(ii) convection.
(b) State two factors that determine the quantity of heat in a body.
(c) Explain the statement: The vecilic latent heat of vaporization of mercury is 2.72 x 10\(^5\) Jkg\(^{-1}\).
(d)A jug of heat capacity 250 Jkg\(^{-1}\) contains water at 28°C. An electric heater of resistance 35\(\Omega\) connected to a 220 V source is used to raise the temperature of the water until it boils at 100°C in 4 minutes. After. another 5 minutes, 300 g of water has evaporated. Assuming no heat is lost to the surroundings, calculate the:
(ii) specific latent heat of vaporization of steam. [Specific heat capacity of water = 4200 kg\(^{-1}\)K\(^{-1}\)]
(a) Factors affecting heat loss
(i) By radiation:
- The temperature of the body (and of its surroundings).
- The nature and colour of the surface (dull/black surfaces radiate more than shiny/white ones) and its surface area.
(ii) By convection:
- The temperature difference between the body and the surrounding fluid.
- The exposed surface area and the freedom of the fluid to circulate (draught/movement of the fluid).
(b) Factors determining the quantity of heat in a body
- The mass of the body.
- The temperature (and the nature/specific heat capacity of the material).
(c) The statement "the specific latent heat of vaporization of mercury is \(2.72\times10^{5}\ \text{J kg}^{-1}\)" means that \(2.72\times10^{5}\) joules of heat is required to change 1 kilogram of mercury from liquid to vapour at its boiling point without any change in temperature.
(d) The jug of water
Power of heater: \(P = \dfrac{V^{2}}{R} = \dfrac{220^{2}}{35} = \dfrac{48400}{35} = 1382.9\ \text{W}\).
(i) Mass of water before heating. In the first \(t_{1} = 4\text{ min} = 240\text{ s}\) the temperature rises from \(28^{\circ}\text{C}\) to \(100^{\circ}\text{C}\), so \(\Delta\theta = 72\ \text{K}\).
Heat supplied \(= P t_{1} = 1382.9\times240 = 3.319\times10^{5}\ \text{J}\).
Taking heat capacity of jug \(C = 250\ \text{J K}^{-1}\) and \(c_{water} = 4200\ \text{J kg}^{-1}\text{K}^{-1}\):
\((m c_{water} + C)\Delta\theta = P t_{1}\)
\((4200m + 250)\times 72 = 331886\)
\(4200m + 250 = 4609.5 \Rightarrow 4200m = 4359.5 \Rightarrow m = 1.04\ \text{kg}\).
The mass of water before heating is about 1.04 kg.
(ii) Specific latent heat of vaporization of steam. In the next \(t_{2} = 5\text{ min} = 300\text{ s}\) the water boils at constant temperature and \(0.3\ \text{kg}\) evaporates.
Heat supplied \(= P t_{2} = 1382.9\times300 = 4.149\times10^{5}\ \text{J}\).
\(L = \dfrac{P t_{2}}{m_{evap}} = \dfrac{414857}{0.3} = 1.38\times10^{6}\ \text{J kg}^{-1}\).
