Physics / Electrical Equivalent of Heat
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EXPERIMENT 50 · PHYSICS LABORATORY

Electrical Equivalent of Heat lab report

A worked electrical equivalent of heat lab report with simulated observation tables, graph, calculations, uncertainty and results.

WORKED EXAMPLE · SIMULATED OBSERVATIONS

A complete example report.

Simulated demonstration data.

This report shows the method and arithmetic. It is not evidence of a physical experiment. Replace its observations with your own measurements when preparing a real lab report.

Student: __________________Student ID: __________________Course: __________________Date: __________________Partner: __________________Instructor: __________________

1. Title and aim

The Heat Equivalent of Electrical Energy: Joule's Law of Electrical Heating. To determine the electrical and mechanical equivalent of heat (Joule's constant J) by electrical calorimeter heating, verify Joule's law of electric power dissipation (W = V I t), and evaluate radiation cooling corrections.

2. Abstract

Five observations were generated from the ideal teaching model over the settings listed below. A straight-line fit gave an estimate of electrical equivalent of heat of 4.1858 J cal⁻¹, differing by 0.004% from the model value. The reported scatter comes from scale rounding and any stated model approximation, not measured laboratory noise.

3. Apparatus and fixed settings

4. Theory and graph choice

Joule's law of electrical heating establishes that the electrical energy delivered by a power supply is converted quantitatively into thermal internal energy. In an electrical calorimeter, a submerged heating coil of resistance R carries current I under potential difference V for duration t, dissipating electrical energy W = V · I · t. This energy is absorbed by the surrounding water and calorimeter vessel, raising their temperature by ΔT according to Q = C_sys · ΔT. By measuring the temperature rise across six distinct heating durations and applying linear regression, students determine Joule's constant (J = W / Q), verify the electrical equivalent of heat (J = 4.1860 J/cal), and calculate the heat equivalent of electrical energy (a = 1/J = 0.2389 cal/J).

W = V · I · t = J · Q + b; J = m_slope, a = 1 / J

The calculation fits y = mx + b in SI units. The graph may display scaled axes for readability; those display factors are removed before calculating the physical result. The gradient has units J cal⁻¹.

5. Method

  1. Weigh the clean, dry inner copper calorimeter vessel on a balance, add exactly 250.0 g of distilled water, and re-weigh to verify total mass. Place the inner vessel inside the outer insulating air jacket, seat the phenolic lid, and insert the immersion heating coil, motorized stirrer, and fractional-degree thermometer.
  2. Activate the motorized stirrer at gentle speed and allow the calorimeter system 5 minutes to reach thermal equilibrium with the ambient laboratory. Read and record the initial quiescent temperature T₀ = 20.00 °C on the fractional-degree thermometer.
  3. Connect the regulated DC power supply in series with the ammeter and immersion heating coil. Connect the digital voltmeter directly across the coil terminal posts (four-wire Kelvin sensing) to exclude lead resistance drops. Energize the supply momentarily to set V = 10.00 V and I = 2.500 A (P = 25.00 W), then open the main switch.
  4. Simultaneously close the power switch and start the digital chronometer. Maintain continuous gentle stirring while monitoring the voltmeter and ammeter to verify constant power delivery. Run the heating for an initial duration of 120.0 s (2.0 min, delivering W = 3000 J), open the switch, and continue stirring until the thermometer reaches its maximum stabilized reading T.
  5. Repeat the timed heating runs for cumulative durations of 240.0 s (6000 J), 360.0 s (9000 J), 480.0 s (12000 J), 600.0 s (15000 J), and 720.0 s (18000 J), recording the peak stabilized temperature T at each duration. Compute the temperature rise ΔT = T − T₀ and the corresponding heat absorbed Q = C_sys · ΔT in calories.
  6. Plot heat absorbed Q (in calories) on the horizontal axis and electrical energy W = V · I · t (in Joules) on the vertical axis. Compute ordinary least-squares linear regression to determine the slope J = ΔW / ΔQ (in J/cal) and the vertical intercept b. Calculate the heat equivalent of electrical energy a = 1 / J (in cal/J) and compare against standard reference values.

6. Observations and calculation table

All observations use fixed model settings. Values are rounded only at the instrument reading stage; further calculations use full precision. Repeated physical trials are not claimed.

Table 1. Five simulated measurements; x and y are in SI units.
No.Heating duration t (s)Calorimeter temperature T (°C)x (SI)y (SI)Fitted y (SI)Residual (SI)
112022.76715.668003000.000002996.385053.61495
224025.531433.929006000.000006002.89615-2.89615
336028.292149.597009000.000008998.553421.44658
448031.062867.8580012000.0000012005.06452-5.06452
560033.823583.5260015000.0000015000.72179-0.72179
672036.584299.1940018000.0000017996.379063.62094

Download the sample observations (CSV) ↓

7. Worked calculation for one observation

Heating duration t = 120 s; observed calorimeter temperature T = 22.76 °C (temperature rise ΔT = 2.76 °C); electrical work W = V · I · t = 10.00 × 2.500 × 120 = 3000.0 J; heat absorbed Q = C_sys · ΔT = 259.3 × 2.76 = 715.67 cal. Single-run Joule equivalent estimate J_single = W / Q = 3000.0 / 715.67 = 4.1919 J cal⁻¹. Linear regression of W against Q across all six durations yields the reported slope J and intercept b.

8. Graph, gradient and final calculation

1500.00357.326000.001432.3710500.002507.4315000.003582.4919500.004657.55Heat absorbed Q (cal)Electrical energy W (J)

Dots show the sample observations. The line is the ordinary least-squares fit.

x̄ = 2508.29533; ȳ = 10500.00000
Σ(x − x̄)² = 8.9892e+06
Σ(x − x̄)(y − ȳ) = 3.7627e+07
m = Σ(x − x̄)(y − ȳ) / Σ(x − x̄)² = 4.18581978
b = ȳ − mx̄ = 0.72777736
Slope S = J = 4.1858 J cal⁻¹; Intercept b = 0.73 J; Electrical equivalent of heat J = 4.1858 J cal⁻¹ (reference: 4.1860 J cal⁻¹); Heat equivalent of electrical energy a = 1 / J = 0.238902 cal J⁻¹ (theoretical: 0.23889 cal J⁻¹)

9. Uncertainty and comparison

Percentage difference = |estimate − model value| / |model value| × 100 = 0.0043%. Percentage difference is not experimental uncertainty.

For n = 5 observations, Σresidual² = 62.82970; s² = Σresidual²/(n − 2). Gradient standard error = √[s²/Σ(x − x̄)²] = 0.00132. Intercept standard error = √[s²(1/n + x̄²/Σ(x − x̄)²)] = 3.68939.

Propagating only the relevant fit coefficient gives a fit-only standard error of 0.00132 J cal⁻¹. This is a one-standard-error quantity under the regression assumptions. Deterministic rounded model data do not supply a complete physical uncertainty budget.

Calorimeter convective and radiative heat loss to the ambient environment (mitigated by starting the calorimeter water at an initial temperature T₀ below ambient room temperature so that ambient heat absorption during the first half cancels radiative loss during the second half according to Rumford’s equal temperature differential principle), thermometer thermal lag and stem emergence cooling (minimized by continuous gentle mechanical stirring with a spiral stirrer and using a finely graduated fractional-degree thermometer), uncertainty in the water equivalent of the calorimeter vessel and immersion accessories (W_eq = 9.3 ± 0.3 g, contributing ~0.1% to C_sys = 259.3 cal/°C), DC power supply voltage ripple and drift (regulated within ±0.01 V and ±0.002 A, with four-wire Kelvin potential sensing directly at the heater coil binding posts bypassing lead resistance drops), and chronometer switching synchronization (±0.2 s over 120 s to 720 s durations, <0.1%) govern the uncertainty envelope. Ordinary least-squares regression of electrical work W against heat absorbed Q averages out thermometer meniscus quantization and minor residual cooling deviations, with the slope isolating the pure electrical equivalent of heat J and the vertical intercept isolating any baseline thermal offset.

10. Result and discussion

J = 4.1860 J cal⁻¹ = 4.1858 J cal⁻¹
Model comparison value: 4.1860 J cal⁻¹

The observations follow the stated model relationship. Check whether the intercept and residual pattern suggest a systematic offset or a missing approximation. Agreement with the model does not independently verify its input constants.

11. Precautions and limitations

Calculations assume uniform thermal mixing without fluid stagnation, negligible heat loss to the ambient environment under Rumford radiation compensation, and negligible mechanical energy added by the stirrer. Real calorimeters exhibit small convection losses governed by Newton’s law of cooling (dQ/dt = k(T − T_env)) and slight temperature lag in the thermometer bulb during rapid heating.

12. Conclusion

The worked example illustrates how to obtain electrical equivalent of heat from measurements, graph variables and a fitted coefficient. The estimate is 4.1858 J cal⁻¹. A real experiment must additionally evaluate the apparatus-specific uncertainties and corrections described above.