Specific heat capacity
The specific heat capacity $c$ of a substance is the energy needed to raise the temperature of 1 kg of it by 1 °C (or 1 K — same thing for a difference).
$\Delta E = mc\Delta\theta$
- $\Delta E$ in J.
- $m$ in kg.
- $c$ in J/kg/K.
- $\Delta\theta$ in °C or K.
Common values
- Water: 4200 J/kg/K (very high — water is a great heat reservoir).
- Aluminium: 900 J/kg/K.
- Iron: 450 J/kg/K.
- Copper: 385 J/kg/K.
- Air: 1010 J/kg/K (per kg of air).
The high value for water explains why coastal climates are mild (the sea absorbs/releases heat slowly), why hot-water bottles work well, and why water is the standard coolant.
✦Worked example— Worked example 1
A 0.40 kg copper kettle base ($c = 385$) cools from 90 °C to 25 °C. Energy released?
- $\Delta\theta = 65$ K.
- $\Delta E = mc\Delta\theta = 0.40 \times 385 \times 65 = 10,010$ J ≈ 10 kJ.
✦Worked example— Worked example 2
A 2.0 kg block of aluminium is heated by a 100 W heater for 5.0 minutes. The block's temperature rises by 16.7 °C. Find experimental $c$ and compare to the textbook value (900 J/kg/K).
- $\Delta E = Pt = 100 \times 300 = 30,000$ J.
- $c = \Delta E /(m\Delta\theta) = 30,000/(2.0 \times 16.7) \approx 898$ J/kg/K.
- Very close to 900 J/kg/K — small discrepancy is due to heat lost to surroundings.
Required practical 1 — measuring SHC
Apparatus: insulated metal block (or beaker of liquid), 12 V immersion heater, joulemeter (or ammeter, voltmeter and timer), thermometer, balance.
Method:
- Measure mass $m$ of the block on a balance.
- Insert thermometer and heater into the block.
- Record initial temperature $\theta_1$.
- Switch on heater for measured time $t$; note current $I$ and p.d. $V$ (or read joulemeter).
- Record maximum temperature $\theta_2$.
- Energy supplied $E = VIt$ (or read directly).
- $c = E/(m \Delta\theta)$, where $\Delta\theta = \theta_2 - \theta_1$.
Sources of error: heat loss to surroundings (largest); thermometer lag; uneven heating in the block.
⚠Common mistakes
- Using mass in grams — convert to kg.
- Forgetting to insulate (gives an answer too high if energy escapes).
- Confusing $c$ with $L$ (latent heat) — $c$ is for temperature change, $L$ is for state change.
- Reading temperature too early — let the block equilibrate.
➜Try this— Quick check
How much energy raises 200 g of water from 25 °C to 75 °C?
- $m = 0.20$ kg, $\Delta\theta = 50$ K, $c = 4200$.
- $\Delta E = 0.20 \times 4200 \times 50 = 42,000$ J = 42 kJ.
AI-generated · claude-opus-4-7 · v3-deep-physics