3.1. Conservation of Energy#
“The total energy of the world, kinetic plus potential, is a constant when we look closely enough… [If we see energy not conserved, then] this is due to a lack of appreciation of what it is that we see.”
— Richard Feynman
Overview#
In this section, we introduce the principle of conservation of energy and the First Law of Thermodynamics, which unifies the ideas of heat and work into one fundamental statement.
Conservation of Mechanical Energy#
Between 1732 and 1736, Bernoulli and Euler combined the discoveries of Newton (laws of motion) and Leibniz (the connection between weight × vertical displacement and weight × velocity squared) into an early form of the law of conservation of mechanical energy. A simple example is the interchange between potential and kinetic energy:
When an object (mass \(m\)) of height \(h\) is dropped, its gravitational potential energy \(mgh\) is converted into kinetic energy \(\tfrac{1}{2}mv^2\), assuming negligible air resistance.
Apollo 15
During the Apollo 15 mission, astronaut David Scott famously dropped a hammer and a feather on the Moon. With negligible air resistance, both hit the ground simultaneously. This illustrates that in free fall, all objects (no matter their mass) accelerate at the same rate, and thus gain the same velocity over the same drop distance:
Mechanical Equivalent of Heat#
Fig. 23 Joule’s apparatus for measuring the mechanical equivalent of heat.#
In 1847, James Prescott Joule measured how mechanical work converts into heat. He famously found that dropping a total of \(778\) lb\(\cdot\)ft of weight (e.g., by turning a paddle in water) raised the temperature of \(1\) lb of water by \(1\,^\circ\text{F}\). Converting to SI units:
Here, \(4.18\,\text{J}\,\text{g}^{-1}\,{^\circ\text{C}}^{-1}\) is the specific heat of water—i.e., the heat capacity per gram.
First Law of Thermodynamics#
Joule’s work revealed that heat and mechanical energy are interconvertible. To include heat (\(q\)) and work (\(w\)) in one statement of energy conservation, we use:
where
\(\Delta U\) is the change in internal energy,
\(q\) is the heat absorbed by the system,
\(w\) is the work done on the system.
Differential Form#
In differential form,
The notation \(\delta\) is used to remind us that \(q\) and \(w\) are path functions, not state functions.
Types of Work#
In Module 1.1, we defined work as “energy transferred when a force acts over a distance.” Mathematically:
Many physical processes fit a “generalized force” \(\times\) “generalized displacement” pattern:
Generalized “Force” |
Generalized “Displacement” |
\(\delta w\) |
Example |
|---|---|---|---|
Mechanical \(F\) |
\(x\) |
\(F \,dx\) |
Lifting a weight |
Linear tension \(k\) |
\(l=x - x_0\) |
\(k \,dl\) |
Stretching a spring |
Surface tension \(\gamma\) |
\(A\) |
\(\pm\,\gamma \, dA\) |
Blowing a soap bubble |
Pressure \(P\) |
\(V\) |
\(-P\,dV\) |
Compressing a gas |
Chemical \(\mu\) |
\(N\) or \(n\) |
\(\mu\,dN\) |
Forming a molecule |
Electrical \(E\) |
Charge \(q_{\text{el}}\) |
\(E\,dq_{\text{el}}\) |
Moving an electric charge |
Sign convention in chemistry:
\(w>0\) when work is done on the system (system’s energy goes up).
\(w<0\) when work is done by the system (system’s energy goes down).
Example 1. Expanding Against Constant Pressure#
Calculate the work necessary to expand an ideal gas from \(20\,\mathrm{L}\) to \(85\,\mathrm{L}\) against a constant external pressure of \(2.5\,\mathrm{bar}\).
Solution
Convert Units to SI
\[P_{\mathrm{ext}} = 2.5\,\mathrm{bar} = 2.5\times10^5\,\mathrm{Pa},\quad V_i = 20\,\mathrm{L} = 20\times10^{-3}\,\mathrm{m^3},\quad V_f = 85\,\mathrm{L} = 85\times10^{-3}\,\mathrm{m^3}.\]Set Up the Work Integral
For a constant external pressure,\[\delta w = -P_{\mathrm{ext}}\,dV \quad \Longrightarrow\quad w = -\int_{V_i}^{V_f} P_{\mathrm{ext}}\,dV.\]Perform the Integration
\[w = -P_{\mathrm{ext}}\int_{V_i}^{V_f} dV = -P_{\mathrm{ext}}\bigl(V_f - V_i\bigr).\]Numerical Result
\[w = -\bigl(2.5\times10^5\,\mathrm{Pa}\bigr) \Bigl[\bigl(85\times10^{-3}\bigr) - \bigl(20\times10^{-3}\bigr)\Bigr]\mathrm{m^3} = -16{,}250\,\mathrm{J} = \boxed{-16\,\mathrm{kJ}}.\]Interpretation
The negative sign indicates the system (gas) does \(16\,\mathrm{kJ}\) of work on the surroundings.
According to the convention \(\Delta U = q + w\), if no heat is exchanged (\(q=0\)), the gas would lose \(16\,\mathrm{kJ}\) in internal energy because it expands.
Example 2. Expanding a Water Bubble#
Calculate the work necessary to expand a water “bubble” (two surfaces) with surface tension \(\gamma = 72\,\mathrm{J/m^2}\) from a radius of \(1\,\mathrm{cm}\) to \(3.25\,\mathrm{cm}\).
Solution
Convert Units
\[\gamma = 72\,\mathrm{J/m^2}, \quad r_i = 1\times10^{-2}\,\mathrm{m}, \quad r_f = 3.25\times10^{-2}\,\mathrm{m}.\]Identify the Surface Area Change
For a single spherical surface, \(A = 4\pi r^2\), so \(dA = 8\pi r\,dr\).
A bubble has two surfaces (inner and outer), so its total area is \(2 \times 4\pi r^2 = 8\pi r^2\). Hence,
\[\text{(bubble area)} = 8\pi r^2 \quad\Longrightarrow\quad dA = 16\pi r\,dr.\]
Write the Work Expression
Since \(\delta w = -\gamma\,dA\) (negative sign for expansion work done by the system):\[w = - \int_{A_i}^{A_f} \gamma \,dA = - \gamma \bigl(A_f - A_i\bigr).\]Perform the Integral
Substituting \(A = 8\pi r^2\) for the bubble:\[\Delta A = A_f - A_i = 8\pi \bigl(r_f^2 - r_i^2\bigr).\]Thus,
\[w = -\,\gamma\,\Delta A = -\,\gamma \,8\pi\bigl(r_f^2 - r_i^2\bigr).\]Calculate Numerically
\[w = -\,72\,\mathrm{J/m^2} \,\times\,8\pi \Bigl[\bigl(3.25\times10^{-2}\bigr)^2 - \bigl(1\times10^{-2}\bigr)^2\Bigr]\mathrm{m^2}.\]Evaluate numerically (you should find):
\[w \approx \boxed{-1.7\,\mathrm{J}}.\]Interpretation
Expanding the bubble (from smaller to larger radius) does work on the surroundings, giving \(w<0\) from the system’s perspective.
If no heat is supplied, the system’s internal energy would decrease by \(1.7\,\mathrm{J}\).
Example 3. Stretching a Hookean Fiber#
Calculate the work required to stretch a fiber obeying Hooke’s law, with \(k=100\,\mathrm{N/cm}\), by \(0.15\,\mathrm{cm}\).
Solution
Convert Units
\[k = 100\,\frac{\mathrm{N}}{\mathrm{cm}} = 100\times10^2\,\frac{\mathrm{N}}{\mathrm{m}} = 10^4\,\frac{\mathrm{N}}{\mathrm{m}}, \quad l = 0.15\,\mathrm{cm} = 0.15\times10^{-2}\,\mathrm{m}.\]Write the Work Expression
Hooke’s law for tension: \(F = k\,\Delta l\). The infinitesimal work is:\[\delta w = F\,dx = k(x - x_0)\,dx.\]Integrate
If we stretch from \(x_0\) to \(x = x_0 + l\):\[w = \int_{x_0}^{x_0 + l} k\,(x - x_0)\,dx = \frac{1}{2} \,k\,(l)^2.\]Substitute Numbers
\[w = \tfrac{1}{2}\,(10^4)\, \Bigl(0.15\times10^{-2}\,\mathrm{m}\Bigr)^2 = 0.5\times10^4 \times\bigl(0.15\times10^{-2}\bigr)^2\,\mathrm{J}.\]Numerically,
\[w \approx 1.1\times10^{-2}\,\mathrm{J}.\]Interpretation
This is the energy required (work done on the system) to stretch the fiber by \(0.15\,\mathrm{cm}\).
The positive sign indicates the system absorbs energy (an external force pulls on the fiber).