(a) List two factors each that affect heat loss by:
(i) radiation;
(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:
(i) mass of water in the jug before heating;
(ii) specific latent heat of vaporization of steam. [Specific heat capacity of water = 4200 kg(^{-1})K(^{-1})]
Explanation
(a) (i) Radiation:
Surface area, Temperature.
(ii) Convection:
Nature, Density/Viscosity of the fluid, Thermal conductivity of fluid, Specific heat capacity of fluid and Exposed surface area.
(b) Factors that determine the quantity of heat in a body: Heat/thermal capacity and. Temperature
(c) It means that 2.72 x 105J of heat energy is required/ needed to change I kg of mercury at its boiling point to vapour (without temperature change)
(d) (i) heat supplied by heater = ivt = (frac{v^2}{R})t
heat gained by water = M(_w)C(_w)((theta_2) – (theta_1))
where M(_w) = mass of water
C(_w) = s.h.c of water
(theta_2) = final temp. of water
(theta_1) = initial temp. of water
Heat gained by jug = C(_j) ((theta_2) – (theta_1))
(frac{v^2t}{R}) = M(_w)C(_w) ((theta_2) – (theta_1))
= (frac{220^2 times 4 times 60}{35})
= M(_w) x 4200[100 – 28] + 250[100 – 28]
M(_w) = 1.038kg