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(a) Define boiling point of a liquid. (b) Describe how water in a round bottom flask ...

(a) Define boiling point of a liquid.
(b) Describe how water in a round bottom flask could be made to boil without heating it. [diagram not necessary]
(c) State three applications of expansion of metals.
(d) A room with floor measurements 7m x 10 m contains air of mass 250 kg at a temperature of 34°C. The air is cooled until the temperature falls to 24°C. Calculate the: (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\(^{-1}\)K\(^{-1}\); density of air = 1.25 kg m\(^{-3}]
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    Correct Answer: Option n
    Explanation:
    (a) Boiling point of a liquid is the temperature at which saturated vapour pressure of a liquid equals the external atmospheric pressure.
    (b) Fill a round bottom flask partially with water, then pump out the air gradually with a vacuum pump. At a state when the saturation vapour pressure equal atmospheric pressure, the water is observed to be boiling.
    (c) Application of i expansion of metals are in:
    - Riveting two or more metal plates.
    - Fixing of metal rims on metal wheels.
    - Fusing of platinum wire through walls of glass vessels.
    - Automatic fire alarms and
    - Thermostat.
    (d)(i) Volume = \(\frac{Mass}{Density} = \frac{250}{1.25}\)
    = 200\(m^2\)
    But volume = Ibh
    h = \(\frac{v}{Ib} = \frac{200}{7 \times 10}\)
    = 2.86m
    (ii) Q = mc\(\theta\)
    V = 250 x 1010(34 - 24)
    = 252500 or 2.525 x 10\(^6\)
    (iii) The actual energy value will be higher in value than the calculated value because --Part of heat is extracted. from the walls in the room and other materials. he room also is not airtight, there are heat leakages in and out of the room.

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