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The Timpson Lyceum · University Physics I (2025)

Temperature and Heat - Lecture 17 (University Physics I)

Overview

This lecture opens the thermodynamics unit by building the macroscopic concepts of temperature and heat from the ground up. We start with the operational idea of temperature (something measurable, e.g. by the length of an expanding rod), formalize it with the zeroth law of thermodynamics and the standard temperature scales (Fahrenheit, Celsius, Kelvin, and absolute zero). We then study how matter responds to temperature changes through thermal expansion (linear and volumetric) and the thermal stress that arises when expansion is prevented. Finally we define heat as energy in transit, quantify it with specific heat and latent heat for phase changes, and survey the three modes of heat transfer: conduction, convection, and radiation. These ideas are foundational for the entire thermodynamics sequence.

Key Concepts

Temperature and the zeroth law

Temperature scales

Thermal expansion

Thermal stress

Quantity of heat, specific heat, and moles

Phase changes and latent heat

Heat transfer

Worked Example

A steel surveyor's tape reads short when warm. A steel tape is calibrated to read exactly correct at 20 ∘C20\,{}^\circ\text{C}. On a 35 ∘C35\,{}^\circ\text{C} day a surveyor reads a distance of 35.794 m35.794\ \text{m} off the tape. Because the tape itself has expanded, each marked interval now spans a slightly greater true length, so the actual distance is longer than the reading. With αsteel=1.2×10−5 K−1\alpha_{\text{steel}} = 1.2\times10^{-5}\ \text{K}^{-1} and ΔT=15 K\Delta T = 15\,\text{K}:

The true length is the reading scaled up by the tape's fractional expansion,

x=x0 (1+α ΔT)x=35.794 (1+(1.2×10−5)(15))x≈35.80 m.\begin{aligned} x &= x_0\,(1 + \alpha\,\Delta T) \\ x &= 35.794\,\bigl(1 + (1.2\times10^{-5})(15)\bigr) \\ x &\approx 35.80\ \text{m}. \end{aligned}

A thermal-stress force. An aluminum spacer (α=2.4×10−5 K−1\alpha = 2.4\times10^{-5}\ \text{K}^{-1}, Y=7.0×1010 PaY = 7.0\times10^{10}\ \text{Pa}, area A=20 cm2=0.0020 m2A = 20\ \text{cm}^2 = 0.0020\ \text{m}^2) just fits between two rigid steel walls at 17.2 ∘C17.2\,{}^\circ\text{C}. Find the stress and the force on each wall when it warms to 22.3 ∘C22.3\,{}^\circ\text{C} (ΔT=5.1 K\Delta T = 5.1\ \text{K}):

FA=−Y α ΔT=−(7.0×1010)(2.4×10−5)(5.1)≈−8.6×106 PaF=(FA)A=(8.6×106)(0.0020)≈1.7×104 N.\begin{aligned} \frac{F}{A} &= -Y\,\alpha\,\Delta T = -(7.0\times10^{10})(2.4\times10^{-5})(5.1) \approx -8.6\times10^{6}\ \text{Pa} \\ F &= \left(\frac{F}{A}\right)A = (8.6\times10^{6})(0.0020) \approx 1.7\times10^{4}\ \text{N}. \end{aligned}

A mere 5 °C rise generates about 17,000 N — in practice the spacer simply pushes slightly into the walls.

Summary