Physics / Mechanical Equivalent of Heat
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EXPERIMENT 30 · PHYSICS LABORATORY

Mechanical Equivalent of Heat theory

Understand mechanical equivalent of heat: the equations, assumptions, supporting concepts and measurement method.

THE WORKING RELATIONSHIP
W = 2π r M_load g N = J · C_sys (T_f − T_0) = J · Q ⇔ Y = J · X ⇔ J = S

Keep the physical quantities, graph variables and units consistent.

01The Downfall of the Caloric Theory and Rumford’s Cannon-Boring Experiments

Throughout the eighteenth century, thermal physics was dominated by the caloric theory pioneered by Antoine Lavoisier and Joseph Black. Caloric was conceived as an elastic, weightless, self-repulsive fluid that permeated all matter, flowing spontaneously from warmer bodies to colder ones. In this framework, heat absorbed or released during physical processes was simply caloric changing its spatial concentration, and friction was explained as mechanical rubbing 'squeezing out' caloric from the interstices between metal particles, akin to squeezing water from a sponge. In 1798, Benjamin Thompson, Count Rumford, delivered a fatal experimental blow to this concept while supervising the boring of brass cannons at the military arsenal in Munich. Rumford submerged a cannon cylinder and a dull boring tool inside a wooden box containing 18.75 pounds of cold water. Driven by two horses, the blunt borer rubbed continuously against the brass bottom without cutting away any metal shavings. Within thirty minutes, the water warmed noticeably, and after two and a half hours, the water boiled vigorously. Rumford pointed out that the supply of heat was entirely inexhaustible as long as mechanical rotation was maintained. Because a finite quantity of brass could not possess an infinite store of caloric fluid, Rumford reasoned that heat could not be a material substance; it had to be a form of microscopic mechanical motion excited by the macroscopic work done by the horses.

\text{Caloric Fluid Theory (Lavoisier)} \implies \Delta Q = 0 \quad \text{vs.} \quad \text{Kinetic Heat Theory (Rumford)} \implies W \to Q \quad (\text{Inexhaustible})
How did Count Rumford’s cannon-boring experiments disprove the caloric theory of heat?

Rumford showed that a blunt boring tool rubbing against brass generated an inexhaustible, continuous supply of heat that boiled gallons of water without producing metal chips. Because a finite mass of brass could not contain an infinite amount of caloric fluid, heat could not be a material substance; it had to be a mode of motion generated by mechanical work.

02The First Law of Thermodynamics and Joule’s Principle of Equivalence

Between 1843 and 1850, the English physicist James Prescott Joule placed the kinetic nature of heat on an exact mathematical and experimental foundation. Using a falling-weight apparatus that drove a set of brass paddle wheels inside an insulated copper container filled with water, Joule measured the mechanical work performed by the descending weights (W = M g h) and the resulting temperature rise ΔT of the water. Over hundreds of meticulous trials, Joule established the principle of mechanical equivalence: whenever mechanical work is dissipated in a closed system to produce thermal effects, the ratio of mechanical work expended to thermal heat absorbed is an invariant universal physical constant: J = W / Q. This empirical equivalence led directly to Rudolf Clausius's and William Thomson's (Lord Kelvin's) mathematical formulation of the First Law of Thermodynamics: for any thermodynamic system, energy is conserved, and heat Q and work W are two mutually convertible forms of energy transfer across the system boundary: ΔU = Q − W (in modern SI notation where both are measured in joules). Historically, because thermal energy was quantified using the calorie (the heat required to raise 1 g of water by 1 °C), Joule’s constant J served as the conversion factor between mechanical units (foot-pounds, ergs, or joules) and thermal units (calories). In modern SI, Joule’s constant is recognized as the conversion factor J = 4.1860 J cal⁻¹ (or 4186 J kcal⁻¹).

W = J \cdot Q \iff J = \frac{W}{Q} \approx 4.1860 \text{ J cal}^{-1} \quad \text{and} \quad \Delta U = Q - W
What is the physical significance of Joule’s constant J in the context of the first law of thermodynamics?

Joule’s constant J represents the mechanical equivalent of heat—the conversion factor relating mechanical work (in joules) to thermal energy (in calories). It demonstrated that heat is a form of energy, establishing the First Law of Thermodynamics (energy conservation).

03The Callendar and Rowland Friction Calorimeter Mechanics

In 1879, Henry Rowland improved upon Joule's paddle-wheel method by introducing continuous mechanical rotation of a calorimeter drum, later refined into the elegant bench apparatus devised by Hugh Longbourne Callendar. The apparatus consists of a thin-walled copper drum of outer radius r mounted horizontally and driven at a constant angular velocity ω. A flexible woven belt is wrapped around the drum circumference, carrying a heavy suspended load of mass M_load on the downward-running side, while the upward-running side is connected to a light spring balance. When the drum rotates, kinetic friction between the rotating drum surface and the stationary belt produces a tangential frictional traction force F_f. By adjusting the rotational speed, the student achieves a dynamic equilibrium state where the suspended weight is lifted slightly off the bench rest and hovers steadily in mid-air. In this floating state, the downward gravitational force M_load g is balanced by the upward frictional traction plus the small spring tension T_2: F_f = M_load g − T_2. For a drum rotating through N complete cycles, the peripheral contact surface moves through a linear distance s = 2π r N. The total mechanical work dissipated against sliding friction is therefore: W = F_f · s = (M_load g − T_2) · 2π r N = τ · (2π N), where τ = F_f r is the frictional torque exerted by the drum. If the spring tension T_2 is made negligible (or subtracted explicitly), W = 2π r M_load g N.

W = F_f \cdot s = (M_{\text{load}} g - T_2) \cdot 2\pi r N = \tau \cdot (2\pi N) \quad \text{where } s = 2\pi r N
How is the mechanical work calculated in the Callendar rotating drum friction calorimeter?

The work is calculated as the product of the net frictional force (F = M_load g − T_spring) and the total distance traversed by the drum surface (s = 2π r N): W = (M_load g − T_spring) · 2π r N = τ · (2π N).

04Belt Friction Mechanics and the Capstan Equation

The mechanical equilibrium of the flexible friction belt wrapped around the rotating cylinder is governed by Euler's capstan (or belt friction) equation: T_1 / T_2 = exp(μ θ), where T_1 is the tension on the tight side (supporting the hanging load M_load g), T_2 is the tension on the slack side (the spring balance), μ is the coefficient of kinetic friction between the canvas belt and copper drum, and θ is the total angular wrap in radians. For a belt wrapped 1.5 complete turns around the drum, the wrap angle is θ = 1.5 × 2π = 3π radians ≈ 9.42 rad. If the kinetic friction coefficient between woven canvas and smooth copper is μ ≈ 0.25, the exponential amplification factor evaluates to exp(0.25 × 3π) = exp(2.356) ≈ 10.55. Consequently, the tension T_1 on the load side is over ten times larger than the tension T_2 on the spring balance: T_2 = T_1 / 10.55. For a hanging load of M_load = 3.00 kg (T_1 = 3.00 × 9.807 = 29.42 N), the spring balance experiences a counter-tension of only T_2 ≈ 2.79 N. The net effective frictional force driving thermal generation is F_f = T_1 − T_2 = 29.42 − 2.79 = 26.63 N. The capstan relation explains why a small holding force on the spring end can comfortably support and stabilize a heavy hanging weight without requiring massive clamping mechanisms.

\frac{T_1}{T_2} = e^{\mu \theta} \implies T_2 = M_{\text{load}} g \, e^{-\mu \theta} \quad \text{and} \quad F_f = T_1 - T_2 = M_{\text{load}} g (1 - e^{-\mu \theta})
What does the capstan equation describe in the context of the friction calorimeter belt?

The capstan equation T1/T2 = exp(μθ) describes how tension increases exponentially around a curved friction surface. It explains why a small tension T2 on the spring balance side can balance a much larger load T1 = Mg on the hanging side over a wrap of 1.5 turns (θ = 3π).

05Heat Capacity of the Calorimeter Assembly and Water Equivalent

The thermal heat Q absorbed during friction cycling is evaluated from the temperature elevation of the calorimeter assembly: Q = C_sys (T_f − T_0), where C_sys is the total heat capacity expressed in calories per kelvin. By historical definition, one calorie is the thermal energy required to elevate the temperature of one gram of pure liquid water by one degree Celsius (from 14.5 °C to 15.5 °C), meaning the specific heat capacity of water is c_w = 1.000 cal g⁻¹ K⁻¹. The copper drum, however, also absorbs sensible heat as its temperature rises. For a copper vessel of mass m_cal = 200.00 g and specific heat capacity c_cu = 0.0924 cal g⁻¹ K⁻¹, the heat capacity of the metal drum alone evaluates to C_cal = m_cal c_cu = 200.00 g × 0.0924 cal g⁻¹ K⁻¹ = 18.48 cal K⁻¹. In calorimetric terminology, this is called the water equivalent of the calorimeter: W_cal = 18.48 g of water. When charged with m_w = 100.00 g of distilled water, the aggregate system capacity in calories evaluates to C_sys = m_w c_w + m_cal c_cu = 100.00 + 18.48 = 118.48 cal K⁻¹. In SI units, multiplying C_sys by 4.1860 J cal⁻¹ gives C_sys,J = 495.96 J K⁻¹. Multiplying C_sys by the measured temperature rise ΔT = T_f − T_0 quantifies the total thermal energy in calories absorbed by the apparatus.

C_{\text{sys}} = m_w c_w + m_{\text{cal}} c_{\text{cu}} = 100.00 + 18.48 = 118.48 \text{ cal K}^{-1} \quad (W_{\text{cal}} = 18.48 \text{ g})
What is meant by the water equivalent of the calorimeter drum, and why must it be included in the calculation of heat absorbed?

The water equivalent W_cal is the mass of water that would absorb the identical quantity of thermal energy as the metal calorimeter drum for a 1 °C temperature rise (W_cal = m_cal · c_cu). It must be included because the copper drum warms alongside the water, absorbing a substantial fraction (~15%) of the total heat generated.

06Newton’s Cooling Corrections and Ambient Heat Loss Management

Because friction cycling requires between 2 and 6 minutes of continuous rotation, the calorimeter drum experiences heat exchange with the ambient room via convection and radiation. In accordance with Newton's law of cooling, the rate of environmental heat loss is proportional to the temperature excess over the room: dQ_loss/dt = k (T − T_ambient). If the drum begins at room temperature T_0 = T_ambient = 20.0 °C and warms to T_f = 35.0 °C, the drum is hotter than the ambient air throughout the entire rotation interval, continuously leaking thermal energy to the room. If uncorrected, the observed final temperature T_f is systematically lower than the ideal adiabatic temperature, causing the calculated heat Q = C_sys (T_f − T_0) to be underestimated and driving the apparent mechanical equivalent J = W / Q to be falsely inflated. Two techniques minimize this systematic defect: (1) Symmetrical thermal swing (Rumford's method): if the water is pre-cooled to 7 °C below room temperature (T_0 = 13 °C) and rotated until it reaches 7 °C above room temperature (T_f = 27 °C), the heat absorbed from the room during the cold half cancels the heat lost to the room during the hot half; (2) Graphical rate correction: tracking the post-rotation cooling rate for 5 minutes and extrapolating back to the midpoint of the rotation run restores the true adiabatic peak.

\frac{dQ_{\text{loss}}}{dt} = k (T - T_{\text{room}}) \implies T_{f,\text{corrected}} = T_{f,\text{obs}} + \int_0^{t_{\text{run}}} k (T - T_{\text{room}}) dt
How does uncorrected heat loss to the room affect the experimental value of Joule’s constant J?

Heat lost to the room during rotation reduces the measured temperature rise ΔT, causing the calculated heat absorbed Q to be too low. Because J = W / Q, dividing the work by an underestimated Q causes the calculated value of J to be systematically too high.

07Thermodynamic Definition of the Calorie and Modern SI Energy Units

Historically, the calorie was defined in terms of water thermometry: the 15 °C calorie (cal_15) was defined as the quantity of thermal energy required to warm 1 gram of air-free water from 14.5 °C to 15.5 °C at standard atmospheric pressure (101.325 kPa). Because the specific heat capacity of liquid water c_w(T) varies slightly with temperature (exhibiting a broad minimum near 35 °C and rising toward 0 °C and 100 °C), different historical calorie definitions yielded slightly different mechanical equivalents: the 15 °C calorie corresponds to J = 4.1855 J; the 20 °C calorie corresponds to J = 4.1818 J; and the mean calorie (1/100 of the heat required to warm water from 0 °C to 100 °C) equals J = 4.1897 J. In 1929, the International Steam Table conference defined the International Table calorie as exactly 1 cal_IT ≡ 4.1868 J, while in 1948 the Ninth General Conference on Weights and Measures (CGPM) adopted the joule (1 J = 1 N m = 1 kg m² s⁻²) as the sole standard SI unit for all forms of energy, work, and heat. The thermochemical calorie was fixed by convention as exactly 1 cal_th ≡ 4.184 J. Today, the mechanical equivalent of heat is not treated as an empirical conversion factor, but rather as an essential historical milestone confirming that heat and mechanical work are dimensionally identical manifestations of energy.

1 \text{ cal}_{15} = 4.1855 \text{ J}, \quad 1 \text{ cal}_{\text{IT}} \equiv 4.1868 \text{ J}, \quad 1 \text{ cal}_{\text{th}} \equiv 4.1840 \text{ J}
Why did different historical definitions of the calorie yield slightly different values for Joule’s constant J?

Because the specific heat capacity of liquid water varies slightly with temperature (by about 1% between 0 °C and 100 °C). Defining the calorie over different temperature intervals (e.g. 14.5–15.5 °C vs 19.5–20.5 °C vs mean 0–100 °C) produced minor differences in the exact mechanical equivalent.

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Connect the equations to a measurement.

The simulation evaluates mechanical work using W = 2π r M_load g N, where r = 0.0500 m is the drum radius, M_load = 3.000 kg is the suspended load, g = 9.80665 m s⁻² is gravitational acceleration, and N is the revolution count. Heat absorbed is evaluated using Q = C_sys (T_f − T_0) in calories, where C_sys = 118.48 cal K⁻¹ is the total thermal capacity (100.00 g water + 18.48 g water equivalent of the copper drum). Equilibrium temperatures are calculated using T_f = T_0 + W / (C_sys · J), with J = 4.1860 J cal⁻¹ and rounded to 0.01 °C to simulate precision laboratory thermometry. Ordinary least-squares regression of W against Q across five revolution settings yields the reported mechanical equivalent of heat.