Difficulty: Introductory. Prerequisites: first law of thermodynamics concepts (see companion notes), basic algebra, familiarity with percentage error.
This set of notes focuses on the experimental side: Joule's paddle-wheel apparatus, how to extract specific heat from lab data, and the kinds of post-lab and exam questions you will face. It pairs with the First Law of Thermodynamics core concepts notes.
Joule's paddle-wheel experiment converts the gravitational potential energy of a falling block into thermal energy in water, letting you measure specific heat directly. You record the temperature change and the energy input, then calculate c = ΔQ / (m × ΔT). By varying block mass, water mass and starting temperature across multiple runs, you confirm that specific heat is a constant property of the substance, independent of experimental conditions.
Joule's apparatus (paddle-wheel experiment)
A device in which a falling mass, connected by a string over pulleys, rotates a paddle inside an insulated water container. The gravitational PE of the mass converts to thermal energy in the water. Think of it as a controlled way to turn "falling" into "warming."
Insulated system
A system designed so that no heat escapes to the surroundings. In the idealised simulation, this means 100% of the block's energy ends up as thermal energy in the water. In a real lab, insulation is never perfect.
Block mass
The mass of the weight that falls to drive the paddle. A heavier block stores more gravitational PE at the same height, so it delivers more energy to the water.
Water mass
The quantity of water in the insulated container. More water means the same energy is spread over more material, producing a smaller temperature change.
Temperature change per unit of heat energy (ΔT/ΔQ)
The ratio recorded in Table 1. It tells you how many degrees the water rises for each joule of energy delivered. A smaller ratio means the substance is harder to heat (higher heat capacity). Units: °C/J.
Percentage error
A measure of how far your experimental value is from the accepted (textbook) value, expressed as a percentage. Formula: % error = |experimental - accepted| / accepted × 100. In simple terms, it tells you how close your experiment got to the "right answer."
A block of known mass is released from a known height. Its gravitational PE (mblock × g × h) is converted into kinetic energy as it falls.
The falling block, via a string and pulley, spins a paddle inside an insulated water container.
The paddle churns the water, converting ordered kinetic energy into disordered thermal energy.
The simulation is idealised: no energy lost to friction, the environment, or the apparatus itself. Every joule of PE becomes a joule of thermal energy in the water.
The purpose of Part 1 is to build intuition about the relationship ΔQ = c × m × ΔT by changing one variable at a time.
Run 1 (baseline): block mass = 2.0 kg, water mass = 5.0 kg, initial temp = 20.0°C. The simulation delivers 1,960 J and the water temperature rises by 0.094°C.
Run 2 (heavier block): block mass increased to 5.0 kg. More PE, more energy delivered, larger ΔT. Everything else is held constant so you can isolate the effect of energy input.
Run 3 (less water): water mass halved to 2.5 kg, block mass stays at 5.0 kg. Same energy input as Run 2, but half the water to absorb it. ΔT roughly doubles.
Run 4 (different starting temperature): initial temperature dropped to 10°C, water mass and block mass unchanged from Run 3. The ΔT should be the same as Run 3, confirming that specific heat does not depend on starting temperature.
Part 2 uses the formula c = ΔQ / (m × ΔT) to extract the specific heat of water from each run.
Run A (default settings): standard setup, water mass 5.0 kg, height 100 m, block mass 5.0 kg. Record ΔT and ΔQ, then compute c.
Run B (less water): water mass reduced to 1.0 kg. The temperature change will be larger, but c should come out the same.
Run C (lower height): height halved to 50 m, water mass back to 5.0 kg. Less PE, so less energy delivered and a smaller ΔT, but c should still match.
Run D (higher starting temperature): height back to 100 m, starting temperature raised to 50°C. Again, c should be unaffected.
The point: if the experiment is clean, every run yields c ≈ 4,190 J/(kg·°C) regardless of which variable was changed.
Read ΔT (temperature change) and ΔQ (total heat delivered) from the simulation.
Divide: c = ΔQ / (m × ΔT). Use the water mass for m, not the block mass.
Record to three significant figures after the decimal point, as the lab instructions require.
Average all four values of c from Table 2.
Compare to the accepted value (4,190 J/(kg·°C)) using percentage error.
c = \frac{\Delta Q}{m \times \Delta T}\frac{\Delta T}{\Delta Q} \quad \text{(units: °C/J)}For Run 1: ΔT/ΔQ = 0.0940 / 1960 = 4.796 × 10⁻⁵ °C/J.
\% \text{ error} = \frac{|c_{\text{exp}} - c_{\text{accepted}}|}{c_{\text{accepted}}} \times 100Worked example: if your average experimental c = 4,180 J/(kg·°C), then % error = |4,180 - 4,190| / 4,190 × 100 = 0.239%.
\Delta Q = m_{\text{block}} \times g \times hWorked example for Run 1: ΔQ = 2.0 × 9.8 × 100 = 1,960 J (matches the simulation output).
Worked example for a 5.0 kg block: ΔQ = 5.0 × 9.8 × 100 = 4,900 J.
Students often plug the block mass into the specific heat formula where the water mass should go. The block determines ΔQ (through PE = mgh). The water mass goes into c = ΔQ / (m × ΔT).
Students sometimes report ΔT/ΔQ (Table 1) in the wrong units or confuse it with specific heat. ΔT/ΔQ has units of °C/J and is the inverse relationship: it tells you temperature rise per joule. Specific heat (c) has units of J/(kg·°C) and includes the mass.
Students occasionally expect the specific heat to change when they change the starting temperature. It does not (at least within the range of this experiment). If your Run D value differs from Run A, the issue is measurement precision, not physics.
Students sometimes average the ΔT/ΔQ values from Table 1 and treat that as specific heat. Those ratios depend on the water mass used in each run and cannot be averaged meaningfully across runs with different masses.
⚠️ Post-lab questions 1 and 2 ask you to explain what happened when a variable changed. The answer pattern is always the same: identify which term in ΔQ = c × m × ΔT changed, state the direction (increased/decreased), and explain the effect on ΔT using the rearranged formula.
⚠️ Question 3 (explain the role of each variable) is asking for a conceptual walkthrough, not just the formula. Name each variable, state what it represents physically, and describe what happens to ΔT when that variable increases while the others stay constant.
⚠️ Percentage error calculation (question 6) is a gimme if you know the formula. Do not forget the absolute value: it does not matter whether your value is above or below the accepted one.
⚠️ Sources of experimental error (question 7) will appear in some form on nearly every physics lab exam. Memorise three to four real sources: heat loss, friction, energy absorbed by apparatus, and measurement precision of thermometers.
Fill in the blank: the energy input in Joule's experiment equals ______ × g × h. (mass of the block)
True or false: to calculate specific heat, you divide ΔQ by the block mass times ΔT. (False. You divide by the water mass times ΔT.)
Fill in the blank: if your experimental specific heat is 4,200 J/(kg·°C) and the accepted value is 4,190 J/(kg·°C), the percentage error is ______%. (0.239%)
True or false: the ratio ΔT/ΔQ from Table 1 is the same as specific heat. (False. ΔT/ΔQ has units °C/J and does not include mass. Specific heat has units J/(kg·°C).)
Fill in the blank: in an idealised Joule experiment, the insulation ensures that ______% of the block's PE becomes thermal energy in the water. (100)
Q: In Run A, the simulation reports ΔT = 0.234°C and ΔQ = 4,900 J for 5.0 kg of water. Calculate the specific heat of water from this data.
A: c = ΔQ / (m × ΔT) = 4,900 / (5.0 × 0.234) = 4,900 / 1.170 = 4,188 J/(kg·°C).
Q: In Run B, the water mass is reduced to 1.0 kg while the energy input stays at 4,900 J. Predict the temperature change.
A: ΔT = ΔQ / (c × m) = 4,900 / (4,190 × 1.0) = 1.169°C. With five times less water, the temperature change is about five times larger than Run A.
Q: Your four experimental values of c from Table 2 are 4,188, 4,191, 4,185, and 4,192 J/(kg·°C). Calculate the average and the percentage error from the accepted value of 4,190 J/(kg·°C).
A: Average = (4,188 + 4,191 + 4,185 + 4,192) / 4 = 4,189 J/(kg·°C). Percentage error = |4,189 - 4,190| / 4,190 × 100 = 0.024%. The values agree very closely, which is expected in an idealised simulation.
Q: A classmate claims that lowering the initial water temperature from 20°C to 10°C should change the specific heat value they measure. Is this correct? Explain.
A: No. Specific heat is a property of the substance (water), not of the experiment's starting conditions. The initial temperature shifts where on the thermometer the final reading lands, but the size of the temperature change (ΔT) for a given energy input remains the same, so the calculated c is unchanged.
Q: In a real (non-idealised) version of this experiment, your measured specific heat comes out lower than 4,190 J/(kg·°C). What might explain this?
A: If some energy is lost to the environment or absorbed by the apparatus (paddle, container walls), the water receives less energy than you assume. You still divide the full theoretical ΔQ by (m × ΔT), but the actual ΔQ reaching the water is smaller. This makes the real ΔT smaller than expected, and when you compute c = ΔQ / (m × ΔT) with the theoretical ΔQ, the result is higher, not lower. A lower-than-expected c would suggest the thermometer is over-reading ΔT, perhaps due to calibration error or rounding.
This lab method connects to calorimetry experiments you may encounter later, where you mix hot and cold substances and use the same formula to find unknown specific heats or final temperatures. The error analysis skills (percentage error, identifying sources of uncertainty) apply to every physics lab you will do. The concept of converting mechanical energy to thermal energy also connects to heat engines and the second law of thermodynamics, where you study the reverse process and its inherent limitations.
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