First Law of Thermodynamics, PHYS – Study Notes
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Difficulty: Introductory. Prerequisites: basic mechanics (potential energy, kinetic energy, conservation of energy).

This topic sits at the boundary between mechanics and thermal physics. It extends the conservation of energy idea you already know into the domain of heat, and it underpins everything from engine design to climate science. If you are comfortable with gravitational potential energy and the idea that energy is neither created nor destroyed, you have enough to follow along.

TL;DR

The first law of thermodynamics says energy is conserved: heat added to a system either raises its internal energy or does work, and nothing is lost or created. In Joule's paddle-wheel experiment, gravitational potential energy converts into thermal energy in water, and the temperature rise depends on how much energy arrives and how much water is there to absorb it. The constant that links energy, mass and temperature change for a given substance is its specific heat capacity.


Key Terms

First law of thermodynamics

The total energy of an isolated system is constant. Any heat energy added to the system (Q) equals the change in its internal energy (ΔU) plus any work the system does (W). In simple terms, energy does not appear from nowhere or vanish: it just changes form.

Heat energy transfer

The flow of thermal energy between objects or systems at different temperatures. Think of it as energy moving from something hotter to something cooler until both reach the same temperature.

Thermal equilibrium

The state reached when two objects in thermal contact stop exchanging net heat because they are at the same temperature. In simple terms, the flow stops when there is no temperature difference left to drive it.

Internal energy (U)

The total microscopic kinetic and potential energy of the particles in a substance. Think of it as the "hidden" energy inside a material, related to how fast its molecules are moving and how they interact.

Specific heat capacity (c)

The amount of heat energy required to raise the temperature of one kilogram of a substance by one degree Celsius. For water, c = 4,190 J/(kg·°C). In simple terms, it tells you how stubborn a material is about warming up: a high specific heat means it takes a lot of energy to change its temperature.

Mechanical equivalent of heat

The principle, demonstrated by Joule, that a measurable amount of mechanical work produces an equivalent, predictable amount of heat. This confirmed that heat is a form of energy, not a separate substance.

Gravitational potential energy (PE)

The energy an object has because of its height above a reference point, calculated as PE = mgh. In Joule's experiment, this is the energy source: a falling mass converts its PE into thermal energy in the water.


Core Content

Energy Conservation and the First Law

  • The first law is a bookkeeping rule: energy in = energy stored + energy out. No exceptions.

  • For a system where no work is done (like water sitting in an insulated container), all the heat that goes in raises the internal energy, which we observe as a temperature increase.

  • The equation form used in this lab: ΔQ = c × m × ΔT. All the gravitational PE of the falling block converts to thermal energy in the water.

Heat Transfer Mechanisms

  • Conduction: energy passes through direct contact between particles. A metal spoon in hot soup gets warm this way.

  • Convection: bulk movement of a heated fluid carries energy. This is how a radiator warms a room.

  • Mechanical work converting to heat: rubbing your hands together, or Joule's paddle churning water. The key insight is that ordered mechanical energy becomes disordered thermal energy.

The Role of Each Variable in ΔQ = c × m × ΔT

  • ΔQ (heat energy transferred, in joules): the total energy delivered to the water. In the lab, this equals the gravitational PE of the falling block (mgh). More energy in means a bigger temperature rise, all else being equal.

  • c (specific heat capacity): a property of the substance itself. Water's value (4,190 J/(kg·°C)) is high compared to most materials, which is why water is so effective at absorbing heat without a dramatic temperature change.

  • m (mass of the water, in kg): more water means the same energy is spread across more material, so the temperature rise is smaller. Halving the water mass doubles the temperature change for the same energy input.

  • ΔT (temperature change, in °C): the outcome you measure. Rearranging: ΔT = ΔQ / (c × m). Temperature change is directly proportional to energy input and inversely proportional to both mass and specific heat.

What the Lab Data Shows

  • Runs 1 vs 2 (block mass increased, water mass constant): a heavier block has more gravitational PE, so more energy is transferred to the water. The temperature change is larger.

  • Run 3 (water mass halved): the same energy goes into less water, so the temperature change roughly doubles. The ratio ΔT/ΔQ changes because m has changed.

  • Run 4 (different starting temperature): the temperature change ΔT stays the same as Run 3 because specific heat does not depend on the starting temperature (at least in this idealized simulation). The initial temperature shifts the final temperature but not the size of the change.


Formulas and Diagrams

Primary Formula

\Delta Q = c \times m \times \Delta T

Where ΔQ is in joules (J), c is in J/(kg·°C), m is in kilograms (kg), and ΔT is in degrees Celsius (°C).

Rearranged Forms

To find specific heat:

c = \frac{\Delta Q}{m \times \Delta T}

To find temperature change:

\Delta T = \frac{\Delta Q}{c \times m}

To find mass:

m = \frac{\Delta Q}{c \times \Delta T}

Gravitational PE to Thermal Energy (Joule's Experiment)

The energy input equals the gravitational potential energy of the falling block:

\Delta Q = m_{block} \times g \times h

Where g = 9.8 m/s² and h is the height the block falls. In the idealised simulation, 100% of this PE becomes thermal energy in the water.

Units Check

A quick dimensional check keeps you out of trouble: J = (J/(kg·°C)) × kg × °C. The kg and °C units multiply out to leave joules on both sides.


Real-World Applications

Water's high specific heat is the reason coastal cities have milder climates than inland ones: the ocean absorbs and releases enormous amounts of energy with relatively small temperature swings. It is also why car engines use water-based coolant in their radiators, and why you heat water on the stove for a long time before it boils, compared to how quickly a dry pan gets hot.


Common Misconceptions

  • Students often think that temperature and heat are the same thing. They are not. Heat (Q) is energy in transit; temperature (T) is a measure of average kinetic energy of particles. You can add a large amount of heat to a large body of water and barely change its temperature.

  • Students often think a higher starting temperature means a larger temperature change for the same energy input. The starting temperature shifts the final reading but does not affect ΔT when specific heat is constant over that range.

  • Students often confuse the mass of the falling block with the mass of the water in the formula ΔQ = c × m × ΔT. The block's mass determines how much energy is available (via PE = mgh). The water's mass is what goes into the specific heat equation.

  • Students often forget that specific heat is a property of the substance, not the experiment. Changing the block mass, the water mass, or the starting temperature should all yield the same value of c for water if the measurements are clean.


Why It Matters / Exam Flags

  • ⚠️ Rearranging ΔQ = c × m × ΔT to solve for any one variable is a near-certain exam question. Be fluent with all three rearranged forms.

  • ⚠️ Expect a question asking you to explain why doubling the water mass halves the temperature change for a given energy input. The inverse proportionality between m and ΔT is a favourite.

  • ⚠️ Percentage error calculations comparing your experimental specific heat to the accepted value (4,190 J/(kg·°C)) are standard post-lab fare. Know the formula: % error = |experimental - accepted| / accepted × 100.

  • ⚠️ Be ready to list sources of experimental error in a real (non-idealised) version of this experiment: heat loss to surroundings, friction in pulleys, energy absorbed by the stirring mechanism, measurement precision.


Quick Self-Test

  1. True or false: heat and temperature are the same physical quantity. (False. Heat is energy in transit; temperature is a measure of average molecular kinetic energy.)

  1. Fill in the blank: the specific heat capacity of water is approximately ______ J/(kg·°C). (4,190)

  1. True or false: if you double the mass of the water and keep everything else the same, the temperature change will also double. (False. It will halve.)

  1. Fill in the blank: in Joule's experiment, the energy source is the ______ of the falling block. (gravitational potential energy)

  1. True or false: changing the initial water temperature from 20°C to 10°C changes the specific heat of water. (False. Specific heat is a material property, not dependent on starting temperature.)


Practice Q&A

Q: A 5.0 kg block falls from 100 m in Joule's apparatus. The insulated container holds 5.0 kg of water. Calculate the expected temperature change of the water. (Use g = 9.8 m/s², c = 4,190 J/(kg·°C).)

A: PE = 5.0 × 9.8 × 100 = 4,900 J. ΔT = ΔQ / (c × m) = 4,900 / (4,190 × 5.0) = 0.234°C.

Q: Explain why increasing the block mass from 2.0 kg to 5.0 kg increases the temperature change of the water.

A: A heavier block has more gravitational potential energy (PE = mgh). More energy is converted to thermal energy in the water, so ΔQ is larger. Since ΔT = ΔQ / (c × m), a larger ΔQ produces a larger ΔT when c and m are unchanged.

Q: In Run 3, the water mass was halved from 5.0 kg to 2.5 kg while the block mass stayed at 5.0 kg. What happened to the temperature change, and why?

A: The temperature change roughly doubled. The same energy (ΔQ) was distributed across half the water mass. Since ΔT = ΔQ / (c × m), halving m doubles ΔT.

Q: Your experimental runs in Table 2 each produced a value for specific heat. The accepted value is 4,190 J/(kg·°C). If your average measured value is 4,180 J/(kg·°C), what is the percentage error?

A: % error = |4,180 - 4,190| / 4,190 × 100 = 10 / 4,190 × 100 ≈ 0.239%.

Q: List three possible sources of experimental error in a real (non-idealised) version of Joule's experiment.

A: (1) Heat loss from the water to the surrounding environment. (2) Friction in the pulleys absorbing some of the block's energy. (3) The stirring paddle and container absorbing some thermal energy rather than all of it going into the water.


Connections to Other Topics

This material connects directly to the second law of thermodynamics, which governs the direction of heat flow and introduces entropy. It also links back to conservation of energy in mechanics: Joule's experiment is the bridge between the kinetic energy and potential energy you studied in mechanics and the thermal energy you will encounter throughout thermodynamics. If your course covers calorimetry or heat engines later, the specific heat concept from this lab is the foundation for both.


Related Terms / Search Tags

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