Aim
To determine the specific heat capacity of aluminium by supplying a measured amount of electrical energy and measuring the temperature rise.
Variables
- Manipulated: Heating time, t
- Responding: Temperature of the block, θ
- Constant: Mass of the aluminium block and the power of the heater
Apparatus & materials
- Aluminium block (1 kg) with holes for the heater and thermometer
- Immersion heater
- Thermometer
- Power supply
- Ammeter
- Voltmeter
- Stopwatch
- Electronic balance
- Insulating jacket
- A little oil
Procedure
- Measure the mass m of the aluminium block with the balance.
- Insert the immersion heater and, with a drop of oil for good contact, the thermometer into the block, then wrap it in the insulating jacket.
- Record the initial temperature θ₀, then switch on the heater and the stopwatch together.
- Record the ammeter reading I and the voltmeter reading V while the heater is on.
- Record the temperature θ every minute for several minutes.
- Switch off and note the highest steady temperature reached.
Tabulating results
Record the fixed values m, V and I, and a table of heating time t in s (or min) against temperature θ in °C. Note the initial and final temperatures to find the temperature rise Δθ.
The graph
Plot temperature θ (y-axis) against time t (x-axis). The straight portion has a gradient equal to the rate of temperature rise, P / (mc).
Analysis
The electrical energy supplied equals the heat gained: VIt = mcΔθ. So c = VIt / (mΔθ). Using the graph, c = P / (m × gradient), where P = VI.
Precautions
- Wrap the block in insulation to reduce heat loss to the surroundings.
- Put a drop of oil in the thermometer hole for good thermal contact.
- Let the heater warm the block before taking the first steady reading.
Sample results and what they show
These are example readings, not a mark scheme. With a 1.0 kg aluminium block, a heater at V = 12 V and I = 4.0 A (power P = VI = 48 W), the temperature rose steadily: at t = 0, 1, 2, 3, 4, 5 min the block read about 28.0, 31.2, 34.4, 37.6, 40.8 and 44.0 °C.
The rise is close to a steady 3.2 °C each minute, which tells you the heater delivers energy at a nearly constant rate and the insulation is doing its job. The first minute often lags slightly because the heater and block are still warming through, so the very first interval can be a little smaller, take readings over several minutes and use the steady middle portion.
The total rise here is Δθ ≈ 16.0 °C over 300 s. A curve that flattens towards the end is a sign of increasing heat loss to the surroundings once the block is much hotter than the room.
Reading the graph and finding the answer
Plot θ / °C on the y-axis against t / s on the x-axis. The steady middle section is a straight line whose gradient is the rate of temperature rise.
Take the gradient from a large triangle on the straight portion. Using (0 s, 28.0 °C) and (300 s, 44.0 °C): gradient = (44.0 − 28.0) °C ÷ (300 − 0) s = 16.0 °C ÷ 300 s = 0.0533 °C s⁻¹.
Since P = mc × (rate of rise), the gradient equals P/(mc), so c = P ÷ (m × gradient) = 48 W ÷ (1.0 kg × 0.0533 °C s⁻¹) = 900 J kg⁻¹ °C⁻¹. Using the gradient rather than a single before-and-after Δθ is better because it uses all the readings and sidesteps the slow start.
The value sits close to the accepted 900 J kg⁻¹ °C⁻¹ for aluminium.
Marks examiners look for
State the variables: manipulated t / s, responding θ / °C, with the block mass and the heater power kept constant. The precautions carry real marks here: lag the block in an insulating jacket to cut heat loss, and put a drop of oil in the thermometer hole so the thermometer reads the block's temperature quickly.
Record the fixed values m / kg, V / V and I / A above a table of t / s against θ / °C, with consistent decimals. Waiting until the heater has warmed the block before taking the first steady reading is the fair-test detail that keeps the gradient honest.
On the graph, use only the straight portion for the gradient and draw a single best-fit line. Quote c with its unit, J kg⁻¹ °C⁻¹, and note that the true value is likely a little high because some heat still escapes, a sensible evaluation point.
Do not treat the block as gaining heat only from the heater if you have not insulated it.
Source: DSKP KSSM Physics Form 4 and 5 (Versi English) (Bahagian Pembangunan Kurikulum (BPK), KPM)