(a) Define boiling point of a… | Physics

Question 1 Report

(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]

(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}]

Answer

(a) Boiling point

The boiling point of a liquid is the constant temperature at which the liquid boils, i.e. at which its saturated vapour pressure equals the external atmospheric pressure.

(b) Boiling water without heating it

Put warm water in a round-bottomed flask, cork it, and connect it to a vacuum pump. As the pump lowers the pressure above the water, the boiling point falls; when the reduced pressure equals the water's vapour pressure at that temperature, the water boils without any further heating. (Alternatively, boil the water, cork the flask, invert it and pour cold water over the top: the vapour inside condenses, pressure drops, and the water boils again.)

(c) Three applications of expansion of metals

The bimetallic strip in thermostats and fire alarms.
Fitting (shrink-fitting) of metal tyres/rims onto wheels and rivets.
Expansion gaps/rollers left in railway lines and bridges.

(d) Cooling the room air from 34°C to 24°C

Floor area \(= 7\times10 = 70\,\text{m}^{2}\); air mass \(= 250\,\text{kg}\); \(c = 1010\,\text{Jkg}^{-1}\text{K}^{-1}\); density \(= 1.25\,\text{kg m}^{-3}\).

(i) Height of the room: volume of air \(= \dfrac{m}{\rho} = \dfrac{250}{1.25} = 200\,\text{m}^{3}\).

\[ \text{height} = \frac{\text{volume}}{\text{floor area}} = \frac{200}{70} = 2.86\,\text{m} \]

(ii) Energy extracted:

\[ Q = mc\,\Delta\theta = 250\times1010\times(34-24) = 250\times1010\times10 = 2.525\times10^{6}\,\text{J} \]

(iii) Which is higher? The actual energy that must be extracted is higher than the calculated value. The calculation accounts only for cooling the air; in practice heat must also be removed from the walls, furniture and occupants, and heat continually leaks in from the warmer surroundings, so more energy is needed than the value found above.

Read lesson note on Temperature And Its Measurement (JAMB)

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