Skip to main content
United StatesPhysics 2Syllabus dot point

Why do different materials need different amounts of energy to heat up, and how fast does heat conduct through a slab?

Topic 9.5 Specific Heat and Thermal Conductivity: apply Q = mc(delta T) for heating and the conduction rate equation for steady heat flow.

A focused answer to AP Physics 2 Topic 9.5, covering specific heat capacity and the relation Q = mc(delta T), calorimetry with conservation of energy, the rate of heat conduction through a material, and the role of thermal conductivity, with full worked examples.

Generated by Claude Opus 4.811 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section
  1. What this topic is asking
  2. Specific heat capacity
  3. Calorimetry: conservation of energy
  4. The rate of heat conduction
  5. Try this

What this topic is asking

The College Board (Topic 9.5) wants you to use specific heat to find the heat needed to change a substance's temperature, Q=mc ΔTQ = mc\,\Delta T, to solve calorimetry problems with conservation of energy, and to describe the rate of heat conduction through a material.

Specific heat capacity

Specific heat captures how "thermally sluggish" a material is. Water's unusually high value, 41864186 J/(kg K), is why oceans moderate climate and why water is a good coolant: it soaks up a lot of energy for a small temperature rise. From Q=mc ΔTQ = mc\,\Delta T, the temperature change is ΔT=Q/(mc)\Delta T = Q/(mc), so for equal heat and mass, a larger specific heat gives a smaller temperature change. This inverse relationship is a frequent exam point.

Calorimetry: conservation of energy

Calorimetry is the first law (Topic 9.4) with no work done: all the energy transfer is heat, and in an insulated container it simply moves from hot to cold until both reach a common temperature. The strategy is always the same: write "heat lost = heat gained," express each side as mc ΔTmc\,\Delta T with the correct signs, and solve for the equilibrium temperature. The body with the larger mcmc dominates the mixture's final temperature, which is why a small hot object barely warms a large tank of water.

The rate of heat conduction

Conduction transfers energy through a material by atomic collisions, and the College Board asks you to reason about its rate:

Qt=kA ΔTL\dfrac{Q}{t} = \dfrac{k A\,\Delta T}{L}

Here kk is the thermal conductivity (large for metals, small for insulators), AA the cross-sectional area, ΔT\Delta T the temperature difference across the slab, and LL the thickness. Heat flows faster through a better conductor, a larger area or a bigger temperature difference, and slower through a thicker slab. This is why metal feels colder than wood at the same temperature (it conducts heat from your hand quickly) and why insulation is made thick and of low-conductivity material. The strategic thread of this topic is the split between two questions: Q=mc ΔTQ = mc\,\Delta T answers "how much energy to change the temperature," while the conduction equation answers "how fast does energy flow through." Both are applications of energy conservation that feed the first-law reasoning of the unit.

Try this

Q1. Calculate the heat needed to raise 2.02.0 kg of a substance (c=900c = 900 J/(kg K)) by 1515 K. [2 points]

  • Cue. Q=mc ΔT=(2.0)(900)(15)=2.7×104Q = mc\,\Delta T = (2.0)(900)(15) = 2.7 \times 10^4 J.

Q2. State how the rate of heat conduction through a slab changes if its thickness is doubled (all else equal). [1 point]

  • Cue. It halves, because the rate is inversely proportional to thickness LL.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2024 (style)6 marksSection II (short FRQ). A 0.400.40 kg block of metal at 9090 degrees Celsius is dropped into 0.500.50 kg of water at 2020 degrees Celsius in an insulated cup. The specific heat of water is 41864186 J/(kg K) and of the metal is 390390 J/(kg K). (a) Write the energy-conservation statement linking the heat lost by the metal and gained by the water. (b) Calculate the final equilibrium temperature. (c) Explain why the temperature change of the water is small compared with that of the metal.
Show worked answer →

A 6-point FRQ on calorimetry.

(a) Conservation (2 points): heat lost by metal equals heat gained by water (insulated cup): mmcm(90−T)=mwcw(T−20)m_m c_m (90 - T) = m_w c_w (T - 20).
(b) Final temperature (3 points): (0.40)(390)(90−T)=(0.50)(4186)(T−20)(0.40)(390)(90 - T) = (0.50)(4186)(T - 20). Left: 156(90−T)156(90 - T); right: 2093(T−20)2093(T - 20). So 14040−156T=2093T−4186014040 - 156T = 2093T - 41860, giving 55900=2249T55900 = 2249T, T=24.9T = 24.9 degrees Celsius.
(c) Small water change (1 point): water has a much larger mass times specific heat (mcmc), so it absorbs the energy with only a small temperature rise.

Markers reward the energy-balance equation, solving for the equilibrium temperature, and explaining the small water change through the large mcmc of water.

AP 2023 (style)1 marksSection I (multiple choice). Two blocks of equal mass, one copper and one aluminum, absorb the same amount of heat. Aluminum has the larger specific heat. Which block warms more? (A) the copper block (B) the aluminum block (C) both warm equally (D) it cannot be determined. Justify your reasoning.
Show worked answer →

A 1-point MCQ on specific heat. The answer is (A).

From Q=mc ΔTQ = mc\,\Delta T, with the same QQ and mm, the temperature change is inversely proportional to specific heat: ΔT=Q/(mc)\Delta T = Q/(mc). The copper block has the smaller specific heat, so it has the larger temperature change. The trap is (B): the larger specific heat means a smaller, not larger, temperature rise.

Related dot points

Sources & how we know this