Aim
To determine the specific latent heat of fusion of ice by melting it with a measured amount of electrical energy, using a control to correct for the surroundings.
Variables
- Manipulated: Whether the heater is switched on (experiment) or off (control)
- Responding: Mass of ice melted, m
- Constant: Heating time and the power of the heater
Apparatus & materials
- Immersion heater
- Crushed ice
- Two filter funnels
- Two beakers
- Electronic balance
- Power supply
- Ammeter
- Voltmeter
- Stopwatch
- Retort stands
Procedure
- Pack crushed ice into two funnels and place an empty beaker under each; one funnel is the experiment and the other is the control.
- Insert the immersion heater into the experiment funnel and connect it in series with the ammeter and power supply, with the voltmeter across it.
- Let both funnels drip until the ice is at melting point, then empty and replace the beakers.
- Switch on the heater and the stopwatch together, and record the readings V and I.
- After a fixed time t, switch off, and measure the mass of water collected in each beaker with the balance.
- Find the mass melted by the heater alone, m = m(experiment) − m(control).
Tabulating results
Record the values V, I and the heating time t, and the mass of water collected in the experiment and the control funnels. The corrected mass melted by the heater is m = m(experiment) − m(control).
The graph
A graph is not required. The control funnel provides the correction for ice melted by heat from the surroundings during the same time.
Analysis
The electrical energy supplied melts the corrected mass of ice: VIt = mL, where m is the corrected mass. So the specific latent heat of fusion L = VIt / m.
Precautions
- Use a control funnel to correct for ice melted by the surroundings.
- Start only when the ice is dripping at its melting point (0 °C).
- Make sure all the melted water drips into the beaker and none is lost.
Sample results and what they show
These are example readings, not a mark scheme. With the heater at V = 12 V and I = 4.0 A for t = 5.0 min (300 s), the experiment funnel collected about 48.0 g of melted water while the control funnel, heater off, collected about 5.0 g in the same time.
The control tells you that about 5.0 g melted from the warmth of the room, not from the heater. Subtracting gives the mass melted by the heater alone: m = 48.0 − 5.0 = 43.0 g = 0.0430 kg.
Without the control you would credit all 48.0 g to the heater and get a value of L about 12% too low. Both funnels must start dripping steadily at 0 °C before you begin, so the only difference between them during the timed run is the heater, and every drop must reach the beaker so no melted water is lost.
Getting the answer from the readings
This experiment does not need a graph, the answer comes from the corrected mass and the electrical energy. Convert the collected masses first: corrected mass m = m(experiment) − m(control) = 48.0 g − 5.0 g = 43.0 g = 0.0430 kg.
The electrical energy supplied melts that mass: VIt = mL. So L = VIt ÷ m = (12 V × 4.0 A × 300 s) ÷ 0.0430 kg = 14400 J ÷ 0.0430 kg = 3.35 × 10⁵ J kg⁻¹.
That is close to the accepted specific latent heat of fusion of ice, about 3.34 × 10⁵ J kg⁻¹. If you had skipped the control and used 48.0 g, you would get 3.0 × 10⁵ J kg⁻¹, noticeably low, which shows why the control correction is the heart of the method.
Marks examiners look for
The manipulated 'variable' here is really the comparison: heater on for the experiment funnel and off for the control, with the same heating time t / s and the same power kept constant. Naming the control funnel and explaining what it corrects for is the single most valuable point.
Record V / V, I / A, t / s, and the masses m(experiment) / g and m(control) / g, then the corrected m / kg, with consistent decimals and units in the headings. Fair-test details worth credit: start timing only when both funnels drip at the melting point, keep both funnels identical apart from the heater, and make sure all the melt drips into the beaker.
Quote L with its unit, J kg⁻¹, and note that a small heat loss or a little water clinging in the funnel keeps the value slightly off the accepted one. Do not compare an experiment run at one time with a control run at another, they must be timed together.
Source: DSKP KSSM Physics Form 4 and 5 (Versi English) (Bahagian Pembangunan Kurikulum (BPK), KPM)