Cooling Curve of a Sealed Sample
A technician heats a 250 g sealed metal canister containing an unknown pure liquid in a well-insulated calorimeter. Power input is held at 120 W. Starting at 18 °C, the temperature rises at a steady 0.40 K/s until it reaches 78 °C, where it stays constant for 305 s despite continued heating. After that the temperature rises again at 0.40 K/s. The canister's own heat capacity is stated as 60 J/K and is assumed constant. The technician must decide whether the plateau is a phase change or a fault, and what the liquid is.
1. Determine the specific heat capacity of the liquid from the first heating stage, and state whether the plateau is consistent with boiling.
Read the worked answer
During the first stage the sample absorbs all 120 W. The temperature rate is 0.40 K/s, so the total heat capacity of (liquid + canister) is C = P/(dT/dt) = 120/0.40 = 300 J/K. Subtracting the canister's 60 J/K leaves 240 J/K for the liquid. With m = 0.250 kg, c = 240/0.250 = 960 J/(kg·K). That is close to the specific heat of ethanol (~2400) - no; it is close to that of a light organic such as benzene (~1700) - also no. In fact 960 J/(kg·K) is unremarkable for many liquids and does not by itself identify the substance, so identification must come from the plateau. The plateau at constant power means energy is going into latent heat, not temperature, which is exactly the signature of boiling at constant pressure. So the plateau is consistent with boiling; a sensor fault would not normally hold the temperature so flat for over five minutes.
2. Use the plateau to estimate the latent heat of vaporisation per unit mass, and comment on whether the value is physically plausible.
Read the worked answer
Energy supplied during the plateau: E = P·t = 120 × 305 = 36 600 J. If this is the latent heat of vaporisation of the whole 0.250 kg, then L = E/m = 36 600/0.250 = 146 400 J/kg ≈ 1.5 × 10^5 J/kg. Typical values for organic liquids are 2–9 × 10^5 J/kg (water is 2.26 × 10^6 J/kg), so 1.5 × 10^5 J/kg is on the low side but not absurd for a volatile liquid. The key judgement is that the calculation assumes all the liquid vaporises during the plateau. If the canister is sealed and only partly filled, vapour forms and the pressure rises; the plateau then reflects boiling at rising pressure, and the measured L is only an effective average, not a true latent heat at one pressure. The technician should not quote 1.5 × 10^5 J/kg as a precise material constant without checking fill fraction and pressure.
3. The canister is later found to have a small slow leak. Explain how this changes interpretation of the plateau.
Read the worked answer
With a leak, vapour escapes and the mass of liquid falls during the plateau. The energy balance becomes P·t = m_initial·L + (heat to warm escaping vapour), and the apparent L computed as E/m_initial is an overestimate of the true latent heat because some energy leaves with the escaping vapour and some liquid mass is lost. Also, a leak at constant external pressure means boiling occurs at essentially the same temperature, so the flat plateau is still expected - the leak does not destroy the phase-change interpretation, it corrupts the quantitative value. The correct judgement is: the plateau remains good evidence of boiling, but the numerical L must be treated as an upper-bound estimate and flagged for re-measurement in a sealed, weighed system.