IB Physics SL Topic B.1 — Heat & Thermal Transfer Paper 1 & 2 Specific Heat Capacity ~7 min read

Specific Heat Capacity

Heat up equal masses of water and iron by the same amount, and one takes far longer than the other. Specific heat capacity is the number that explains the difference.

📘 What you need to know

What Determines the Thermal Energy Needed?

Changing an object’s temperature always takes energy, and how much depends on three things:

Thermal energy transferred Q = mcΔT

Rearranging this equation for c explains the definition of specific heat capacity directly:

c = Q⁄(mΔT)
Water 4200 J/(kg K)Aluminium 900 J/(kg K)Iron 450 J/(kg K)Lead 130 J/(kg K)Approximate specific heat capacity
Water’s unusually high specific heat capacity means it takes far more energy to heat up (or release, to cool down) than most metals

Why Specific Heat Capacity Matters

Water’s high specific heat capacity is why coastal regions tend to have milder climates than inland areas at the same latitude — large bodies of water absorb and release huge amounts of thermal energy without their temperature swinging very much. A metal like lead, by contrast, has a very low specific heat capacity, so it heats up and cools down almost instantly by comparison.

🧭 Recipe: Solving a Thermal Equilibrium Mixing Problem

  1. List the knowns for both substances — mass, specific heat capacity, and starting temperature
  2. State the energy balance — energy lost by the hotter substance equals energy gained by the cooler one
  3. Write out mcΔT for each substance, using the same final temperature for both, since they reach thermal equilibrium together
  4. Solve the resulting equation for the unknown, usually the final equilibrium temperature
Quick recap: Q = mcΔT. Higher specific heat capacity means slower heating and cooling for the same mass and energy input.
WE 1

A 120 g block of aluminium is heated from 18 °C to 95 °C. The specific heat capacity of aluminium is 900 J kg⁻¹ K⁻¹. Calculate the thermal energy required.

Step 1 — List the known quantities m = 0.120 kg, c = 900 J kg⁻¹ K⁻¹, ΔT = 95 − 18 = 77 °C Step 2 — Substitute into Q = mcΔT Q = 0.120 × 900 × 77 ≈ 8320 J ≈ 8.3 kJ
WE 2

A 40 g piece of lead at 150 °C is dropped into 150 g of water at 20 °C. The specific heat capacity of lead is 130 J kg⁻¹ K⁻¹, and of water is 4200 J kg⁻¹ K⁻¹. Determine the final temperature of the water and lead, assuming no energy is lost to the surroundings.

Step 1 — State the energy balance Energy lost by lead = Energy gained by water −m_Pb c_Pb (T_f − 150) = m_w c_w (T_f − 20) Step 2 — Substitute the known values −0.040 × 130 × (T_f − 150) = 0.150 × 4200 × (T_f − 20) Step 3 — Expand and solve for T_f 780 − 5.2T_f = 630T_f − 12 600 → 13 380 = 635.2T_f T_f ≈ 21.1 °C Because the water’s mass and specific heat capacity are both much larger than the lead’s, the final temperature barely shifts from the water’s starting point.

💡 Top tips

⚠ Common mistakes

Up next: Specific Latent Heat — where we look at the energy needed to change state, rather than to change temperature.

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