Vaporization energy
Determine the specific heat of vaporization of water by continuously tracking the mass of water evaporating during boiling with a FizziQ Connect mini-balance.
Why does it take so much energy to boil water? When water reaches 100°C, its temperature stops rising despite continuous heat input: all the energy goes into breaking the hydrogen bonds that hold water molecules together in the liquid phase. This energy, called the latent heat of vaporization, is remarkably large — about 2260 kJ/kg. By tracking the mass of boiling water with a connected balance while supplying a known power, you can measure this fundamental quantity directly.
Learning objectives:
The student tracks the mass of boiling water on a connected balance while supplying known power, calculating the vaporization energy from the linear mass decrease.
Level:
High school
Serge paupy
Author:
Duration (minutes) :
85
What students will do :
- Measure mass loss during boiling with a connected balance
- Calculate the energy supplied from power and time
- Determine the specific heat of vaporization from L = E/Δm
- Compare with the theoretical value of 2260 kJ/kg
- Understand the concept of latent heat and phase change
Scientific concepts:
- Specific heat of vaporization
- Phase change (liquid to gas)
- Hydrogen bonds
- Energy conservation
- Power and energy (E = P × t)
- Linear mass loss during constant-power boiling
Sensors:
- FizziQ Connect mini-balance (force sensor)
What is required:
- Smartphone or tablet with FizziQ Connect
- M5 Stack with force sensor (balance)
- Calorimeter or insulated beaker
- Immersion heater
- Power meter (wattmeter)
- Electric kettle for initial boiling
- Lab stand and clamp
Experimental procedure:
Open FizziQ Connect and select 'External sensors'. Connect to the M5 Stack and select the mass (force) measurement.
Boil water in the kettle. Pour the boiling water into the calorimeter (or insulated beaker).
Place the calorimeter on the mini-balance and tare (zero) the reading.
Place the immersion heater in the water, attached to the lab stand with a clamp. The heater must not rest on the balance (only the water and calorimeter contribute to the mass).
Connect the power meter between the outlet and the heater. Note the power P displayed (in watts).
Turn on the heater and wait for a vigorous boil to establish.
Once boiling is steady, start the recording in FizziQ Connect (REC) and start the timer.
Record for 15 minutes. The mass should decrease steadily on the FizziQ screen.
Stop the recording (STOP). Note the mass of water evaporated Δm and the total time Δt.
Calculate the energy supplied E = P × Δt, then the specific heat of vaporization L_vap = E / Δm. Compare with the theoretical value of 2260 kJ/kg.
Expected results:
Mass decreases linearly over time, confirming constant vaporization rate at constant power. Over a typical 240-second interval with a 500W heater, about 50 g of water evaporates, giving L ≈ 500 × 240 / 0.050 = 2400 kJ/kg. This is close to but slightly above the theoretical 2260 kJ/kg due to heat losses.
Scientific questions:
- Why does the temperature remain constant during boiling even though energy is being supplied?
- What happens to the hydrogen bonds during vaporization?
- Why is the experimental value typically higher than the theoretical one?
- How does atmospheric pressure affect the boiling temperature and vaporization energy?
- Why does the mass decrease linearly rather than exponentially?
- How much energy would it take to completely evaporate 1 liter of water from room temperature?
Scientific explanations:
The specific heat of vaporization L_vap represents the energy needed to convert 1 kg of liquid to gas at constant temperature and pressure. For water at 100°C, L_vap ≈ 2260 kJ/kg.
This remarkably high value compared to other liquids is explained by the need to break hydrogen bonds between water molecules. Each molecule forms 3-4 hydrogen bonds that must be overcome during vaporization.
During boiling, the temperature remains constant at 100°C (at standard atmospheric pressure). All the supplied energy goes into the phase change, not into raising the temperature.
The energy supplied by the heater is calculated by E = P × Δt, where P is the power in watts (measured by the wattmeter) and Δt the time in seconds.
The experimental value is generally overestimated because some of the heater's energy does not go into vaporizing water: it heats the calorimeter, compensates for heat losses to the environment, and warms the air above the water.
Continuous mass tracking by the FizziQ Connect mini-balance shows that mass decreases linearly with time. This confirms that the vaporization rate is constant when the heating power is constant.
Extension activities:
- Why does the temperature remain constant during boiling even though energy is being supplied?
- What happens to the hydrogen bonds during vaporization?
- Why is the experimental value typically higher than the theoretical one?
- How does atmospheric pressure affect the boiling temperature and vaporization energy?
- Why does the mass decrease linearly rather than exponentially?
- How much energy would it take to completely evaporate 1 liter of water from room temperature?
Frequently asked questions:
Q: The mass reading is noisy during boiling.
R: Boiling creates vibrations that shake the balance. Use a heavy, stable calorimeter and ensure the heater does not touch the balance or calorimeter.
Q: My value of L_vap is too high.
R: Heat losses to the environment are the main cause. Use a well-insulated calorimeter and a lid to reduce losses.
Q: The mass decrease is not linear.
R: Ensure the water is at a steady boil throughout. The initial phase (before full boiling) will show non-linear behavior. Use only the data from the steady boiling phase.
Q: Can I use a microwave instead of an immersion heater?
R: No, because you cannot easily measure the power delivered to the water. An immersion heater with a wattmeter gives a precise power measurement.