(a) Define the boiling point of a liquid. (b) With the aid of a sketch diagram, describe an experiment to determine the boiling point of a small quantity of...

(a) Boiling point of a liquid

The boiling point of a liquid is the constant temperature at which the liquid changes into vapour throughout its bulk, that is, the temperature at which the saturated vapour pressure of the liquid becomes equal to the external (atmospheric) pressure.

(b) Experiment to determine the boiling point of a small quantity of a liquid

Place the small quantity of the liquid in a clean ignition (test) tube and lower it into a beaker of water so that the liquid is well below the water surface. Support a thermometer with its bulb immersed in the liquid, and place a stirrer in the surrounding water. Stand the beaker on a wire gauze and tripod above a Bunsen burner.

Heat the water bath gently and stir the water continuously so that the heat reaches the small tube evenly. Watch the thermometer: its reading rises steadily and then becomes constant while the liquid is boiling vigorously. Record this steady temperature. The constant temperature at which the thermometer reading remains fixed while the liquid boils is the boiling point of the liquid.

(c) Mass of water that boils away

Data: mass of copper \(m_{Cu}=300\,\text{g}=0.300\,\text{kg}\), specific heat capacity of copper \(c_{Cu}=4.0\times10^{2}\,\text{J kg}^{-1}\text{K}^{-1}\), mass of water \(m_w=250\,\text{g}=0.250\,\text{kg}\), specific heat capacity of water \(c_w=4.2\times10^{3}\,\text{J kg}^{-1}\text{K}^{-1}\), specific latent heat of vaporization \(L=2.26\times10^{6}\,\text{J kg}^{-1}\).

Heat lost by the copper as it cools from \(950^{\circ}\text{C}\) to the final temperature \(100^{\circ}\text{C}\):

\[ Q_{Cu}=m_{Cu}\,c_{Cu}\,\Delta\theta = 0.300\times(4.0\times10^{2})\times(950-100)=0.300\times400\times850=102000\,\text{J} \]

Heat needed to raise the \(250\,\text{g}\) of water from \(25^{\circ}\text{C}\) to \(100^{\circ}\text{C}\):

\[ Q_{1}=m_{w}\,c_{w}\,\Delta\theta = 0.250\times(4.2\times10^{3})\times(100-25)=0.250\times4200\times75=78750\,\text{J} \]

By conservation of energy, the heat remaining after warming the water is used to boil some of it into steam:

\[ Q_{2}=Q_{Cu}-Q_{1}=102000-78750=23250\,\text{J} \]

If \(m_v\) is the mass of water boiled away, then \(Q_{2}=m_v L\):

\[ m_v=\frac{Q_{2}}{L}=\frac{23250}{2.26\times10^{6}}=1.03\times10^{-2}\,\text{kg}\approx 10.3\,\text{g} \]

The mass of water that boils away is about \(10.3\,\text{g}\) (\(1.03\times10^{-2}\,\text{kg}\)).