Physics 2nd Paper · Classes 11-12 · Chapter 1 · 1.2

First Law of Thermodynamics.

Pump a bicycle tyre hard and the barrel turns warm, though you lit no fire. Where did your work go? Into heat. Boil water in a covered pot and the lid is pushed up: heat turned into work. The first law of thermodynamics keeps the account for both directions in a single line.

1. System, surroundings and internal energy

The branch of physics that studies the interconversion of heat and mechanical energy and the relation between them is called thermodynamics. Count Rumford, Humphry Davy and James Prescott Joule showed by experiment that work, or mechanical energy, produces heat, and that heat is a form of motion. Thermodynamics grew out of that idea.

A fixed quantity of matter bounded by a surface or enclosure, in which variables like pressure, volume and temperature can be measured, is a thermodynamic system. Gas shut in a piston cylinder or a balloon is a system. Water boiling in an uncovered pot is not, because the vapour escapes. Everything around the system is its surroundings: anything able to exchange energy with it counts. The air around a piston and cylinder is therefore its surroundings. Tap the four points in the picture.

The quantities whose values fix the state of a system are its thermodynamic coordinates. For gas in a cylinder they are pressure, volume and absolute temperature. The final steady state of an isolated system is thermodynamic equilibrium: pressure, volume and temperature are then the same at every point. Any change in the coordinates is a thermodynamic process.

The energy that lies hidden or latent inside a system and shows itself in suitable conditions is its internal energy, written u. It rises when heat is applied to the system, and the book says it stays constant when no heat is applied. How work changes it you will see in the adiabatic process.

2. The first law: where did the heat go

The first law has two statements. Joule's statement: when work is wholly changed into heat, or heat wholly into work, the two are proportional, W ∝ Q, or W = JQ. J is the mechanical equivalent of heat, or Joule's equivalent. Its unit is J cal⁻¹, and the book's examples take 4.2 J cal⁻¹, so 1 cal of heat fully changed into work gives 4.2 J.

Clausius's statement: when heat is supplied to a system, part of it raises the internal energy, that is the temperature, and the rest is spent on external work. Heat supplied = rise in internal energy + external work. This is a special form of the law of conservation of energy: whether heat turns into another energy or another energy into heat, the total stays the same and nothing is lost.

Two things follow. First, the law ties heat and work together: they are equivalent. A fixed amount of work needs a fixed amount of heat, and a fixed amount of heat needs a fixed amount of work. Second, you cannot get work or energy without spending something, so a perpetual motion machine that runs without fuel or energy cannot be built.

The mathematical form is dQ = du + dW, or du = dQ − dW (1.2). Here dQ is the heat absorbed, du the change in internal energy and dW the work done by the system. It applies to every substance. In worked examples the same statement is written ΔQ = Δu + ΔW.

Gas in a cylinder. Put heat in, let the piston move, and watch the account below always balance: dQ = du + dW.

The sign rules are written from the system's side. dQ is positive when the system takes in heat and negative when it loses heat. du is positive when internal energy rises and negative when it falls. dW is positive when the system does work on the surroundings and negative when the surroundings do work on the system. Drop the six cases into the right boxes.

Water at the foot of a waterfall is slightly warmer than at the top. While falling, its potential energy becomes kinetic energy, and on reaching the ground part of that becomes heat and sound. If all the potential energy became heat, mgh = mSΔθ, so Δθ = gh/S. For a height of 500 m, g = 10 m s⁻² and S = 4200 J kg⁻¹ K⁻¹, Δθ = 10 × 500/4200 = 1.19°C.

3. One equation, four processes

In an isothermal process the temperature is constant, so du = 0 and dQ = dW (1.3). The work done by the system equals the heat supplied or taken in.

For n moles of ideal gas going from V1 to V2 isothermally, the work is W = ∫PdV = nRT ln(V2/V1) = Q. On the P-V graph it equals the area between the curve and the volume axis.

An example: if a system gives out 500 J of heat at constant volume, dQ = −500 J and dV = 0, so du = −500 J. The minus sign says the internal energy of the system has fallen.

In an adiabatic process no heat is exchanged, so dQ = 0 and dW = −du. In expansion the work comes out of the internal energy, so u2 < u1 and the system cools. In compression du = dW and u2 > u1, so the system warms.

Why does compression warm it? Work is done on the system with energy from outside, and the gas cannot give out heat, so both the internal energy and the temperature rise. That is why the bicycle pump turns warm.

In an adiabatic change with PV^γ = K, the work done is W = [1/(γ − 1)](P1V1 − P2V2) = R(T1 − T2)/(γ − 1), the last form being for one mole of gas.

At constant volume dV = 0, so the work dW = PdV = 0 and dQ = du: all the heat supplied raises the internal energy. On the P-V graph it is a line parallel to the y axis.

At constant pressure, with volume going from V1 to V2, the work is W = P(V2 − V1) = PΔV, that is pressure × change in volume. On the P-V graph it is a line parallel to the x axis. Now tap through the four processes.

4. The book's worked examples

Every numerical in the book is the same balance ΔQ = Δu + ΔW with different signs. Example 1.2(1): a system absorbs 800 J of heat and its internal energy rises by 500 J, so ΔW = ΔQ − Δu = 800 − 500 = 300 J. With the sign flipped: a system absorbs 1500 J and 300 J of work is done on it, so ΔW = −300 J and Δu = 1500 − (−300) = 1800 J. In the box below, put in the two you know and find the third. Whatever you type goes nowhere. The notebook is yours.

Example 1.2(2): the pressure is fixed at 400 Pa, and 800 J of heat supplied gives 1200 J of work. From 1200 = 400(V2 − V1) the volume rises by 3 m³. And 800 = Δu + 1200 gives Δu = −400 J: the work exceeds the heat, so the shortfall came out of the internal energy.

Example 1.2(3): at 1 × 10⁵ N m⁻² a gas goes from 0.05 m³ to 0.06 m³. The work is W = PΔV = 1 × 10⁵ × 0.01 = 1000 J. The new temperature is T2 = V2T1/V1 = 0.06 × 298/0.05 = 357.6 K, with T1 = 273 + 25 = 298 K.

Example 1.2(4): when a lead bullet strikes a target and stops, its kinetic energy becomes heat. To raise the temperature by 1.12°C, ½mv² = J × m × s × Δθ. With s = 30 cal kg⁻¹ °C⁻¹ and J = 4.2 J cal⁻¹, v = √(2 × 4.2 × 30 × 1.12) = 16.8 m s⁻¹.

Example 1.2(5): half the fall energy becomes heat and melts a quarter of a block of ice. Then mgh/2J = (m/4)L, so h = JL/2g = (4.2 × 80000)/(2 × 9.8) ≈ 17.14 km, with L = 80000 cal kg⁻¹.

Example 1.2(6): a hot 10 g iron nail is dropped into 100 g of water at 10°C and the water reaches 20°C. The heat the water takes in is 0.1 × 1 × 10 = 1 kcal. The nail gives out the same: 1 = 0.01 × 0.11 × (θ − 20), so θ = 929.09°C.

5. Check yourself

Three questions in the exam's style. Start with the first.

The second puts the sign rules to work.

Think the last one through before you answer.

The whole note on seven cards.

Source: Physics 2nd Paper, Classes 11-12 (NCTB), Chapter 1 “Thermodynamics”, section 1.2.