Required Practical 1: Specific heat capacity

This is one of the eight required practicals you can be examined on. The skill being tested is measuring an unknown property accurately, controlling variables, and using $\Delta E = m c \Delta\theta$ to calculate $c$.

Aim

Measure the specific heat capacity of a metal block (or a beaker of water/oil).

Apparatus

• 1 kg metal block with two cylindrical holes (one for an immersion heater, one for a thermometer)
• 12 V immersion heater
• Joulemeter (or low-voltage power supply, ammeter and voltmeter, plus a stopwatch)
• Digital thermometer (or –10 to 110 °C glass thermometer)
• Top-pan balance
• Thick insulating wrap (cotton wool or foam jacket)
• Stopwatch

Method (block version)

1. Use the balance to record the mass $m$ of the metal block.
2. Place the heater into one hole and the thermometer into the other. Put a few drops of oil into the thermometer hole to ensure good thermal contact.
3. Wrap the block in insulation to reduce energy lost to surroundings.
4. Record the starting temperature $\theta_0$.
5. Switch the joulemeter on, start the stopwatch and run the heater for 10 minutes (600 s). If using V and I, record voltage and current and calculate $E = VIt$.
6. After switching off, continue stirring/waiting until the temperature stops rising — this is your final temperature $\theta_1$.
7. Calculate $\Delta\theta = \theta_1 − \theta_0$ and $c = \Delta E / (m \Delta\theta)$.

Variables

• Independent (you change): energy supplied (or time of heating).
• Dependent (you measure): temperature rise.
• Control: mass of substance, insulation, starting temperature.

Sample data

A 1.00 kg aluminium block warms from 22.0 °C to 32.0 °C when 9000 J is supplied:

$c = 9000 / (1.00 \times 10) = 900\text{ J/kg °C}$ — matches the textbook value for aluminium.

Sources of inaccuracy

• Heat lost to surroundings — biggest source. Use insulation and read temperature after the heater is off so the heat has time to spread evenly through the block.
• Heater takes time to warm up — some energy is stored in the heater itself, not the block.
• Thermometer resolution (typically ±0.5 °C) — use a digital probe or a Beckmann thermometer for higher accuracy.
• Uneven heating — the spot near the heater is hotter than the far edge until thermal equilibrium is reached.

Improving the experiment

• Wrap with thicker insulation.
• Use a higher-power heater for shorter time so less is lost.
• Use a temperature probe with logging software for many measurements.
• Repeat and take a mean.

How they ask it in exams

Common question types:

1. "Why was the block insulated?" — to reduce energy transferred to the surroundings via heating, so all measured energy goes into the block.
2. "Why is oil placed in the thermometer hole?" — to give good thermal contact between the block and thermometer (air is a poor conductor).
3. "Suggest one source of inaccuracy and how to reduce it." — see list above.
4. "Calculate $c$ from the data." — straightforward $\Delta E / (m \Delta\theta)$.

Calculation walk-through

If you have V = 12.0 V, I = 4.0 A, t = 300 s, mass 0.80 kg, $\Delta\theta = 12.5$ °C:

• Energy supplied $E = VIt = 12.0 \times 4.0 \times 300 = 14{,}400$ J.
• $c = 14{,}400 / (0.80 \times 12.5) = 1440$ J/kg °C.

That tells you the substance is probably water (4200) — wait, no: 1440 is closest to oil. Always sanity-check against the textbook value to make sure your method was reasonable.