 5 to 6 different beakers or cups with diameters from about 5 cm to 10 cm, giving areas from 20 cm² to 80 cm². Two runs each.
What you measure: Temperature logged every 30 s for 30 minutes with a temperature probe (±0.1 °C). The constant k is found from the gradient of ln(T − T_room) against time.
Controlled variables: Water volume: use 200 ml in each, which changes the depth, so consider also a set with equal depth. Starting temperature: 80 °C. Room temperature: measured with a second probe, no draughts. Container material: same thin-walled glass or metal, insulated on the sides and base.
Physics and graph: dT/dt = −k(T − T_env), so ln(T − T_env) = −kt + constant. Plot ln(T − T_env) against t, gradient −k. Then plot k against area A: if evaporation and convection from the top dominate, k should be roughly proportional to A/(mc).
SL and HL: SL: extract k from each run and plot k against area. Top band or HL depth: separate the evaporation contribution using lids, and explain the change from the exponential model at high temperature difference.
Where marks are lost: Research design: changing the water depth and mass together with the area. Data analysis: fitting a single exponential to the whole curve, including the early part where evaporation makes it non-linear. Conclusion: claiming proportionality without checking the fit. Evaluation: not considering the changing room temperature.
Data: A temperature probe with a data logger is helpful; without one, manual readings every minute work but are noisier. The main uncertainty is evaporation and draughts.

Cooling constant of water in cups of different surface area

Research question. How does the open top area A of a beaker (A = 20 to 80 cm², five sizes) filled with 150 g of water at 80 °C affect the cooling constant k over 20 minutes?

B.1 Thermal energy transfers
SL and HL
Easy data
Rarely listed

My take. A good sound choice if only one thing changes. Choose area with a fixed material, then repeat with lids on to isolate evaporation, which makes it your own.

Method, physics and where marks are lost+

Independent variable: Open surface area A of five similar cylindrical containers, 20, 32, 45, 60, 80 cm², from the measured diameters, each run three times.
What you measure: Water temperature every 30 s from a digital thermometer or logger; k from the gradient of ln(T − T_room) against time.
Controlled variables: Mass of water: 150 g on a balance. Starting temperature: 80 °C by the kettle and thermometer. Room temperature: recorded each run, drafts avoided by working in one closed room. Container material: all glass or all the same plastic, with the same lid condition.
Physics and graph: Newton's law of cooling: dT/dt = −k(T − T_room), so ln(T − T_room) against t is a straight line with gradient −k. Then plot k against A to find whether it is proportional. Evaporation from an open surface adds a second loss path.
SL and HL: SL students can complete the two step analysis. To reach the top band, separate evaporation by covering the cup with a lid and comparing, and use the linearity of the ln graph residuals to state where Newton's law stops holding at large temperature differences.
Where marks are lost: Research design: shape, material and area varied at once. Data analysis: k found from a curve fit with no check of the linear plot. Evaluation: ignoring evaporation, so the area effect is not just conduction and radiation.
Data: A kettle, thermometer or logger and a set of beakers are enough; room drafts and evaporation are the main uncertainties.

Cooling curves of water, oil and salt solution

Research question. How does the mass fraction of salt in water (0 to 20 % in 5 steps) affect the initial cooling rate of 150 g samples starting at 70 °C in identical lidded beakers?

B.1 Thermal energy transfers
SL and HL
Easy data
Rarely listed

My take. Better than comparing random liquids, because salt fraction is a continuous variable with a clear expected trend. Use lids and equal mass to get a fair test.

Method, physics and where marks are lost+

Independent variable: Salt mass fraction, 0, 5, 10, 15, 20 %, made up on a balance, three runs each. Cooking oil can be added as an extra comparison liquid.
What you measure: Temperature every 30 s over 15 minutes from a digital thermometer; cooling rate from the initial gradient of the T against t graph, and k from ln(T − T_room) against t.
Controlled variables: Sample mass: 150 g on a balance. Start temperature: 70 °C. Container and lid: same beaker and lid for each run. Room temperature: measured each run, no drafts. Stirring: same gentle stir before each reading.
Physics and graph: Rate of energy loss P = m c dT/dt, and Newton's law of cooling gives dT/dt = −k(T − T_room). The specific heat capacity of the solution falls with salt content, so plot k or the initial rate against mass fraction, and compare a calculated c with the data value.
SL and HL: Suitable for SL. Top band work relates the change in cooling rate to the change in specific heat capacity and to the surface loss, and separates evaporation by using lids.
Where marks are lost: Research design: different liquids vary density, volume and specific heat at once. Data analysis: reading cooling rate from a single point. Evaluation: evaporation and convection differences not addressed.
Data: A kettle, thermometer or logger, beakers and salt are enough; evaporation and non-uniform temperature in the liquid are the main uncertainties.

Cooling of salt solutions of different density

Research question. How does the mass of potassium chloride dissolved in 200 g of hot water, from 0 to 15 g in steps of 3 g, affect the initial cooling rate of the solution over 10 minutes?

B.1 Thermal energy transfers
SL and HL
Needs care
Rarely listed

My take. Fine but the expected effect is small and easily lost in noise. Better to frame it around heat capacity and check your prediction, not density.

Method, physics and where marks are lost+

Independent variable: Mass of KCl dissolved in 200 g of water: 0, 3, 6, 9, 12 and 15 g. Two or three repeated cooling runs for each.
What you measure: Temperature from a digital thermometer or logger every 30 s from about 70 °C. Initial cooling rate found from the gradient of the first few minutes of the temperature time graph. Density checked with a measuring cylinder and balance.
Controlled variables: Starting temperature: 70 °C for every run. Water mass: 200 g weighed. Container: the same insulated cup with lid. Room temperature and draughts: same bench, recorded at start and end.
Physics and graph: Newton's law of cooling, dT/dt = −k(T − T_room). Plot ln(T − T_room) against t and use the gradient as −k. Salt changes the specific heat capacity and mass, so compare k with a prediction from c of the solution.
SL and HL: SL: cooling curves and gradients, with a clear link to heat capacity. Top band: linearise with ln, predict k from the changed heat capacity and separate the effect of evaporation. HL: nothing extra needed.
Where marks are lost: Research design: calling density the variable when mass of salt is what is changed, and dissolving not being complete. Data analysis: rate taken from two points. Evaluation: not addressing that evaporation and lid losses dominate over a small effect.
Data: Thermometer or logger, cups and a balance; the effect is small, so repeats and a lid are essential to see any trend above scatter.

Cooling of warm water in wrapped containers

Research question. How does the initial cooling rate of 200 mL of water, starting at 60 °C in a wrapped beaker, depend on the thickness of a wool layer, from one to six layers, over 15 minutes?

B.1 Thermal energy transfers
SL and HL
Easy data
Rarely listed

My take. Easy and reliable, but common as a topic. Fix the thickness question or the material choice so it is measurable. The link to animals is only an analogy. Testing your own gear, such as a jacket fabric or a hot water bottle cover, would make it personal.

Method, physics and where marks are lost+

Independent variable: Number of wool layers wrapped round the beaker: 1, 2, 3, 4, 5 and 6 (6 values), plus an unwrapped control. Each value is repeated 3 times.
What you measure: Water temperature every 30 s for 15 minutes, using a digital thermometer or a temperature probe. Cooling rate is the gradient of the temperature against time in the first 5 minutes, or the fitted decay constant of the whole curve.
Controlled variables: Same beaker and 200 mL of water, measured in a measuring cylinder. Starting temperature 60 °C, set with a kettle and a thermometer. Room temperature logged and draughts avoided, with a lid on the beaker. Same wool, with the layer thickness measured with a ruler and the wrap tight each time.
Physics and graph: Newton's law of cooling: dT/dt = −k(T − Troom), so ln(T − Troom) against t is a straight line with gradient −k. Rate of energy loss P = mc dT/dt. Plot k or P against the number of layers, or against the inverse of thickness, to test whether conduction through the layer, where P ∝ A ΔT/d, matches the data. Comparing different materials, as in the original idea, is better done with equal thickness.
SL and HL: SL students plot cooling curves and compare rates for each thickness. Top band work linearises with ln(T − Troom), links k to thermal conductivity, and discusses heat lost through the lid and base. HL students can estimate effective thermal conductivity of the wool from the gradient and compare it with a published value.
Where marks are lost: Research design: comparing materials of different thickness mixes two variables, so fix one. Data analysis: reading a hand held thermometer with no uncertainty and using one run only. Conclusion: claiming a material is better without a numerical value. Evaluation: heat loss from the top and base, and uneven wrapping, are seldom addressed.
Data: Needs beakers, a kettle, a thermometer or probe, a stopwatch and wool. The main uncertainty is uneven wrapping and evaporation from the water surface.

Cooling rate of hot water in containers of different surface area

Research question. How does the exposed surface area of hot water (from 20 cm² to 80 cm², 6 values) in cylindrical beakers affect the initial rate of cooling in °C per minute, starting at 80 °C?

B.1 Thermal energy transfers
SL and HL
Easy data
Rarely listed

My take. Worth doing if you flip it to cooling, which is far easier to measure than heating. Twist: use real objects like a mug shape you actually own and link the result to why soup cools faster in a wide bowl.

Method, physics and where marks are lost+

Independent variable: Open surface area of the water, changed by using 6 cylindrical containers of different diameter (about 5 to 10 cm), each with the same mass of water. Three repeats per container.
What you measure: Temperature against time with a digital thermometer or temperature probe and datalogger, every 30 s for 10 min. The initial cooling rate is found from the gradient of a tangent, or from a fit of the first few minutes.
Controlled variables: Starting temperature: heat every sample to 80 °C and start timing at the same reading. Mass of water: weigh on a balance. Room conditions: same bench, no draughts, door closed. Container material and lid: use identical thin metal or plastic and either no lid throughout or a lid throughout.
Physics and graph: Newton's law of cooling and the rate of energy loss P = hAΔT (with evaporation as an extra effect). Plot initial cooling rate against surface area; a straight line through the origin supports proportionality, and the gradient links to h and to the mass and specific heat capacity of the water. Note that the original wording heats objects, so cooling is the cleaner measurable version.
SL and HL: SL: measure rates, plot rate against area, state whether it is proportional. Top band or deeper: separate evaporation from convection by comparing lidded and open runs, and fit exponential decay to extract a cooling constant per container.
Where marks are lost: Research design: depth of water changes with area if volume is not fixed, and this is missed. Data analysis: tangent gradients drawn by eye with no uncertainty. Conclusion: claiming proportionality without comparing the fit to the uncertainty. Evaluation: ignoring evaporation and the heat lost through the sides and base.
Data: Beakers or cans, thermometer or probe, balance and stopwatch are enough; the main uncertainty is evaporation and draughts changing the loss between runs.

Correcting heat loss in a specific heat capacity block

Research question. How does the thickness of foam insulation around a 1.0 kg aluminium block, varied from 0 cm to 5 cm in 1 cm steps, affect the specific heat capacity calculated from a 10 minute electrical heating run?

B.1 Thermal energy transfers
SL and HL
Needs care
Rarely listed

My take. Turns a routine practical into an inquiry about systematic error. Worth choosing if you do the cooling curve correction, otherwise it is just a familiar experiment.

Method, physics and where marks are lost+

Independent variable: Thickness of insulation (0 cm to 5 cm in 1 cm steps, 6 values, 3 repeats each).
What you measure: Joulemeter or voltmeter and ammeter give the electrical energy E = VIt. A thermometer or temperature probe gives ΔT. Apparent c = E/(mΔT), then corrected using the cooling curve measured after heating stops.
Controlled variables: Heater power: a stabilised supply and the same 12 V, 50 W heater. Heating time: 600 s with a timer. Starting temperature: block at room temperature, checked before each run. Room conditions: away from draughts, with ambient temperature recorded.
Physics and graph: Energy balance E = mcΔT + heat lost. Plot the apparent c against insulation thickness and show how it tends towards the true value of about 900 J kg⁻¹ K⁻¹. Alternatively, plot temperature against time on the cooling curve and use the loss rate to add back the lost thermal energy.
SL and HL: SL: measure and plot apparent c against thickness, and compare with the accepted value. Top band: model the loss with Newton's cooling law from the cooling curve and correct each result. HL is not needed.
Where marks are lost: Research design: thermometer not in good contact, with no oil in the hole. Data analysis: no percentage difference from the accepted value or uncertainty propagated. Conclusion: claiming insulation removes all loss. Evaluation: ignoring the thermal energy in the heater and the lag in temperature.
Data: A block with a heater and probe hole is standard school kit; the main uncertainty is temperature lag in the block and heat loss.

Counting glass panes in a model window

Research question. How does the number of glass panes, from 1 to 5 with 5 mm air gaps, affect the power lost from hot water at 60 °C through a model window?

B.1 Thermal energy transfers
SL and HL
Needs care
Rarely listed

My take. Fine but close to the gap width idea, so pick one. The series resistance graph gives you a clean model to test, which is the real strength here.

Method, physics and where marks are lost+

Independent variable: Number of panes in the window, 1 to 5 (5 values), stacked with equal 5 mm spacers, 3 repeats each.
What you measure: Water temperature against time with a thermometer or logger, giving the initial cooling rate. Power lost is found from mcΔθ/Δt and thermal resistance from R = ΔT/P.
Controlled variables: Same water mass and start temperature. Same window area and gap width. Box walls insulated identically. Same room with no draughts, and the room temperature recorded each run.
Physics and graph: Thermal resistances of layers in series add, so total resistance should grow linearly with the number of gaps. Plot thermal resistance (ΔT/P) against number of panes; the gradient is resistance per pane and the intercept gives the box's other losses.
SL and HL: SL students show that heat loss falls as panes are added and explain why. Top band work tests the series resistance model and finds diminishing returns from a fixed total thickness.
Where marks are lost: Research design: gaps not kept equal as panes are added. Data analysis: not converting to resistance so the linear model cannot be tested. Conclusion: ignoring the intercept from other heat paths. Evaluation: not addressing edge leaks and seal quality.
Data: Uses microscope slides or acrylic sheets, spacers and a logger; the main uncertainty is leaks and stacking alignment.

Cylinder shape and Newton cooling of hot water

Research question. How does the ratio of exposed surface area to volume, varied from about 0.5 cm^-1 to 1.5 cm^-1 using six metal cylinders holding 200 cm³ of water each, affect the initial cooling constant k of water starting at 80 °C?

B.1 Thermal energy transfers
SL and HL
Easy data
Rarely listed

My take. Simple and safe, and it works well if you keep the material fixed and analyse ln(T - T_room) rather than just comparing final temperatures. Personal twist: choose real objects such as a mug, a flask and a saucepan and ask which shape a coffee shop should use to keep drinks warm longest.

Method, physics and where marks are lost+

Independent variable: Ratio A/V of the container, changed by using six thin-walled aluminium or steel cans of different radius and height, all filled with the same 200 cm³ of water. A/V is calculated from measured diameter and water height, including the open top or a fitted lid. Range about 0.5 to 1.5 cm^-1, three trials per shape.
What you measure: Water temperature read every 30 s for 20 minutes with a digital temperature probe and data logger (or thermometer plus stopwatch). Ln(T - T_room) is plotted against time for the first 10 to 15 minutes, and the gradient magnitude gives the cooling constant k in s^-1.
Controlled variables: Starting temperature of 80 °C, checked with the probe before each run. Water volume of 200 cm³, measured with a measuring cylinder. Room temperature, recorded throughout and kept steady by a closed room with no draught. Lid material and stirring, using the same insulating lid and one gentle stir before each reading, and containers standing on the same insulating mat.
Physics and graph: Rate of heat loss is proportional to surface area and to the temperature difference with the surroundings, so dT/dt = -k(T - T_room) with k proportional to A/(mC), that is to A/V for the same water. Plot ln(T - T_room) against t: the gradient is -k. Then plot k against A/V: a straight line through the origin supports proportionality, and its gradient links to the heat transfer coefficient and water's specific heat capacity.
SL and HL: An SL student can measure cooling curves, extract k for each shape and show that k rises with A/V, with uncertainty bars from repeats. Top band work separates the surface losses (convection, evaporation, radiation) by using lids, tests whether the line really passes through the origin, and explains any intercept as heat lost through the base or the lid. HL depth can add a discussion of radiative loss with the T⁴ law and a comparison of the fitted heat transfer coefficient with a literature value.
Where marks are lost: Research design: A/V changed while wall material, thickness and lid are also different, so several variables change together; open top ignored in the area calculation. Data analysis: fitting the whole curve including the late stage near room temperature, where the reading noise dominates, and not propagating uncertainty in A/V. Conclusion: claiming proportionality without checking the intercept or comparing k values with the spread from repeats. Evaluation: not discussing evaporation and uneven temperature inside the water as systematic errors.
Data: Needs only cans, a thermometer or probe, a kettle and a stopwatch, and the main uncertainty is evaporation and draughts changing the loss from run to run.

Does water's specific heat capacity drift between 20 and 70 °C?

Research question. Does the specific heat capacity of water, measured with a 50 W immersion heater, change when its starting temperature is 20, 30, 40, 50, 60 and 70 °C?

B.1 Thermal energy transfers
SL and HL
Needs care
Rarely listed

My take. Good if you accept that the true answer is nearly flat and make heat loss the real subject. Twist: try a liquid you use at home, such as olive oil, where c does change measurably.

Method, physics and where marks are lost+

Independent variable: Starting water temperature, six values from 20 °C to 70 °C, each repeated three times.
What you measure: Temperature rise over a fixed 120 s of heating, from a digital thermometer or temperature probe reading to 0.1 °C. Energy from joulemeter or from V·I·t. Calculate c = E/(mΔT).
Controlled variables: Water mass 200 g by balance each time. Same heating time and same power, checked with a voltmeter and ammeter. Same insulated polystyrene cup with lid. Stirring at a constant rate before every reading.
Physics and graph: Q = mcΔT and E = VIt. Plot calculated c against starting temperature and see if the gradient is zero within uncertainty. Heat loss grows with temperature, so a correction from a cooling curve is expected.
SL and HL: SL students can get a flat line and a value near 4.18 kJ kg⁻¹ K⁻¹, then argue why. Stronger work measures heat loss with a heater-off cooling run at each temperature and subtracts it. Accurate handling of the fact that real c varies by under 1% is what pushes it into top band.
Where marks are lost: Conclusion: claiming c changes when the trend is just heat loss at higher temperatures. Data analysis: no propagated uncertainty on ΔT, which is small. Evaluation: no quantified heat loss correction.
Data: Needs an immersion heater and a good thermometer; the dominant uncertainty is heat loss to the air at high starting temperatures.

Evaporation rate against open surface area of water

Research question. How does the evaporation rate of water at 22 °C vary with exposed surface area, using circular dishes of diameter 4, 6, 8, 10, 12 and 14 cm over 24 hours?

B.1 Thermal energy transfers
SL
Easy data
Rarely listed

My take. An easy and safe choice that can score well if you record the humidity and test for proportionality. Twist: use a dish from your own kitchen and place a control dish beside each run.

Method, physics and where marks are lost+

Independent variable: Open surface area, six dishes of known diameter with area πd²/4, from about 13 to 154 cm².
What you measure: Mass lost over 24 hours, measured with a 0.01 g balance at the start and end. Rate of evaporation is mass lost divided by time. Repeat each dish three times.
Controlled variables: Same starting water depth in every dish (not volume), because depth might matter. Same room, shelf and temperature, with a thermometer and humidity reading. No draughts, dishes kept in a closed cupboard or away from windows. Same water source.
Physics and graph: Evaporation as loss of the highest energy molecules, and rate of mass loss expected proportional to area. Plot mass loss rate against area, expecting a straight line through the origin, with gradient as mass flux per unit area. Link to latent heat by calculating the energy removed, L·Δm.
SL and HL: SL students can do a straight line fit and interpret the gradient. To reach top band, compare with a model using vapour pressure, or show edge effects make the line curve. HL adds nothing specific.
Where marks are lost: Research design: controlling volume instead of depth, so dishes of different area have different depths. Data analysis: no uncertainty on area or drifting room humidity. Evaluation: missing that air over small dishes is affected by rim effects.
Data: Needs a balance to 0.01 g and patience; the main uncertainty is humidity and air movement changing between days.

Ice melting in warm water: latent heat by mixing

Research question. How does the initial temperature of 150 g of water, varied from 30 °C to 70 °C in steps of 10 °C, affect the value of the specific latent heat of fusion of ice, found from the final mixing temperature?

B.1 Thermal energy transfers
SL and HL
Easy data
Rarely listed

My take. Common school experiment, but the twist of testing whether L depends on starting temperature turns it into a proper test of the method. Worth choosing if you are careful with the ice mass and treat the flat trend as a result.

Method, physics and where marks are lost+

Independent variable: Initial water temperature: 30, 40, 50, 60, 70 °C (5 values), 3 repeats each, using a fresh ice cube of about 20 g every time.
What you measure: Final equilibrium temperature read with a digital thermometer (±0.1 °C) and masses from a balance (±0.01 g). Latent heat L is calculated from energy lost by the water and calorimeter equalling energy gained by the melting ice and its meltwater.
Controlled variables: Ice mass: weighed after the cube is dried and by the rise in cup mass at the end. Ice starting temperature: use ice sitting in melting ice water at 0 °C, then blotted. Calorimeter: same polystyrene cup and lid every time. Stirring: constant gentle stirring until the minimum temperature is reached.
Physics and graph: Q = mcΔT and Q = mL. Energy balance: m_w c_w (T_i − T_f) + C_cal (T_i − T_f) = m_ice L + m_ice c_w (T_f − 0). Plot m_w(T_i − T_f) against m_ice, or plot energy lost by the warm water against ice mass melted at fixed T_i. Gradient gives L, and a flat trend of L against T_i shows the method is consistent.
SL and HL: SL: calculate L for each temperature, compare with 334 kJ/kg and comment on the trend. Top band or HL depth: measure the calorimeter heat capacity in a separate run, propagate uncertainties, and model heat gain from the room to explain a drift of L with starting temperature.
Where marks are lost: Research design: forgetting to dry the ice, so surface water adds mass and gives too low an L. Data analysis: not including the calorimeter or the warming of meltwater from 0 °C. Conclusion: claiming a physical dependence of L on temperature when the spread is within uncertainty. Evaluation: not quantifying heat exchange with the room.
Data: Polystyrene cup, thermometer, balance and ice are enough; the main uncertainty is water clinging to the ice and heat gained from the room.

Insulating a hot can with household materials

Research question. How does the thickness x of a wool wrapping (x = 0 to 2.0 cm in 5 steps) round a 250 mL aluminium can affect the time for 200 g of water to cool from 80 °C to 50 °C?

B.1 Thermal energy transfers
SL and HL
Easy data
Rarely listed

My take. Fine, but a which is best comparison is thin. Vary thickness of one material and extract a conductivity, and it becomes a real investigation.

Method, physics and where marks are lost+

Independent variable: Thickness x of a single insulator, 0, 0.5, 1.0, 1.5, 2.0 cm (measured with callipers), three runs each. A second material can be repeated at the same thicknesses for comparison.
What you measure: Time to cool by 30 K with a thermometer or logger, and the average power loss P = m c ΔT / t calculated from it.
Controlled variables: Water mass: 200 g on a balance. Start temperature: 80 °C. Lid: same lid, insulated in the same way. Room temperature: recorded and drafts avoided.
Physics and graph: Conduction through a layer: P = k A ΔT / x, though for a cylinder the log form is more accurate. A plot of 1/P against x is expected to be straight with gradient related to 1/(kA ΔT); the conductivity of the material is found and compared with a published value. Average power from P = m c ΔT / t.
SL and HL: Entirely SL. Top band work notes the cylindrical geometry, uses a mean temperature difference over the run, and compares the thermal conductivity found with the data value for wool or for still air.
Where marks are lost: Research design: different materials compared with different thickness and packing. Data analysis: no conversion to a quantity that can be compared with the literature. Evaluation: heat lost through the lid and base not treated.
Data: Only cans, a thermometer and materials are needed; heat lost through the uninsulated top and base is the main uncertainty.

Linear expansion coefficients of metal rods heated by steam

Research question. How does the extension of a 60 cm rod of aluminium, brass and steel change as its temperature rises from 20 °C to 95 °C in steps of about 15 °C, and what are the expansion coefficients?

B.1 Thermal energy transfers
SL and HL
Needs care
Rarely listed

My take. A good practical because there is a clear linear model, but only with a dial gauge, since a ruler is too coarse. The twist is to compare three metals and check the values against the data booklet.

Method, physics and where marks are lost+

Independent variable: Temperature rise of the rod (about 6 values from 20 °C to 95 °C), with three rod materials, so the metal is a second variable; rods heated in a steam jacket or a hot water tube.
What you measure: Extension ΔL measured with a dial gauge or a micrometer against a fixed stop (mm, resolution 0.01 mm), with the temperature from a digital thermometer at the middle of the rod; α is calculated from the gradient.
Controlled variables: Initial length: same 60.0 cm, measured with a rule at 20 °C. Rod cross-section: rods of the same diameter. Heating: same steam generator and time to reach equilibrium. Fixing: same clamp arrangement so the rod pushes only on the gauge.
Physics and graph: ΔL = αL₀ΔT. Plot ΔL (y) against ΔT (x): the gradient is αL₀, so α = gradient / L₀. Compare α for the three metals with the data booklet.
SL and HL: SL: one linear graph per metal and a comparison with data values. Top band: repeat on heating and cooling to test for lag, treat the extension of the gauge itself, and use the fractional uncertainty. Plastics and ceramics from the original are not practical in school and add nothing.
Where marks are lost: Research design: the input idea is too vague, since 'material type' and temperature are both varied with no method for measuring small changes. Data analysis: the extension is only about 1 mm, so the relative uncertainty is large and often not shown. Conclusion: no comparison with the accepted values. Evaluation: the temperature along the rod is not uniform and the thermometer reads the steam, not the metal.
Data: Needs a steam generator, expansion rods and a dial gauge; the main uncertainty is the small extension and uneven rod temperature.

Metal specific heat capacity by method of mixtures

Research question. How does the specific heat capacity of aluminium, brass, copper, iron and zinc samples of about 100 g each, measured by dropping them from boiling water at 100 °C into 150 g of water in a polystyrene cup, compare with data book values, in J kg⁻¹ K⁻¹?

B.1 Thermal energy transfers
SL and HL
Easy data
Rarely listed

My take. Safe and doable, but well used, so it only stands out if you treat heat loss quantitatively. Twist: use the same metals as your school's own kitchen or workshop offcuts and test whether alloys differ from pure metals.

Method, physics and where marks are lost+

Independent variable: Metal type: 5 different metals, each with 5 repeated trials. An extension could vary the mass of one metal over 5 values from 50 g to 250 g.
What you measure: Water temperature rise read with a digital thermometer (±0.1 °C) or temperature probe. Specific heat is calculated from energy lost by the metal equal to energy gained by the water and cup.
Controlled variables: Water mass, weighed on a balance to ±0.1 g each trial. Starting water temperature, checked with the same thermometer. Metal start temperature, held by leaving it in boiling water for 5 minutes. Transfer time and lid use, kept short and consistent to limit heat loss.
Physics and graph: m_metal c_metal (T_hot − T_final) = m_water c_water (T_final − T_cold) plus a cup correction. Plot energy gained by the water against m_metal(T_hot − T_final) for a single metal with varying mass, so the gradient equals c_metal. Otherwise compare calculated c with accepted values.
SL and HL: SL students calculate c for each metal and compare percentage differences with book values. Top band work models heat losses, for example by plotting a cooling curve of the cup and extrapolating back to the mixing time, and uses uncertainty propagation carefully. HL adds nothing syllabus-wise, so depth comes from the treatment of systematic error.
Where marks are lost: Research design: not accounting for the cup's heat capacity or heat lost during transfer. Data analysis: uncertainty on a small temperature rise is large and often ignored. Conclusion: claiming agreement without a quantitative comparison to the accepted value. Evaluation: blaming vague 'heat loss' with no estimate of its size or direction.
Data: Needs a kettle, polystyrene cups, a balance and a thermometer. The main uncertainty is heat loss during transfer and the small temperature rise for low-mass samples.

Rate of heat conduction through different sheet materials

Research question. How does the rate of heat flow through 5 mm thick sheets of cork, wood, acrylic, glass and aluminium compare, using the temperature rise of 100 g of water over 10 minutes with a 60 °C source on the other side?

B.1 Thermal energy transfers
SL and HL
Needs care
Rarely listed

My take. Fine if you tie it to P = kAΔT/d and get real k values. Give it a personal angle such as testing building or clothing insulation you can actually buy.

Method, physics and where marks are lost+

Independent variable: Sheet material (five types) with identical thickness, plus optionally thickness of one material at 2, 4, 6, 8, 10 mm.
What you measure: Temperature of a water-filled aluminium calorimeter on the cool side recorded every 30 s with a digital thermometer or probe (°C); heat flow rate P = mcΔT/Δt from the early linear section.
Controlled variables: Heat source temperature: water bath held at 60 °C with a thermometer. Contact area: same cut circle, same clamp pressure. Thickness: measured with a micrometer or calipers. Starting temperature of water and room conditions the same, with lagging around the sides.
Physics and graph: P = kAΔT/d. Plot P against 1/d for one material: the gradient equals kAΔT and gives k. For different materials, compare k with database values. Note that ΔT changes during the run, so use initial gradients.
SL and HL: SL: rank materials and estimate k for each. Top band: correct for heat losses with a blank control, account for contact resistance, and test the linear relation with thickness.
Where marks are lost: Research design: 'heat transfer rate' left undefined and thickness not matched. Data analysis: using the whole cooling curve rather than the initial rate. Conclusion: no comparison with tabulated conductivity. Evaluation: side losses and poor contact are large and usually not quantified.
Data: Needs a temperature probe, hot water bath, calorimeter and sheets; the main uncertainty is heat loss to the surroundings and contact quality.

Regelation: wire cutting through ice under load

Research question. How does the load on a thin wire (0.5 to 5.0 kg in six steps) affect the speed at which it passes through a block of ice at 0 °C?

B.1 Thermal energy transfers
HL topic
Hard data
Rarely listed

My take. Risky. The original idea of measuring a melting point shift with a thermometer cannot work, but the wire experiment is a real classic if you compare steel with nylon to test the explanation. Only choose it if you enjoy the conclusion being complicated.

Method, physics and where marks are lost+

Independent variable: Hanging mass on a 0.3 mm steel wire, six values from 0.5 kg to 5.0 kg, three runs each.
What you measure: Distance moved by the wire through the ice, measured with a ruler and a stopwatch or video, then speed = distance/time.
Controlled variables: Ice block from one freezer batch and started at 0 °C, shown by a melting ice slush. Wire diameter and length kept constant. Room temperature recorded and the block insulated, or run in a cold room. Same contact width on the ice.
Physics and graph: Pressure p = F/A with A = wire diameter × contact length. The melting point shifts with pressure, dT/dp ≈ −0.0075 K/MPa. Plot cutting speed against pressure, expecting proportionality if pressure melting dominates, though heat conduction through the wire matters too.
SL and HL: Mostly beyond SL. Even an HL student needs to be careful: the real effect of pressure is tiny, and heat flow from the room through the wire probably dominates. Top band work compares the predicted temperature shift with the speed and evaluates that heat conduction through the wire, not pressure alone, explains the result.
Where marks are lost: Research design: measuring melting point with a thermometer cannot detect a shift of a few hundredths of a kelvin. Conclusion: claiming pressure melting is confirmed when the wire is warmer than the ice. Evaluation: not testing a nylon thread for comparison.
Data: Needs a large clear ice block, a thin wire and steady room conditions; heat conduction along the wire is the hard uncertainty to remove.

Rubber band launch range at different starting temperatures

Research question. How does the temperature of a rubber band (5, 15, 25, 35, 45, 55 °C ± 1 °C) affect the horizontal range of a fixed-stretch launch, in m?

B.1 Thermal energy transfers
SL and HL
Needs care
Rarely listed

My take. Interesting because the result may surprise you. The thermodynamics is only qualitative at school level, so add the force-extension measurement at each temperature to make it solid.

Method, physics and where marks are lost+

Independent variable: Band temperature, 6 values from about 5 °C (ice bath) to 55 °C (warm water bath), with 5 launches per temperature and a new band from the same pack for each.
What you measure: Horizontal range from a tape measure and, better, launch speed from a video in Tracker, allowing the kinetic energy ½mv² of a small projectile to be compared with the stretch energy.
Controlled variables: Stretch: fixed with a stop at the same length. Launch angle: set to 45° with a clamp jig. Projectile: same mass, e.g. a small foam pellet. Time out of the bath: 10 s before firing, dried, with temperature checked by a probe.
Physics and graph: Elastic energy stored is related to the force-extension curve, and a rubber band gets stiffer when hot (the entropy effect), so the trend is not what many students predict. For a projectile, range R ∝ v², and v² is proportional to stored energy, so plot R against temperature to test the trend and explain it using kinetic theory of the polymer chains.
SL and HL: SL: measure R, and describe the change with temperature. Deeper: measure force at fixed extension with a newton meter at each temperature, which links to the stored energy, and discuss hysteresis and the cooling during the delay.
Where marks are lost: Research design: temperature changes during the transfer and is not measured at launch. Data analysis: different bands are not identical, so the spread is large. Conclusion: guessing that 'hot is stretchier' rather than looking at data. Evaluation: not testing wet against dry bands, or ageing of the rubber.
Data: Needs a water bath, thermometer and a launch jig; band variation and cooling before launch are the main uncertainties.

Water volume and cooling rate in moving air

Research question. How does the volume of hot water, from 50 cm³ to 250 cm³ at a starting 70 °C, affect its initial cooling rate in a steady airflow of 2 m/s?

B.1 Thermal energy transfers
SL and HL
Easy data
Rarely listed

My take. Simple and safe, so the marks come from the analysis. Adding a lid comparison is the twist that makes it more than a routine cooling experiment.

Method, physics and where marks are lost+

Independent variable: Volume of water in identical beakers, 50, 100, 150, 200, 250 cm³ (5 values), 3 repeats each.
What you measure: Temperature with a thermometer or probe every 30 s for 10 min, giving initial cooling rate in °C per s. Power lost is found from mcΔθ/Δt.
Controlled variables: Fan at a fixed distance and speed, checked with an anemometer. Same beaker shape, so the surface area changes only through depth. Same start temperature and room temperature. Lid on or off, decided and kept the same.
Physics and graph: Energy loss Q = mcΔθ, and rate of loss depends on surface area and temperature difference, with evaporation adding to it. Plot cooling rate against 1/volume or 1/mass; a straight line through the origin suggests roughly constant power loss.
SL and HL: SL students can compare cooling rates and explain in terms of mass and surface area. Top band work separates evaporation from convection with a lid, and fits Newton's law of cooling to extract a constant for each volume.
Where marks are lost: Research design: airflow not uniform across the beaker, and evaporation ignored. Data analysis: using average rates over different temperature ranges. Conclusion: not linking the trend to power loss. Evaluation: not discussing that beaker walls and base also conduct.
Data: Beakers, a fan, an anemometer and a thermometer are enough; the main uncertainty is uneven airflow and evaporation.

Young modulus of a heated copper wire

Research question. How does the temperature of a copper wire (20, 40, 60, 80, 100 °C) affect its measured Young modulus, in GPa, over an elastic extension of under 0.1%?

B.1 Thermal energy transfers
HL topic
Hard data
Rarely listed

My take. Hard and risky. The expected change is only a few percent, so most school setups cannot detect it. Only choose it if your school has a good optical or dial gauge system, and otherwise switch to a wire of different length or diameter at room temperature.

Method, physics and where marks are lost+

Independent variable: Wire temperature, 5 values from room temperature to about 100 °C, heated by passing a small current or by an enclosing hot water tube, with 3 repeated load series at each temperature.
What you measure: Extension for a range of loads, measured with a travelling microscope, a Vernier scale or a dial gauge. Stress and strain are calculated, and E is found as the gradient of stress against strain. Wire diameter is taken with a micrometer.
Controlled variables: Wire length and diameter: the same wire, measured with a metre rule and micrometer. Load range: kept within elastic behaviour. Thermal expansion: the reference length is corrected or a control wire is used. Time at temperature: allowed to settle for 3 minutes before each reading.
Physics and graph: E = stress/strain = (F/A)/(ΔL/L). Plot stress against strain at each temperature, and the gradient is E. Then plot E against temperature. A drop of only a few percent over 80 K is expected, so uncertainty is a key issue. Thermal expansion of the wire is of a similar size to the elastic extension and needs correcting.
SL and HL: Mostly beyond SL because of the precision needed, but a strong SL student could do it with careful method. Top band: separating thermal expansion from load extension, propagating uncertainty in ΔL, and comparing the trend with the data book value.
Where marks are lost: Research design: thermal expansion mixes with the extension, and thin wires kink or yield. Data analysis: very small extensions (tens of micrometres) have a large percentage uncertainty. Conclusion: claiming a trend when the change is within error. Evaluation: temperature not uniform along the wire.
Data: Needs a long thin wire, precise extension measurement and safe heating; the extension resolution and uniform temperature are the main problems.

Ideas were collected from public lists and published IA titles, merged when they are the same investigation, and rewritten from scratch. “On N sites” counts how many public lists carry the same investigation. How this list was made.

Frequently asked questions

What are good IB Physics IA ideas on thermal energy transfers?

+

Good starting points with easy data that few sites list are cooling constant of water in cups of different surface area, cooling curves of water, oil and salt solution and cooling of warm water in wrapped containers. Each one gives a straight-line graph from school equipment, which is what the Data analysis and Conclusion criteria need.

Which thermal energy transfers IA ideas are overdone?

+

specific heat capacity of salt solutions by electrical heating appear on three or more public lists. They still work, but they need a twist that shows your own thinking.

Can I do a thermal energy transfers IA at SL?

+

26 of the 28 ideas use SL physics. The others rely on HL-only content and are marked as HL topics.

Related

Initial assessment

Find out what is costing the marks.

Twenty minutes, one to one, online. No charge, no commitment.

Request an initial assessment