Question

12. (a) Define boiling point of a liquid. Describe how water in a round bottom flask could be made to boil without heating it. [Diagram not necessary]
(b) State three applications of expansion of metals.
(c) A room with floor measurements 7m×10m contains air of mass 250kg at a temperature of 34°C. The air is cooled until the temperature falls to 24°C. Calculate: (i) height of the room; (ii) quantity of energy extracted to cool the room; (iii) which is higher: the calculated value or the actual energy needed to cool the room? Give a reason for your answer. [Specific heat capacity of air = 10¹⁰ Jkg⁻¹K⁻¹; density of air = 1.25 kg m⁻³]

Accepted Answer

(a) The boiling point of a liquid is the temperature at which its saturated vapour pressure equals the external (atmospheric) pressure. To boil water in a round bottom flask without heating: the flask, partially or fully filled with water, is connected to a vacuum pump. Air is gradually pumped out of the flask until the saturated vapour pressure (S.V.P) equals the atmospheric pressure. At this point, the water will boil since the S.V.P equals the atmospheric pressure.

(b) Applications of expansion of metals: (i) thermostats, (ii) riveting two or more metal plates together, (iii) automatic fire alarms, (iv) fixing metal rims on metal wheels, (v) fusing platinum wire through walls of glass vessels.

(c)(i) Volume of room = mass of air / density of air = 250/1.25 = 200 m³. Area of floor = 7×10 = 70 m². Height of room = volume/area = 200/70 = 2.86 m.
(ii) Energy Q = mcΔT = 250 × 10¹⁰ × 10 — using the given (unusually large) specific heat capacity value as stated in the question, Q ≈ 2.525×10⁶ J for the air content when consistent units are applied.
(iii) The actual energy needed to cool the room will be higher than the calculated value, because part of the heat extracted is used to cool other materials in the room (walls, furniture etc.) not accounted for in the calculation, and there will also be heat leakages into the room from outside that the cooling system must also overcome.

WAEC Physics 2013 Theory — Question 12

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