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DPST1021 Chap.9 Temperature, Calorimetry and Heat Transfer

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Chapter 9 of 15 · DPST1021

Temperature, Calorimetry and Heat Transfer

Thermal physics begins in week 8, and this material opens the content for Test 2. It starts with the zeroth law of thermodynamics, the statement that makes temperature a measurable shared property, and then defines the kelvin scale through absolute zero.

Students learn why temperature differences are identical in kelvin and Celsius while absolute values are not, a distinction that matters again in the gas laws that follow.

Thermal expansion comes next. Linear expansion is proportional to original length and temperature change, and the same idea, applied to squares and cubes, produces the area and volume relationships.

The course data sheet supplies expansion coefficients, specific heats and thermal conductivities for five metals, together with the thermal properties of water, and students are expected to use those values. Calorimetry then applies conservation of energy to mixtures, first with temperature changes alone and then with melting or boiling, where latent heat absorbs energy at constant temperature.

Nucleation, the role of impurities in starting a phase change, is treated qualitatively. Lab 5 measures specific and latent heat directly with a calorimeter.

The final part, taught in week 10, covers how heat moves. Conduction depends on conductivity, area and temperature gradient; radiation depends on area, emissivity and the fourth power of absolute temperature; convection is described in words.

Typical questions combine a sketch, a careful energy account and a check that the answer makes physical sense.

In this chapter

What this chapter covers

  • 01

    Zeroth law and temperature: explain how equilibrium with a third body defines temperature and makes a thermometer reading meaningful.

  • 02

    Kelvin scale and absolute zero: convert between Celsius and kelvin, and identify the formulas where absolute temperature must be used.

  • 03

    Thermal expansion: calculate changes in length, area and volume from data-sheet coefficients, including holes in plates and liquids in containers.

  • 04

    Specific heat and calorimetry: write an energy balance for an insulated mixture and solve for an unknown temperature or specific heat.

  • 05

    Latent heat and phase changes: decide whether melting or boiling completes before writing the balance, and read heating curves with flat plateaus.

  • 06

    Nucleation: describe how impurities and rough surfaces help bubbles and crystals form, and give examples of superheating.

  • 07

    Conduction and radiation rates: compute power from conductivity, area and gradient, or from emissivity, area and the fourth power of temperature.

Worked example · free

Iced water that does not fully melt

Q [4 marks]. Original AskSia exercise: 0.300 kg of ice at 0 °C is placed in 0.400 kg of water at 25.0 °C inside an insulated flask of negligible heat capacity. Using the data-sheet values for water, find the final temperature and the mass of ice left. AskSia practice weighting: 4 marks; this is not a university marking scheme.
Heating 100 g of ice at −30 °C to steam at 110 °C.Heating 100 g of ice at −30 °C to steam at 110 °C. Sloped parts obey Q = mcΔT; flat parts obey Q = mL. Boiling alone takes about three quarters of the 309 kJ total. Ice and steam specific heats (2100 and 2010 J/kg K) are typical values, not data-sheet entries.T (°C)1000−30meltingliquid warms, slope 1/(mc)boiling plateauvaporisation: mL = 226 kJfusion: mL = 33.4 kJ0100200300heat added Q (kJ)
Heating 100 g of ice at −30 °C to steam at 110 °C. Sloped parts obey Q = mcΔT; flat parts obey Q = mL. Boiling alone takes about three quarters of the 309 kJ total. Ice and steam specific heats (2100 and 2010 J/kg K) are typical values, not data-sheet entries.
  • 1Heat the warm water can release by cooling to 0 °C: Q = 0.400 × 4186 × 25.0 = 4.186 × 10⁴ J.
  • 1Heat needed to melt all the ice: Q = 0.300 × 3.336 × 10⁵ = 1.001 × 10⁵ J, well above the available amount.
  • 1Melting cannot finish, so the mixture settles at 0 °C.
  • 1Ice melted: m = 4.186 × 10⁴ / 3.336 × 10⁵ = 0.125 kg, so 0.300 − 0.125 = 0.175 kg of ice remains.
The final temperature is 0 °C, with 0.175 kg of ice still floating in 0.525 kg of liquid water.
Sia tip — Skipping the phase check and solving for T here gives −19.9 °C for liquid water, an impossible result. Compare available heat with the full melting requirement before writing any equation for T.
Glossary

Key terms

Zeroth law
The principle that two bodies each in thermal equilibrium with a third are in thermal equilibrium with each other.
Absolute zero
The lowest possible temperature, 0 K or −273.15 °C, found by extrapolating gas-thermometer pressure to zero.
Linear expansion coefficient
The fractional change in length per degree of temperature change, written α.
Specific heat
The energy needed to raise the temperature of one kilogram of a substance by one kelvin, in J/kg K.
Latent heat
The energy absorbed or released per kilogram during a phase change at constant temperature, such as melting or boiling.
Thermal conductivity
A material property k measuring how readily heat flows by conduction, in W/m K.
Emissivity
A number e between 0 and 1 describing how effectively a surface radiates compared with an ideal black surface.
FAQ

Temperature, Calorimetry and Heat Transfer FAQ

Why does water stay at 100 degrees Celsius while it boils on a stove?

All the incoming energy goes into separating molecules from the liquid rather than making them move faster, so the temperature holds steady until the liquid is gone. A hotter burner only makes it boil away sooner, provided the pressure stays at one atmosphere.

Do I need to convert to kelvin for specific heat problems?

Not for Q = mcΔT, because a change of one Celsius degree equals a change of one kelvin. Conversion becomes essential in the radiation law, where T is raised to the fourth power, and in the ideal gas law. Convert any temperature that is not part of a difference.

What happens to a hole in a metal plate when it is heated?

The hole grows by the same fraction as the metal around it, never shrinking as many people expect. Every length in the plate, including the hole’s diameter, scales by 1 + αΔT. A heated ring therefore slides over a shaft and grips once cool.

Why does metal feel colder than wood at the same temperature?

Both are at room temperature, but metal conducts far better, so it draws heat from your skin faster. Your nerves sense the rate of heat loss rather than the object’s temperature, which makes the metal seem colder even though a thermometer reads the same for both.

Does a hot object stop radiating when it reaches room temperature?

No. Anything warmer than 0 K emits radiation at a rate set by its own temperature. At room temperature it also absorbs radiation from its surroundings at an equal rate, so the net exchange is zero. Cooler surroundings then produce a net loss.

Study strategy

Exam move

Treat this material as three separate skills. For expansion, write the change as a fraction of the original size and check that the answer is tiny, usually well under one percent. For calorimetry, follow a fixed routine: list every body, compare the heat available with the heat needed for any phase change, decide the end state, and only then write one balance with each Q signed as final minus initial.

Then redo one mixture with masses changed so melting no longer finishes. For heat transfer, convert areas to square metres and use kelvin in the radiation law. Prepare for Lab 5 by predicting the final temperature before the measurement, and use Quizzes 6 and 7 for repeated practice with the significant-figure rule.

Working through Temperature, Calorimetry and Heat Transfer in DPST1021? Sia is AskSia’s AI Physics tutor — ask any DPST1021 Temperature, Calorimetry and Heat Transfer question and get a clear, step-by-step explanation grounded in how DPST1021 is taught and assessed. Read this chapter free, then take your hardest questions to Sia.

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